PROBLEM SOLVING STRATEGY: Tangent & Normal Lines

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These problems will always specify that you find the tangent or normal (= perpendicular) line at a particular point. We’ll call that point $(x_0, y_0)$. To answer these questions, you will almost always use the Point-Slope form of a line. Recall that if a line has slope *m *and contains the point $(x_0, y_0)$, then you can write its equation as: $$\textbf{Point-Slope form of a line:} \qquad y – y_0 = m(x – x_0)$$ The problem statement typically specifies the point $(x_0, y_0)$, and so really these problems come down to determining the slope *m *of the line — which we’ll address below. You will use that equation again and again; memorize it if you don’t know it already. (It’s just a variant on the definition of slope: $m = \dfrac{y – y_0}{x – x_0}$). **I. Tangent Line to a Curve** *Very* frequently in beginning Calculus you will be asked to find an equation for the line **tangent** to a curve at a particular point. We’re calling that point $(x_0, y_0)$. To find the line’s equation, you just need to remember that the tangent line to the curve has slope equal to the derivative of the function evaluated at the point of interest: $$m_\text{tangent line} = f'(x_0)$$
That is, find the derivative of the function $f'(x)$, and then evaluate it at $x = x_0$. That value, $f'(x_0)$, *is *the slope of the tangent line. Hence we can write the equation for the tangent line at $(x_0, y_0)$ as \begin{align*}
y – y_0 &= m_\text{tangent line}(x – x_0) \\[8px]
y – y_0 &= f'(x_0)(x – x_0)
\end{align*} If those equations look abstract to you, don’t worry. As soon as you work a few problems, the process will make sense — we promise. **II. Normal Line to a Curve** Sometimes instead a question will ask you instead to find the line **normal** to a curve. That’s the same thing as asking for the line that is **perpendicular** to the curve. You will again use the Point-Slope form of a line. But now to compute the slope of the line, recall that the slopes of perpendicular lines are the negative reciprocals of each other ($m_2 = -\dfrac{1}{m_1}$). We want the slope of the line that is perpendicular to the curve at a point, and hence that is perpendicular to the tangent line to the curve at that point: \begin{align*}
m_\text{normal line} &= \frac{-1}{m_\text{tangent line}}\\[12px]
&= \frac{-1}{f'(x_0)}
\end{align*} Hence we can write the equation for the normal line at $(x_0, y_0)$ as \begin{align*}
y – y_0 &= m_\text{normal line}(x – x_0) \\[8px]
y – y_0 &= \frac{-1}{f'(x_0)}(x – x_0)
\end{align*}

We recommend*not* trying to memorize all of the formulas above. Instead, remember the Point-Slope form of a line, and then use what you know about the derivative telling you the slope of the tangent line at a given point. The problems below illustrate. Question 2 illustrates the process of putting together different pieces of information to find the equation of a tangent line. Question 3, and the rest of the questions, require that you find the pieces of information before you can put them together.

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Question 1 (Warm-up): Graph of *f* and its tangent at a point

Use the information in the graph to
replace the question marks with correct values:**(a)** $f(\underline{?}) = \underline{?}$**(b)** $f'(\underline{?}) = \underline{?}$**(a)** $f(3) = 10$**(b)** $f'(3) = 3$

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Solution SummarySolution (a) DetailSolution (b) Detail

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The question asks about the derivative of a point, and the only point on $y =f(x)$ we know anything about is $x = 3$.

The derivative of $f'(3)$ is equal to the slope of the tangent line at $x = 3$, and that line is shown. Hence to answer this question, focus on that line and ignore the curve.

That slope is given by

\begin{align*} f'(3) = \text{tangent line slope} &= \frac{16-10}{5-3} \\ \\ &= \frac{6}{2} = 3 \quad \cmark \end{align*}

The derivative of $f'(3)$ is equal to the slope of the tangent line at $x = 3$, and that line is shown. Hence to answer this question, focus on that line and ignore the curve.

That slope is given by

\begin{align*} f'(3) = \text{tangent line slope} &= \frac{16-10}{5-3} \\ \\ &= \frac{6}{2} = 3 \quad \cmark \end{align*}

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Question 2: Given *g*(5) and *g'*(5), write the equation for the tangent

For a particular function $g$, we know $g'(5) = 2$ and $g(5) = -3$. Write an equation for the line tangent to $g$ at $x = 5$.

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Since $g'(5) = 2$, we know that the slope of the tangent line at $x = 5$ is $m=2$. Furthermore, the tangent line contains the point (5, -3), since it passes through (grazes) that point on the curve.

Then using the point-slope form of a line that contains the point $(x_o, y_o)$ we have

\begin{align*} y – y_o &= m(x – x_o) \\ y – (-3) &= 2(x-5) \\ y &= 2x – 13 \quad \cmark \end{align*}

Then using the point-slope form of a line that contains the point $(x_o, y_o)$ we have

\begin{align*} y – y_o &= m(x – x_o) \\ y – (-3) &= 2(x-5) \\ y &= 2x – 13 \quad \cmark \end{align*}

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Question 3: Given *f(x)*, find tangent and normal lines at a point

Consider the curve given by $y = f(x) = x^3 - x + 5$.**(a)** Find the equation to the line tangent to the curve at the point (1, 5).**(b)** Find the equation of the line normal (perpendicular) to the curve at the point (1,5).**(a)** $y= 2x +3$**(b)** $y = -\dfrac{1}{2} x + \dfrac{11}{2}$

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Solution SummarySolution (a) DetailSolution (b) Detail

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To write the equation of a line, we need its slope $m$ and a point $(x_o, y_o)$ on it. We already know a point, since the line intersects (grazes) the curve at (1,5).

Hence we just need its slope $m$, which is is the same as the slope of the curve at that point ($x=1$). And that slope equals the function’s derivative at that point:

\begin{align*} f(x) &= x^3 – x + 5 \\ f'(x) &= 3x^2 -1 \\ f'(1) &= 3(1) – 1 = 2 \end{align*} Then we can write the equation of this tangent line, using the point-slope form of a line that contains the point $(x_o, y_o)$:

\begin{align*} y – y_o &= m(x – x_o) \\ y – 5 &= (2)(x – 1) \\ y&= 2x +3 \quad \cmark \end{align*}

Hence we just need its slope $m$, which is is the same as the slope of the curve at that point ($x=1$). And that slope equals the function’s derivative at that point:

\begin{align*} f(x) &= x^3 – x + 5 \\ f'(x) &= 3x^2 -1 \\ f'(1) &= 3(1) – 1 = 2 \end{align*} Then we can write the equation of this tangent line, using the point-slope form of a line that contains the point $(x_o, y_o)$:

\begin{align*} y – y_o &= m(x – x_o) \\ y – 5 &= (2)(x – 1) \\ y&= 2x +3 \quad \cmark \end{align*}

Recall that line 2 is normal (perpendicular) to line 1 if their slopes are negative reciprocals:

$$m_2 = -\frac{1}{m_1}$$ From part (a) we know that the line tangent to the curve at the point (1,5) has slope $m_1 = 2$, and so our perpendicular line at that point has slope

\begin{align*} m_2 &= -\frac{1}{m_1} \\ &= -\frac{1}{2} \end{align*} This line passes through the curve at point (1,5), and so its equation is given by

\begin{align*} y – y_o &= m(x – x_o) \\ \\ y-5 &= -\frac{1}{2} (x -1) \\ \\ y &= -\frac{1}{2} x + \frac{11}{2} \quad \cmark \end{align*}

$$m_2 = -\frac{1}{m_1}$$ From part (a) we know that the line tangent to the curve at the point (1,5) has slope $m_1 = 2$, and so our perpendicular line at that point has slope

\begin{align*} m_2 &= -\frac{1}{m_1} \\ &= -\frac{1}{2} \end{align*} This line passes through the curve at point (1,5), and so its equation is given by

\begin{align*} y – y_o &= m(x – x_o) \\ \\ y-5 &= -\frac{1}{2} (x -1) \\ \\ y &= -\frac{1}{2} x + \frac{11}{2} \quad \cmark \end{align*}

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Question 4: Find the normal line to a function's curve

Find an equation of the normal (perpendicular) line to the curve $y = \sqrt{25 - x^2}$ at the point $(3, 4)$. **Answer**: $y = \dfrac{4}{3}x$

To write the equation of a line, we need its slope $m$ and a point $(x_o, y_o)$ on it. We already know a point, since the line intersects (grazes) the curve at (3,4).

Hence we just need the line’s slope $m$. We know line 2 is normal (perpendicular) to line 1 if their slopes are negative reciprocals: $m_2 = -\dfrac{1}{m_1}$. So let’s first find the slope of the tangent line to the curve at this point; we can then easily find the slope of the line normal to the curve at that point.

The slope of the curve at any point is given by its derivative:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}\left(25 -x^2 \right)^\frac{1}{2} \\ \\ &= \frac{1}{2}\left(25 -x^2 \right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left(25 – x^2 \right) \\ \\ &= \frac{1}{2} \frac{1}{\sqrt{25 – x^2}}(-2x) \\ \\ &= \frac{-x}{\sqrt{25 – x^2}} \end{align*} At $x = 3$: \begin{align*} \left.\frac{dy}{dx} \right|_{x=3} &= \frac{-3}{\sqrt{25 – 3^2}} \\ \\ &= \frac{-3}{\sqrt{16}} = -\frac{3}{4} \end{align*} So the tangent line has slope $m_1 = -\dfrac{3}{4}$. The normal line thus has slope

$$m_2 = -\frac{1}{m_1} = \frac{4}{3}$$ The line we’re after therefore contains the point (3,4) and has slope $m_2 = \dfrac{4}{3}$, and so we can write it in point-slope form as:

\begin{align*} y – y_o &= m_2 (x – x_o) \\ \\ y – 4 &= \frac{4}{3}(x – 3)\\ \\ y &= \frac{4}{3}x -4 + 4 \\ \\ &= \frac{4}{3}x \quad \cmark\\ \\ \end{align*}

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To write the equation of a line, we need its slope $m$ and a point $(x_o, y_o)$ on it. We already know a point, since the line intersects (grazes) the curve at (3,4).

Hence we just need the line’s slope $m$. We know line 2 is normal (perpendicular) to line 1 if their slopes are negative reciprocals: $m_2 = -\dfrac{1}{m_1}$. So let’s first find the slope of the tangent line to the curve at this point; we can then easily find the slope of the line normal to the curve at that point.

The slope of the curve at any point is given by its derivative:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}\left(25 -x^2 \right)^\frac{1}{2} \\ \\ &= \frac{1}{2}\left(25 -x^2 \right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left(25 – x^2 \right) \\ \\ &= \frac{1}{2} \frac{1}{\sqrt{25 – x^2}}(-2x) \\ \\ &= \frac{-x}{\sqrt{25 – x^2}} \end{align*} At $x = 3$: \begin{align*} \left.\frac{dy}{dx} \right|_{x=3} &= \frac{-3}{\sqrt{25 – 3^2}} \\ \\ &= \frac{-3}{\sqrt{16}} = -\frac{3}{4} \end{align*} So the tangent line has slope $m_1 = -\dfrac{3}{4}$. The normal line thus has slope

$$m_2 = -\frac{1}{m_1} = \frac{4}{3}$$ The line we’re after therefore contains the point (3,4) and has slope $m_2 = \dfrac{4}{3}$, and so we can write it in point-slope form as:

\begin{align*} y – y_o &= m_2 (x – x_o) \\ \\ y – 4 &= \frac{4}{3}(x – 3)\\ \\ y &= \frac{4}{3}x -4 + 4 \\ \\ &= \frac{4}{3}x \quad \cmark\\ \\ \end{align*}

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Question 5: Find where the curve is perpendicular to a line

Find the equation of the tangent line to the curve $y = x^2 + 4x -2$ at a point where the tangent is perpendicular to the line $-x - 2y + 6 = 0$. **Answer**: $y = 2x -3$

This question may seem a little hard to parse. We’re essentially going to work our way backwards through what was given, following roughly these four steps:

*Step 1.* Determine the slope of the line the question tells us about in the last sentence, $-x – 2y + 6 = 0$. We’ll call that line Line 1, and we’ll then have its slope $m_1$.

*Step 2.* We’re interested in the point on the curve that’s perpendicular to Line 1, so we’ll find the slope $m_2 = -\dfrac{1}{m_1}$.

*Step 3.* We’ll find the point on the curve that has slope equal to $m_2$.

*Step 4.* We’ll write the equation of the line tangent to the curve at that point.

*Step 1.*

We’re given the equation of a line we’re calling Line 1: $-x – 2y + 6 = 0$. We can easily determine*its* slope by putting the equation into slope-intercept form $y = mx + b$:

\begin{align*} -x – 2y + 6 &= 0\\ -2y &= x – 6 \\ y &= -\frac{1}{2}x + 3 \end{align*} So Line 1 has slope $m_1 = -\frac{1}{2}$.

*Step 2.*

The line we’re after is perpendicular to this line, and recall that line 2 is perpendicular to line 1 if their slopes are negative reciprocals: $m_2 = -\dfrac{1}{m_1}$. Our line therefore has slope

\begin{align*} m_2 &= -\frac{1}{m_1} \\ &= 2 \end{align*}

*Step 3.*

Next, we need to find the point on the curve $y = x^2 + 4x -2$ that has slope equal to $m_2 = 2$. The curve’s slope at any point is given by its derivative there:

\begin{align*} y &= x^2 + 4x -2 \\ \frac{dy}{dx} &= 2x + 4 \end{align*} We’re looking for the point where the curve’s slope $\dfrac{dy}{dx}= 2$: \begin{align*} \frac{dy}{dx} = 2 &= 2x + 4\\ -2x &= 2\\ x &= -1 \end{align*} So we’re interested in the tangent line to the curve at $x = -1$.

*Step 4.*

To write the equation of the tangent line through this point, we also need its $y$-value:

\begin{align*} y &= x^2 + 4x -2 \\ y &= (-1)^2 +4(-1) -2 \\ &= 1 -4 -2 = -5 \end{align*} So our tangent line intersects (grazes) the curve at the point (-1, -5). And we determined earlier that our line has slope $m_2$ = 2. Hence, using the the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_0 &= m(x-x_0) \\ y -(-5) &= (2)[x – (-1)] \\ y +5 &= 2x +2\\ y &= 2x -3 \quad \cmark \end{align*}

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This question may seem a little hard to parse. We’re essentially going to work our way backwards through what was given, following roughly these four steps:

We’re given the equation of a line we’re calling Line 1: $-x – 2y + 6 = 0$. We can easily determine

\begin{align*} -x – 2y + 6 &= 0\\ -2y &= x – 6 \\ y &= -\frac{1}{2}x + 3 \end{align*} So Line 1 has slope $m_1 = -\frac{1}{2}$.

The line we’re after is perpendicular to this line, and recall that line 2 is perpendicular to line 1 if their slopes are negative reciprocals: $m_2 = -\dfrac{1}{m_1}$. Our line therefore has slope

\begin{align*} m_2 &= -\frac{1}{m_1} \\ &= 2 \end{align*}

Next, we need to find the point on the curve $y = x^2 + 4x -2$ that has slope equal to $m_2 = 2$. The curve’s slope at any point is given by its derivative there:

\begin{align*} y &= x^2 + 4x -2 \\ \frac{dy}{dx} &= 2x + 4 \end{align*} We’re looking for the point where the curve’s slope $\dfrac{dy}{dx}= 2$: \begin{align*} \frac{dy}{dx} = 2 &= 2x + 4\\ -2x &= 2\\ x &= -1 \end{align*} So we’re interested in the tangent line to the curve at $x = -1$.

To write the equation of the tangent line through this point, we also need its $y$-value:

\begin{align*} y &= x^2 + 4x -2 \\ y &= (-1)^2 +4(-1) -2 \\ &= 1 -4 -2 = -5 \end{align*} So our tangent line intersects (grazes) the curve at the point (-1, -5). And we determined earlier that our line has slope $m_2$ = 2. Hence, using the the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_0 &= m(x-x_0) \\ y -(-5) &= (2)[x – (-1)] \\ y +5 &= 2x +2\\ y &= 2x -3 \quad \cmark \end{align*}

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Question 6: Tangents to some trig graphs

Find the tangent lines as requested.**(a)** Find an equation of the tangent line to the curve $y = \cos^2 x$ at $x = \dfrac{\pi}{4}$.**(b)** Find an equation of the tangent line to the curve $y = \tan^2 x$ at the point $x = \dfrac{\pi}{6}$.**(a)** $y = -x + \dfrac{\pi}{4} + \dfrac{1}{2}$**(b)** $y = \dfrac{8}{3\sqrt{3}}x – \dfrac{4\pi}{9\sqrt{3}} + \dfrac{1}{3}$

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Solution SummarySolution (a) DetailSolution (b) Detail

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To write the equation of a line, we need its slope $m$ and a point $(x_o, y_o)$ on it. We know this line intersects (grazes) the curve at $x = \dfrac{\pi}{4}$, and so it must have the same $y$-value there:

\begin{align*} \left. y\right|_{x = \frac{\pi}{4}} &= \cos^2 \left(\frac{\pi}{4}\right) && \left[\text{Recall }\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\right] \\ &= \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \end{align*} Hence the line contains the point $(x_o, y_o) = (\dfrac{\pi}{4}, \dfrac{1}{2})$.

Now we need the slope of the line, which is the same as the slope of the curve at $x = \dfrac{\pi}{4}$ since the line is tangent there. And the slope of the curve is given by its derivative at that point:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}\left(\cos x \right)^2 \\ \\ &= 2 \cos x \cdot \frac{d}{dx}(\cos x) \\ \\ &= 2 \cos x \cdot (-\sin x) \\ \\ &= -2 \sin x \, \cos x \end{align*} At $x = \dfrac{\pi}{4}$: \begin{align*} \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{4}} &= -2 \sin \left(\frac{\pi}{4} \right) \cos \left( \frac{\pi}{4}\right) \\ \\ &= -2 \left(\frac{1}{\sqrt{2}} \right) \left(\frac{1}{\sqrt{2}} \right) \\ \\ &= -1 \end{align*} So the line has slope $m = -1$. We can finally write its equation using point-slope form:

\begin{align*} y – y_0 &= m(x-x_0) \\ \\ y – \frac{1}{2} & = (-1)\left(x – \frac{\pi}{4} \right) \\ \\ y &= -x + \frac{\pi}{4} + \frac{1}{2}\quad \cmark \end{align*}

\begin{align*} \left. y\right|_{x = \frac{\pi}{4}} &= \cos^2 \left(\frac{\pi}{4}\right) && \left[\text{Recall }\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\right] \\ &= \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \end{align*} Hence the line contains the point $(x_o, y_o) = (\dfrac{\pi}{4}, \dfrac{1}{2})$.

Now we need the slope of the line, which is the same as the slope of the curve at $x = \dfrac{\pi}{4}$ since the line is tangent there. And the slope of the curve is given by its derivative at that point:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}\left(\cos x \right)^2 \\ \\ &= 2 \cos x \cdot \frac{d}{dx}(\cos x) \\ \\ &= 2 \cos x \cdot (-\sin x) \\ \\ &= -2 \sin x \, \cos x \end{align*} At $x = \dfrac{\pi}{4}$: \begin{align*} \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{4}} &= -2 \sin \left(\frac{\pi}{4} \right) \cos \left( \frac{\pi}{4}\right) \\ \\ &= -2 \left(\frac{1}{\sqrt{2}} \right) \left(\frac{1}{\sqrt{2}} \right) \\ \\ &= -1 \end{align*} So the line has slope $m = -1$. We can finally write its equation using point-slope form:

\begin{align*} y – y_0 &= m(x-x_0) \\ \\ y – \frac{1}{2} & = (-1)\left(x – \frac{\pi}{4} \right) \\ \\ y &= -x + \frac{\pi}{4} + \frac{1}{2}\quad \cmark \end{align*}

To write the equation of a line, we need its slope $m$ and a point $(x_o, y_o)$ on it. We know this line intersects (grazes) the curve at $x = \dfrac{\pi}{6}$, and so it must have the same $y$-value there:

\begin{align*} \left. y\right|_{x = \frac{\pi}{6}} &= \tan^2 \left( \frac{\pi}{6}\right) && \left[\text{Recall }\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\right] \\ &= \left(\frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} \end{align*} Hence the line contains the point $(x_o, y_o) = (\dfrac{\pi}{6}, \dfrac{1}{3})$.

Now we need the slope of the line, which is the same as the slope of the curve at $x = \dfrac{\pi}{6}$ since the line is tangent there. And the slope of the curve is given by its derivative at that point:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}\left(\tan x \right)^2 \\ \\ &= 2 \tan x \cdot \frac{d}{dx} (\tan x) \\ \\ &= 2 \tan x \cdot \sec^2 x \end{align*} At $x = \dfrac{\pi}{6}$: \begin{align*} \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{6}} &= 2 \tan \left( \frac{\pi}{6}\right) \cdot \sec^2 \left( \frac{\pi}{6}\right) && \left[\text{Recall }\sec \left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\right] \\ \\ &=2 \left( \frac{1}{\sqrt{3}}\right) \left( \frac{2}{\sqrt{3}} \right)^2 \\ \\ &=2 \left( \frac{1}{\sqrt{3}}\right) \left( \frac{4}{3} \right) \\ \\ &= \frac{8}{3\sqrt{3}} \end{align*} So the line has slope $m = \dfrac{8}{3\sqrt{3}}$. We can finally write its equation using point-slope form:

\begin{align*} y – y_0 &= m(x-x_0) \\ \\ y – \frac{1}{3} &= \left(\frac{8}{3\sqrt{3}} \right)\left(x – \frac{\pi}{6} \right) \\ \\ y &= \frac{8}{3\sqrt{3}}x – \frac{4\pi}{9\sqrt{3}} + \frac{1}{3} \quad \cmark \end{align*}

\begin{align*} \left. y\right|_{x = \frac{\pi}{6}} &= \tan^2 \left( \frac{\pi}{6}\right) && \left[\text{Recall }\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\right] \\ &= \left(\frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} \end{align*} Hence the line contains the point $(x_o, y_o) = (\dfrac{\pi}{6}, \dfrac{1}{3})$.

Now we need the slope of the line, which is the same as the slope of the curve at $x = \dfrac{\pi}{6}$ since the line is tangent there. And the slope of the curve is given by its derivative at that point:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}\left(\tan x \right)^2 \\ \\ &= 2 \tan x \cdot \frac{d}{dx} (\tan x) \\ \\ &= 2 \tan x \cdot \sec^2 x \end{align*} At $x = \dfrac{\pi}{6}$: \begin{align*} \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{6}} &= 2 \tan \left( \frac{\pi}{6}\right) \cdot \sec^2 \left( \frac{\pi}{6}\right) && \left[\text{Recall }\sec \left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\right] \\ \\ &=2 \left( \frac{1}{\sqrt{3}}\right) \left( \frac{2}{\sqrt{3}} \right)^2 \\ \\ &=2 \left( \frac{1}{\sqrt{3}}\right) \left( \frac{4}{3} \right) \\ \\ &= \frac{8}{3\sqrt{3}} \end{align*} So the line has slope $m = \dfrac{8}{3\sqrt{3}}$. We can finally write its equation using point-slope form:

\begin{align*} y – y_0 &= m(x-x_0) \\ \\ y – \frac{1}{3} &= \left(\frac{8}{3\sqrt{3}} \right)\left(x – \frac{\pi}{6} \right) \\ \\ y &= \frac{8}{3\sqrt{3}}x – \frac{4\pi}{9\sqrt{3}} + \frac{1}{3} \quad \cmark \end{align*}

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Question 7: Given the tangent line determine. . . (Based on an actual exam problem)

The equation of the tangent line to the graph of the function $f$ at $x = 1$ is $y = 3x + 9$.**(a)** (i) What is $f(1)$? (ii) What is $f'(1)$?**(b)** Given that $f(x) = ax^3 + b$, find the constants $a$ and $b$.**(a)** (i) $f(1) = 12$; (ii) $f'(1) = 3$**(b)** $a=1$, $b=11$

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Solution SummarySolution (a) DetailSolution (b) Detail

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(i) The line is tangent to the curve at $x=1$, and hence they share that point in common. On the line, at $x=1$ the $y$-value is

$$y = 3(1) + 9 = 12$$ Since the function contains the same point, we must have $f(1) = 12$. $\quad \cmark $

(ii) The line is in the form $y = mx + b$, so we know immediately that its slope is $m = 3$. Since it is tangent to the graph of $f$ at $x=1$, we know they have the same slope there, and hence $f'(1) = 3$. $\quad \cmark $

$$y = 3(1) + 9 = 12$$ Since the function contains the same point, we must have $f(1) = 12$. $\quad \cmark $

(ii) The line is in the form $y = mx + b$, so we know immediately that its slope is $m = 3$. Since it is tangent to the graph of $f$ at $x=1$, we know they have the same slope there, and hence $f'(1) = 3$. $\quad \cmark $

Part (a) gives us two conditions we must meet: (i) $f(1) = 12$, and (ii) $f'(1) = 3$. These two conditions will let us solve for our two unknowns $a$ and $b$:

(i) $f(1) = 12$: \begin{align*} f(x) &= ax^3 + b \\ f(1) &= a(1)^3 + b = 12\\ a + b &= 12 \end{align*} (ii) $f'(1) = 3$: \begin{align*} f(x) &= ax^3 + b \\ f'(x) &= 3ax^2 \\ f'(1) &= 3a(1)^2 = 3 \\ 3a &= 3 \\ a &= 1 \quad \cmark \end{align*} Using the equation above, $a+b=12$, and so $b = 11$. $\quad \cmark$

(i) $f(1) = 12$: \begin{align*} f(x) &= ax^3 + b \\ f(1) &= a(1)^3 + b = 12\\ a + b &= 12 \end{align*} (ii) $f'(1) = 3$: \begin{align*} f(x) &= ax^3 + b \\ f'(x) &= 3ax^2 \\ f'(1) &= 3a(1)^2 = 3 \\ 3a &= 3 \\ a &= 1 \quad \cmark \end{align*} Using the equation above, $a+b=12$, and so $b = 11$. $\quad \cmark$

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Question 8: An actual unversity exam problem

Let $f$ be the real-valued function defined by $f(x) = \sqrt{1 + 6x}$.**(a)** Give the domain and range of $f$.**(b)** Determine the slope of the line tangent to the graph of $f$ at $x = 4$.**(c)** Determine the $y$-intercept of the line tangent to the graph of $f$ at $x = 4$.**(d)** Give the coordinates of the point on the graph of $f$ where the tangent line is parallel to $y = x + 12$.**(a)** Domain: $[-\dfrac{1}{6}, \infty)$; Range: $[0, \infty)$**(b)** $\dfrac{3}{5}$**(c)** $\dfrac{13}{5}$**(d)** $(\dfrac{4}{3}, 3)$

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Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) DetailSolution (d) Detail

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We cannot have a negative value inside the square root, and so

\begin{align*} 1 + 6x &\ge 0 \\ 6x &\ge -1 \\ x &\ge -1/6 \end{align*} Hence the domain is in the interval $[-\dfrac{1}{6}, \infty). \quad \cmark $

The range of the square root function is all non-negative real numbers: $[0, \infty) \quad \cmark $

\begin{align*} 1 + 6x &\ge 0 \\ 6x &\ge -1 \\ x &\ge -1/6 \end{align*} Hence the domain is in the interval $[-\dfrac{1}{6}, \infty). \quad \cmark $

The range of the square root function is all non-negative real numbers: $[0, \infty) \quad \cmark $

The slope of $f(x)$ at $x = 4$ is given by its derivative there:

\begin{align*} f(x) &= (1 + 6x)^{\frac{1}{2}} \\ f'(x) &= \frac{1}{2}(1 + 6x)^{-\frac{1}{2}}(6) \\ &= \frac{3}{\sqrt{1 + 6x}} \end{align*} And so at $x=4$:

$$f'(4) = \frac{3}{\sqrt{1 + 6(4)}} = \frac{3}{\sqrt{25}} = \frac{3}{5} \quad \cmark $$

\begin{align*} f(x) &= (1 + 6x)^{\frac{1}{2}} \\ f'(x) &= \frac{1}{2}(1 + 6x)^{-\frac{1}{2}}(6) \\ &= \frac{3}{\sqrt{1 + 6x}} \end{align*} And so at $x=4$:

$$f'(4) = \frac{3}{\sqrt{1 + 6(4)}} = \frac{3}{\sqrt{25}} = \frac{3}{5} \quad \cmark $$

From part (b) we know that the line tangent to the curve at $x = 4$ has slope $m = \dfrac{3}{5}$. Furthermore, at $x = 4$ the the curve has $y$-value $y = \sqrt{1 + 6(4)} = \sqrt{25} = 5$, and so the line tangent to the curve there contains the point $(4,5)$. Hence writing the equation for that tangent line in the form $y = mx + b$, we know that

\begin{align*} 5 &= \frac{3}{5} (4) + b \\ \\ &= \frac{12}{5} + b \\ \\ b &= \frac{13}{5} \quad \cmark \end{align*}

\begin{align*} 5 &= \frac{3}{5} (4) + b \\ \\ &= \frac{12}{5} + b \\ \\ b &= \frac{13}{5} \quad \cmark \end{align*}

We are looking for a tangent line that is parallel to the line $y = x + 12$. Since the equation of that line is in the form $y = mx + b$, we can see immediately that it has slope $m = 1$. We are thus looking for the point on the graph of $f$ that has slope equal to 1.

Recall from part (b) that $$f'(x) = \frac{3}{\sqrt{1 + 6x}} $$ Let’s call the $x$-value of the point we’re after $a$, such that $f'(a) = 1$:

\begin{align*} f'(a) = 1 &= \frac{3}{\sqrt{1 + 6a}} \\ \sqrt{1 + 6a} &= 3 \\ 1 + 6a &= 9 \\ 6a &= 8 \\ a &= \frac{4}{3} \end{align*} Now that we know the $x$-value of the point we’re after, we need its $y$-value:

$$f\left(\frac{4}{3}\right) = \sqrt{1 + 6\left(\frac{4}{3}\right)} = \sqrt{9} = 3$$ Hence the coordinates of the desired point on the graph are $\left(\dfrac{4}{3}, 3\right). \quad \cmark $

Recall from part (b) that $$f'(x) = \frac{3}{\sqrt{1 + 6x}} $$ Let’s call the $x$-value of the point we’re after $a$, such that $f'(a) = 1$:

\begin{align*} f'(a) = 1 &= \frac{3}{\sqrt{1 + 6a}} \\ \sqrt{1 + 6a} &= 3 \\ 1 + 6a &= 9 \\ 6a &= 8 \\ a &= \frac{4}{3} \end{align*} Now that we know the $x$-value of the point we’re after, we need its $y$-value:

$$f\left(\frac{4}{3}\right) = \sqrt{1 + 6\left(\frac{4}{3}\right)} = \sqrt{9} = 3$$ Hence the coordinates of the desired point on the graph are $\left(\dfrac{4}{3}, 3\right). \quad \cmark $

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Question 9: Line intersects cubic and parabola (actual exam problem)

The line $x = c$ where $c > 0$ intersects the cubic $y = 2x^3 + 3x^2 - 5$ at point P, and the parabola $y = 4x^2 + 4x + 3$ at point Q.**(a)** If a line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q, find the value of $c$ where $c > 0$.**(b)** Write the equations of the two tangent lines described in (a).**(a)** 1**(b)** Cubic tangent: $y = 12x – 12$; Parabola tangent: $y = 12x – 1$

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It’s all free: our only aim is to support your learning via our community-supported and ad-free site. **Tangent to cubic:**

At $x = c = 1$, the cubic has $y$-value equal to

\begin{align*} f(1) &= 2(1)^3 + 3(1)^2 – 5 \\ &= 2 + 3 -5 = 0 \end{align*} The tangent line thus also contains the point (1, 0).

Furthermore, the tangent line has slope equal to \begin{align*} f'(x) &= 6x^2 + 6x \\ f'(1) &= 6(1)^2 + 6(1) = 12 \end{align*} Hence, using the the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_o &= m(x – x_o) \\ y – 0 &= 12 (x- 1) \\ y &= 12x – 12 \quad \cmark \end{align*}

**Tangent to parabola:**

At $x = c$, the parabola has $y$-value given by $y = g(1) = 4(1)^2 + 4(1) + 3 = 11$. The tangent line thus also contains the point (1, 11).

Furthermore, the tangent line has slope equal to \begin{align*} g'(x) &= 8x + 4 \\ g'(1) &= 8(1) + 4 = 12 \end{align*} (Note that this value is the same as $f'(1)$, as expected.) Hence, using the the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_o &= m(x – x_o) \\ y – 11 &= 12 (x- 1) \\ y &= 12x – 1 \quad \cmark \end{align*}

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Solution SummarySolution (a) DetailSolution (b) Detail

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As always, we find it helpful to start by drawing a quick sketch of the information given—so of the cubic, the parabola, the line $x=c$ (where $c > 0$), and the intersection points P and Q.

Let’s call the cubic function $f(x) = 2x^3 + 3x^2 – 5$, and the quadratic function $g(x) = 4x^2 + 4x + 3$.

Then the slope of the tangent line to the cubic at any point $x$ is given by

$$f'(x) = 6x^2 + 6x$$ and the slope of the tangent line to the quadratic at any point $x$ is given by

$$g'(x) = 8x + 4$$ The problem states that the line tangent to the cubic at point P (at $x = c$) is parallel to the line tangent to the parabola at point Q (also at $x = c$). Hence at $x = c$ the two slopes must be equal:

\begin{align*} 6c^2 + 6c &= 8c + 4 \\ 6c^2 -2c -4 &= 0 \\ 2(3c^2 – c -2) &= 0 \\ 2(3c + 2)(c – 1) &= 0 \end{align*} Hence $c = -\dfrac{2}{3}$ or 1, but the problem states that $c > 0$, so

$$c = 1 \quad \cmark$$

Let’s call the cubic function $f(x) = 2x^3 + 3x^2 – 5$, and the quadratic function $g(x) = 4x^2 + 4x + 3$.

Then the slope of the tangent line to the cubic at any point $x$ is given by

$$f'(x) = 6x^2 + 6x$$ and the slope of the tangent line to the quadratic at any point $x$ is given by

$$g'(x) = 8x + 4$$ The problem states that the line tangent to the cubic at point P (at $x = c$) is parallel to the line tangent to the parabola at point Q (also at $x = c$). Hence at $x = c$ the two slopes must be equal:

\begin{align*} 6c^2 + 6c &= 8c + 4 \\ 6c^2 -2c -4 &= 0 \\ 2(3c^2 – c -2) &= 0 \\ 2(3c + 2)(c – 1) &= 0 \end{align*} Hence $c = -\dfrac{2}{3}$ or 1, but the problem states that $c > 0$, so

$$c = 1 \quad \cmark$$

At $x = c = 1$, the cubic has $y$-value equal to

\begin{align*} f(1) &= 2(1)^3 + 3(1)^2 – 5 \\ &= 2 + 3 -5 = 0 \end{align*} The tangent line thus also contains the point (1, 0).

Furthermore, the tangent line has slope equal to \begin{align*} f'(x) &= 6x^2 + 6x \\ f'(1) &= 6(1)^2 + 6(1) = 12 \end{align*} Hence, using the the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_o &= m(x – x_o) \\ y – 0 &= 12 (x- 1) \\ y &= 12x – 12 \quad \cmark \end{align*}

At $x = c$, the parabola has $y$-value given by $y = g(1) = 4(1)^2 + 4(1) + 3 = 11$. The tangent line thus also contains the point (1, 11).

Furthermore, the tangent line has slope equal to \begin{align*} g'(x) &= 8x + 4 \\ g'(1) &= 8(1) + 4 = 12 \end{align*} (Note that this value is the same as $f'(1)$, as expected.) Hence, using the the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_o &= m(x – x_o) \\ y – 11 &= 12 (x- 1) \\ y &= 12x – 1 \quad \cmark \end{align*}

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Question 10: Tangent line intersects *x*-axis (actual exam problem)

Where does the tangent line to the graph of $y = f(x)$ at the point $(x_0,y_0)$ intersect the x-axis?

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This question sounds quite formal, but it just requires the ideas we’ve been using.
Let’s first (I) find an equation for the tangent line. Then we can (II) find where it intersects the x-axis.

(I) To write the equation for the tangent line, we need to know (1) its slope $m$, and (2) a point on the line.

(1) The slope $m$ of the tangent line to the graph of $y = f(x)$ at the point $(x_0, y_0)$ is $m = f'(x_0)$.

(2) The tangent line passes through the point $(x_0,y_0)$.

Hence, using the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_0 = m(x-x_0) \\ y – y_0 = f'(x_0) (x-x_0) \end{align*}

(II) This line intersects the $x$-axis when $y = 0$. Let’s call the particular $x$-value where this intercept occurs $a$, such that: \begin{align*} 0-y_0 &= f'(x_0)(a-x_0) \\ \\ -\frac{y_0}{f'(x_0)} &= a – x_0 \\ \\ a &= x_0 – \frac{y_0}{f'(x_0)} \quad \cmark \end{align*}

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(I) To write the equation for the tangent line, we need to know (1) its slope $m$, and (2) a point on the line.

(1) The slope $m$ of the tangent line to the graph of $y = f(x)$ at the point $(x_0, y_0)$ is $m = f'(x_0)$.

(2) The tangent line passes through the point $(x_0,y_0)$.

Hence, using the point-slope form of a line that contains the point $(x_o, y_o)$, we can write the tangent line as

\begin{align*} y – y_0 = m(x-x_0) \\ y – y_0 = f'(x_0) (x-x_0) \end{align*}

(II) This line intersects the $x$-axis when $y = 0$. Let’s call the particular $x$-value where this intercept occurs $a$, such that: \begin{align*} 0-y_0 &= f'(x_0)(a-x_0) \\ \\ -\frac{y_0}{f'(x_0)} &= a – x_0 \\ \\ a &= x_0 – \frac{y_0}{f'(x_0)} \quad \cmark \end{align*}

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Question 11: Normal line passes through (0, 3/4)

A line normal (perpendicular) to the curve $y = 2x^2$ at a point in the first quadrant also passes through the point $\left(0, \dfrac{3}{4}\right)$. Find an equation for this line.

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As usual, a quick figure helps a lot. Here we’ve shown the curve $y = x^2,$ and the line that passes through the point $(0, \frac{3}{4})$. This line intersects the curve at the point we’re calling $(x_1, y_1).$ **Note that naming this point on the curve with coordinates like this is crucial to our solution.**

There are several pieces of information we have to put together to solve this problem.

(1) The first is that the slope of a line that is normal (perpendicular) to this curve at the point $(x_1, y_1)$ is given by \[m_\text{normal, at $x=x_1$} = -\dfrac{1}{f'(x_1)} \] Since $f(x) = 2x^2,$ we have $f'(x) = 4x.$

For the particular point of interest, $(x_1, y_1),$ where the line intersects the curve, we can write the slope of the normal line as \[m_\text{normal, at $x=x_1$} = -\dfrac{1}{4x_1} \quad \triangleleft \quad (1) \] Keep that in mind for a moment.

(2) We*also* know that the line contains the points $(0, \frac{3}{4})$ and $(x_1, y_1).$ Hence we can write its slope as
\[m_\text{line} = \frac{y_1-\frac{3}{4}}{x_1-0} \quad \triangleleft \quad (2) \]
Now, the magic: the line must actually meet both conditions (1) and (2), and so we must have
\[m_\text{normal, at $x=x_1$} = m_\text{line}\]
This requirement will let us solve for $y_1,$ as you’ll see:
\begin{align*}
m_\text{normal, at $x=x_1$} &= m_\text{line} \\[8px]
-\dfrac{1}{4x_1} &= \frac{y_1-\frac{3}{4}}{x_1} \\[8px]
-\dfrac{1}{4} &= y_1-\frac{3}{4} \\[8px]
– y_1 &= -\frac{3}{4} + \dfrac{1}{4} \\[8px]
-y_1 &= -\frac{1}{2} \\[8px]
y_1 &= \frac{1}{2} \quad \triangleleft
\end{align*}
We now have the *y*-value where the line intersects the curve. That leaves us with a few steps to go, since the question asked for the equation of the line. So let’s next determine $x_1$:
We know that the point $(x_1, y_1)$ lies on the curve $y = 2x^2,$ and so since $y_1 = \frac{1}{2}$ we must have
\begin{align*}
\frac{1}{2} &= 2x_1^2 \\[8px]
x_1^2 &= \frac{1}{4} \\[8px]
x_1 &= \pm \sqrt{\frac{1}{4}} \\[8px]
&= \pm \frac{1}{2}
\end{align*}
Ah, but the problem specifies “at a point in the first quadrant,” and so we choose the *positive* solution:
\[x_1 = \frac{1}{2} \quad \triangleleft \]
And then we immediately know the slope of the normal line, since we decided (1) above
\[m_\text{normal, at $x=x_1$} = -\dfrac{1}{4x_1} \]
we must have
\begin{align*}
m_\text{normal, at $x=1/2$} &= -\dfrac{1}{4\left(\frac{1}{2} \right)} \\[8px]
&= – \frac{1}{2} \quad \triangleleft
\end{align*}
Now we’re actually almost done: we know that the slope of the normal line is $m_\text{normal} = -\dfrac{1}{2},$ and since we were told the line passes through the point $(0, \frac{3}{4})$ we know it has *y*-intercept $b = \frac{3}{4}.$ Hence using the slope-intercept form of a line:
\begin{align*}
y &= m_\text{normal}x + b \\[8px]
y &= -\frac{1}{2}x + \frac{3}{4} \quad \cmark
\end{align*}

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There are several pieces of information we have to put together to solve this problem.

(1) The first is that the slope of a line that is normal (perpendicular) to this curve at the point $(x_1, y_1)$ is given by \[m_\text{normal, at $x=x_1$} = -\dfrac{1}{f'(x_1)} \] Since $f(x) = 2x^2,$ we have $f'(x) = 4x.$

For the particular point of interest, $(x_1, y_1),$ where the line intersects the curve, we can write the slope of the normal line as \[m_\text{normal, at $x=x_1$} = -\dfrac{1}{4x_1} \quad \triangleleft \quad (1) \] Keep that in mind for a moment.

(2) We

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