#  ## Related Rates

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Anonymous
57 minutes ago

What if they only give the rate at which the upper end (point where ladder meets wall) slides down and not the rate at which the bottom moves. Specifically what rate does theta (angle between ground and base of ladder) change when the top of a 30 ft long ladder slides down a wall at .5ft/s and is 18 ft above the ground at that instant?

Matheno
Editor
12 minutes ago

Nice question! In that case, they’re specifying the rate dy/dt — the rate at which the top of the ladder, y, is changing. Specifically, they’re telling you that dy/dt = -0.5 ft/s.

That means in Step 2, the equation you’d want to start with is sin(theta) = y/30, since the ladder is 30 feet long. (That’s the key starting point here. Does that make sense?)

Then the rest of the steps are essentially the same as above: [Step 3] Take the derivative (now of sin instead of cos). [Step 4] Solve for d(theta)/dt. You’ll have to do a subproblem just like above to find the value of cos(theta) at that instant. Any of the three approaches we show above will let you do that. Then you’ll have all the values you need to substitute in and find your final answer.

We hope that helps!

Anonymous
12 days ago

this is poggers 🙂

Matheno
Editor
12 days ago

Thanks. Glad we could help! : )

Julio Ernesto Argueta
7 months ago

COULD IT WORK WITH TANGENT

Matheno
Editor
7 months ago

$\tan \theta = \frac{y}{x}$
instead of using $\cos \theta = x/10$ as we did above?

Assuming that right: great question, and we love that you’re thinking this through!

Writing $\tan \theta = y/x$ in Step 2 would be a correct representation of the situation given the figure we drew in Step 1. So up to that point, yes.

However, you’d run into a problem in the following steps: When you take the derivative in Step 3, because both x and y are changing as the ladder slides down the wall, you would have both $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ terms. And then when you come to Step 4, we don’t know the value of that latter quantity, so you’d be stuck. (Note also that the derivative would be much harder to take in Step 3, since both x and y are variables and so you’d have to use the quotient rule. By contrast, when using $\cos \theta$ the 10 in the denominator on the right is a constant and so the derivative is quite simple to write down.)

So in short: as a practical matter, no, you can’t use $\tan \theta$ to arrive at the solution here. (At least not nearly as easily. You actually find $\dfrac{dy}{dt}$ as its own problem – see “How fast is the ladder’s top sliding?” But then you’re solving an entirely different problem as another sub-part of this problem when it’s not necessary.)

But we want to emphasize that starting your solution attempt with $\tan \theta = y/x$ is perfectly reasonable. It’s only through experience (and heading down a lot of ultimately-unproductive paths) while we were learning this stuff ourselves that we now know more quickly which the best approach to take is. So: keep practicing!!