i. Geometric fact. Typical problems: A circle’s radius changes, a snowball melts, a rectangle’s height and/or width changes.
ii. Trig function. Typical problems: A searchlight rotates, a rocket takes off, a kite travels horizontally.
iii. Similar triangles. Water fills a cone or trough, sand falls onto a conical pile, person walks away from a light pole that casts a shadow.
iv. Pythagorean theorem. Typical problems: Cars/ships/joggers move along 90 degree paths, baseball players run along the diamond, boat is pulled toward a dock.
The snowball is always a sphere.
[Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr in this problem, and we’re looking for $\dfrac{dr}{dt}$.] Most people find that writing the explicit time-dependence V(t) and r(t) annoying, and so just write V and r instead. Regardless, you must remember that r depends on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dr}{dt}$ term.
At what rate does the radius decrease?
then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{\dfrac{1}{200}} \, \dfrac{\text{cm}}{\text{hr}} \quad \checkmark$[Recall $\dfrac{dx(t)}{dt} = .5 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dP(t)}{dt}.$] Most people find that writing the explicit time-dependence $P(t)$ and x(t) annoying, and so just write P and x instead. Regardless, you must remember that both P and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.
[Recall $\dfrac{dx(t)}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dA(t)}{dt}$ at the moment when $A(t) = 25 \text{ m}^2.$] Most people find that writing the explicit time-dependence $A(t)$ and x(t) annoying, and so just write A and x instead. Regardless, you must remember that both A and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.
[Recall $\dfrac{dx(t)}{dt} = .05 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dV(t)}{dt}$ at the moment when $x(t) = 0.4$ m.] Most people find that writing the explicit time-dependence $V(t)$ and x(t) annoying, and so just write V and x instead. Regardless, you must remember that both V and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.
[Recall $\dfrac{dx(t)}{dt} = .05 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dA(t)}{dt}$ at the moment when $x(t) = 0.6$ m.] Most people find that writing the explicit time-dependence $A(t)$ and x(t) annoying, and so just write A and x instead. Regardless, you must remember that both A and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.
[Recall $\dfrac{dy(t)}{dt} = -0.1 \, \tfrac{\text{ft}}{\text{s}}$, and we’re looking for $\dfrac{dV(t)}{dt}.$] Most people find that writing the explicit time-dependence $V(t)$ and y(t) annoying, and so just write V and y instead. Regardless, you must remember that both V and y depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dy}{dt}$ term.
At what rate does the volume decrease?
then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{2.8} \, \dfrac{\text{ft}^3}{\text{s}} \quad \checkmark$[Recall $\dfrac{dx(t)}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{d\theta(t)}{dt}$ at the moment when $\ell = 50$ m.] Most people find that writing the explicit time-dependence $\theta(t)$ and x(t) annoying, and so just write $\theta$ and x instead. Regardless, you must remember that both $\theta$ and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{d\theta}{dt}$ and $\dfrac{dx}{dt}$ terms.
At what rate does the angle decrease?
then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{0.02} \, \dfrac{\text{rad}}{\text{s}} \quad \checkmark$[Recall $\dfrac{dx(t)}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dy(t)}{dt}$ at the moment when $x(t) = 5.0$ m.] Most people find that writing the explicit time-dependence $y(t)$ and x(t) annoying, and so just write y and x instead. Regardless, you must remember that both y and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.
At what rate does the shadow’s height decrease?
then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{0.24} \, \dfrac{\text{m }}{\text{s}} \quad \checkmark$You can support our work with a cup of coffee
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