Rate of Change - AP Calculus AB

Card 0 of 11256

Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its diagonal when its sides have length 3.23?

Answer

Begin by writing the equations for a cube's dimensions. Namely its surface area and diagonal in terms of the length of its sides:

$d=s\sqrt{3}$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:

$12s\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$

$\phi=4s\sqrt{3}$

$\phi=4(3.23)\sqrt{3}$

$\phi =22.38$

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Question

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the radius when the radius is 34?

Answer

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

The rate of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(34)^2=4624\pi$

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Question

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the radius when the radius is 28?

Answer

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

The rate of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(28)^2=3136\pi$

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Question

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length 1.8?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)2\sqrt{3}s\frac{ds}{dt}$

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{1.8\sqrt{6}}{24}=0.18$

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Question

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length 15.6?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)2\sqrt{3}s\frac{ds}{dt}$

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{15.6\sqrt{6}}{24}=1.59$

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Question

A sphere is fixed inside of a cube, such that it is completely snug. If the sides of the cube, which have length , begin to grow at a rate of , what is the rate of growth of the volume of the sphere?

Answer

The volume of sphere, in terms of its radius, is defined as

$V=\frac{4}{3}\pi r^3$

However, in the case of the problem, we're given the lengths of the sides of a cube in which the sphere fits. Since the outside of the sphere is touching the sides walls of the cube, the length of the cube is the diameter of the sphere:

$r=\frac{s}{2}$

$r=\frac{6}{2}=3$

Furthermore, the rate of growth of a sphere's radius will be half the rate of growth of it's diameter

$\frac{dr}{dt}=\frac{1}{2}\frac{ds}{dt}$

$\frac{dr}{dt}=\frac{1}{2}(0.4)=0.2$

Now returning to the volume equation

$V=\frac{4}{3}\pi r^3$

The rate of growth can be found by taking the derivative with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dV}{dt}=4\pi (6^2)(0.2)$

$\frac{dV}{dt}=28.8\pi$

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Question

A regular tetrahedron is shrinking in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length 9.6?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)2\sqrt{3}s\frac{ds}{dt}$

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{9.6\sqrt{6}}{24}=0.98$

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Question

A regular tetrahedron is shrinking in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length 44.4?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)2\sqrt{3}s\frac{ds}{dt}$

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{44.4\sqrt{6}}{24}=4.53$

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Question

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length 10.13?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(10.13)^2}{4}=44.4$

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Question

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length 4.2?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(4.2)^2}{4}=7.64$

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Question

A regular tetrahedron is diminishing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length 3.23?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(3.23)^2}{4}=4.52$

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Question

A regular tetrahedron is diminishing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length 5.6?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(5.6)^2}{4}=13.6$

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Question

The rate of change of the radius of a sphere is $9e^{-t}$ . If the sphere has an initial radius of , what is the rate of change of the sphere's surface area at time ?

Answer

To say that the rate of change of the radius of a sphere is $9e^{-t}$ means

$\frac{dr(t)}{dt}=9e^{-t}$

The equation for the length of the radius can be found by integrating this equation with respect to time:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$r(t)=-9e^{-t}+C$

The constant of integration can be found by utilizing the initial condition:

$r(0)=-9e^{0}+C=6$

$r(t)=15-9e^{-t}$

The surface area of a sphere is given by the equation

$A=4\pi r^2$

The rate of change of this area can be found by taking the derivative of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dA(t)}{dt}=8\pi (15-9e^{-t}) (9e^{-t})$

$\frac{dA(ln(3))}{dt}=8\pi (15-9e^{-ln(3)}) (9e^{-ln(3)})$

$\frac{dA(ln(3))}{dt}=288\pi$

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Question

We can interperet a derrivative as $\frac{dy}{dx}= \lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}$ (i.e. the slope of the secant line cutting the function as the change in x and y approaches zero) but these so-called "differentials" ( $\Delta x$ and $\Delta y$ ) can be a good tool to use for aproximations. If we suppose that $\frac{dy}{dx}=f'(x)$ , or equivalently . If we suppose a change in x (have a concrete value for $\Delta x$ ) we can find the change in with the afore mentioned relation.

Let $f(x)= \sin( \ln x)$ . Find $f'(e^{\frac{\pi}{3}})$ let and $\Delta x =0.1$ . Find $\Delta y$ under such conditions.

Answer

We find the derivative of the function:

$f(x)= \sin( \ln x)$

$\frac{dy}{dx}=\frac{\cos(\ln x)}{x}$

Evaluating at $x=e^{\frac{\pi}{3}}$

$\frac{dy}{dx}|_e^{\frac{\pi}{3}} =\frac{\cos(\ln (e^\frac{\pi}{3}))}{(e^\frac{\pi}{3})}$

$\frac{dy}{dx}=\frac{1}{2}{e^{-\frac{\pi}{3}}} \approx 0.1754$

$dy=\frac{1}{2}{e^{-\frac{\pi}{3}}}dx$

Letting

$dy=\frac{1}{2}{e^{-\frac{\pi}{3}}}*.1 \approx 0.01754$

Which is our answer.

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Question

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length $\sqrt[3]{6}$ ?

Answer

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

$\frac{dV}{dt}=\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}$

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(\sqrt[3]{6})^2}{4}=\frac{3\sqrt[6]{3}}{2\sqrt[3]{2}}$

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Question

Find the rate of change of a function from to .

Answer

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

Plugging our values into the average rate of change formula, we get:

$\frac{(26-21)}{(3-2)}=\frac{5}{1}=5$ .

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Question

Find $\frac{dy}{dx}$ for $f(x)= \frac{ln(x)}{x^8}$ .

Answer

To solve this problem, we can use either the quotient rule or the product rule. For this solution, we will use the product rule.

The product rule states that $\frac{d(u*v)}{dx}=u'v+v'u$ .

In this case, let $u=x^{-8}\Rightarrow u'=-8x^{-9}$ and $v=ln(x)\Rightarrow v'=\frac{1}{x}$ .

Putting both of these together, we get

$\frac{d(x^{-8}ln(x))}{dx}=-8x^{-9}ln(x)+\frac{1}{x}x^{-8}=\frac{1}{x^{9}}+\frac{-8ln(x)}{x^9}=\frac{1-8ln(x)}{x^9}$ .

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Question

A leaky trough is ten feet long with isosceles triangle cross sections. The cross sections have a base of two feet and a height of two feet six inches. The trough is being filled with water at one cubic foot per minute. However, it is also leaking at a rate of two cubic feet per minute.

When the depth of the water is one foot five inches, how fast is the water level falling?

Answer

You know the net volume is decreasing at a rate of -1 ft/min by adding the rates 1 (being added) and -2(leaking from the trough). However, the question asks what the rate of change of the height is. The equation V=1/2blh (because the cross sections are triangles; the trough is a prism) relates height to volume.

The length (l) is a constant 10 feet, and the base needs to be written in terms of something we know the rate of change. Because the cross sections are triangles, the sides are proportional.

Therefore, $\frac{b}{2}=\frac{h}{2.5}$ and b=0.8h.

After plugging the known values into the volume equation,

$V=\frac{1}{2}100.8h*h$ or .

Then differentiate both sides to relate the rates of change.

$\frac{dV}{dt}=8h\frac{dh}{dt}$ .

Finally, plug in the known values for the rate of change of volume(dV/dt) -1ft/min and the instantaneous height (1 ft 5 in = 17/12 ft).

$\ -1=8\left(\frac{17}{12} \right )\frac{dh}{dt}\ \-1=\frac{34}{3}\frac{dh}{dt} \ \=-\frac{3}{34}=\frac{dh}{dt}$

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Question

Determine the average rate of change of the function from the interval $\left[\frac{\pi}{2},\pi\right]$ .

Answer

Write the formula to determine average rate of change.

$\frac{\Delta f}{\Delta x}= \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$

Substitute the values and solve for the average rate of change.

$\frac{-cos(\pi)-(-cos(\frac{\pi}{2}))}{\pi-\frac{\pi}{2}} = \frac{-(-1)+0}{\frac{\pi}{2}}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$

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Question

Find $\frac{dV}{dt}$ if the radius of a spherical balloon is increasing at a rate of per second.

Answer

The volume function, in terms of a radius , is given as

$V(r)=\frac{4}{3}\pi r^3$ .

The change in volume over the change in time, or

$\frac{dV}{dt}$ is given as

$\frac{dV}{dt} = \frac{d}{dt}[V(r)]$

and by implicit differentiation, the chain rule, and the power rule,

$\frac{d}{dt}[V(r)]=\frac{d}{dt}[\frac{4}{3}\pi r^3]=4\pi r^2 \frac{dr}{dt}$ .

Setting $\frac{dr}{dt}=2$ we get

$\frac{d}{dt}[V(r)]= 4\pi r^2(2)=8\pi r^2$ .

As such,

$\frac{dV}{dt} =8\pi r^2$ .

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