Rate of Change - AP Calculus AB

Card 0 of 11256

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 $\frac{1}{2}$ ?

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{1}{2})^2}{4}=\frac{\sqrt{3}}{16}$

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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 $\frac{\sqrt{2}}{2}$ ?

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{2}}{2})^2}{4}=\frac{\sqrt{3}}{8}$

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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 $\frac{\sqrt{3}}{3}$ ?

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{3}}{3})^2}{4}=\frac{\sqrt{3}}{12}$

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Question

The sides of a square shrink at a rate of $0.01\frac{in}{s}$ . What is the rate of growth of the square if its sides have lengths of ?

Answer

The area of a square is given by the formula:

The rate of growth of the area can be related to the rate of growth of sides by differentiating each side with respect to time:

$\frac{dA}{dt}=\frac{d}{dt}l^2=2l\frac{dl}{dt}$

Therefore, the rate of growth of the square is:

$\frac{dA}{dt}=2(10in)(-0.01\frac{in}{s})=-0.2\frac{in^2}{s}$

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Question

A child is breathing into a bubble wand to create a soap bubble. Treating the bubble as an expanding sphere, If the sphere has a volume of $\frac{32}{3}\pi cm^3$ and is growing at a rate of $\pi\frac{cm^3}{s}$ , what is the rate of growth of the sphere's radius?

Answer

Begin this problem by solving for the radius. The volume of a sphere is given as

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

Solving for the radius:

$r=\sqrt[3]{\frac{3}{4\pi}\frac{32}{3}\pi cm^3}=2cm$

Now time rate of change between quantities can be found by deriving each side with respect to time:

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

Solving for the radius rate of time then gives:

$\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$

$\frac{dr}{dt}=\frac{1}{16\pi cm^2}(\pi \frac{cm^3}{s})$

$\frac{dr}{dt}=\frac {1}{16}\frac{cm}{s}$

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Question

A regular tetrahedron is growing in size. What is the volume of the tetrahedron at the time the rate of growth of its volume is a $\sqrt{6}$ times the rate of growth of its surface area?

Answer

To tackle this problem, define a regular tetrahedron's dimensions 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, so given our problem condition, the rate of growth of its volume is a $\sqrt{6}$ times the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

$s^2=(\sqrt{6})4\sqrt{6}s$

$s=(\sqrt{6})4\sqrt{6}$

To find the volume then:

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

$V=\frac{24^3}{6\sqrt{2}}$

$V=1152\sqrt{2}$

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Question

The rate of change of a cylinder's radius is equal to a fifth of the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is a fifth of the height?

Answer

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to a fifth of the rate of change of its height: $\frac{dr}{dt}=\frac{1}{5}\frac{dh}{dt}$
The radius is a fifth of the height: $r=\frac{1}{5}h$

Using these properties, rewrite the rate equations:

$\frac{dV}{dt}=2\pi r(5r)\frac{dr}{dt}+\pi r^2 (5\frac{dr}{dt})=15\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi (5r) )\frac{dr}{dt}+2\pi r (5\frac{dr}{dt})=24\pi r \frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{15\pi r^2 \frac{dr}{dt}}{24\pi r \frac{dr}{dt}}=\frac{5}{8}r$

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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 $\sqrt[4]{39}$ ?

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:

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

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

$\phi=\frac{\sqrt{3}(\sqrt[4]{39})^2}{4}=\frac{3\sqrt{13}}{4}$

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Question

The two legs of a right triangle are growing at a rate of and . What is the rate of growth of the hypotenuse at time ?

Answer

The length of the hypotenuse of a right triangle is given by the Pythagorean Theorem:

$c(t)=\sqrt{t^6+t^4+4t^2+4}$

Therefore the rate of change of the hypotenuse can be found by taking the derivative of the equation:

We will use the power rule which states,

$x^n\ dx=nx^{n-1}$ and the chain rule,

$f(g(x))dx=f'(g(x))\cdot g'(x)$

to find the following derivative.

$\frac{dc(t)}{dt}=\frac{6t^5+4t^3+8t}{2\sqrt{t^6+t^4+4t^2+4}}$

$\frac{dc(2)}{dt}=\frac{192+32+16}{2\sqrt{64+16+16+4}}$

$\frac{dc(2)}{dt}=\frac{240}{2\sqrt{100}}$

$\frac{dc(2)}{dt}=12$

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Question

A rectangle currently has a length of and a width of , though the length is growing at a rate of $0.2\frac{m}{s}$ . The area of the rectangle is not changing at the moment. What is the rate of shrinkage of the width?

Answer

The area of a rectangle is given by the function:

To compare rates of change between each parameter, take the derivative of each side of this equation with respect to time:

$\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$

We are told that , , and $\frac{dl}{dt}=0.2\frac{m}{s}$ . If the area is not changing, it follows that $\frac{dA}{dt}=0$ .

This reduces the equation to:

$0=8m\frac{dw}{dt}+6m(0.2\frac{m}{s})$

$8m\frac{dw}{dt}=-1.2\frac{m^2}{s}$

$\frac{dw}{dt}=-0.15\frac{m}{s}$

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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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Question

Find the rate of change of a function from $x_{1}=3$ to $x_{2}=6$ .

Answer

Write the formula for the average rate of change from the interval $[{x_{1},x_{2}}]$ .

$\frac{f(x_2)-f(x_{1})}{x_2-x_1}$

Solve for and .

Substitute the known values into the formula and solve.

$\frac{f(x_2)-f(x_{1})}{x_2-x_1}=\frac{3-(-3)}{6-3}=\frac{6}{3}=2$

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Question

Suppose the rate of a square is increasing at a constant rate of meters per second. Find the area's rate of change in terms of the square's perimeter.

Answer

Since the question is asking for the rate of change in terms of the perimeter, write the formula for the perimeter of the square and differentiate it with the respect to time.

$\frac{dP}{dt}= 4\frac{ds}{dt}$

The question asks in terms of the perimeter. Isolate the term by dividing four on both sides.

$s=\frac{P}{4}$

Write the given rate in mathematical terms and substitute this value into $\frac{dP}{dt}$ .

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

$\frac{dP}{dt}= 4\frac{ds}{dt}=4(2)=8$

Write the area of the square and substitute the side.

$A=s^2=\left(\frac{P}{4}\right)^2=\frac{P^2}{16}$

Since the area is changing with time, take the derivative of the area with respect to time.

$\frac{dA}{dt}=\frac{P}{8} \cdot \frac{dP}{dT}$

Substitute the value of $\frac{dP}{dt}$ .

$\frac{dA}{dt}=\frac{P}{8} \cdot 8=P$

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Question

You are looking at a balloon that is away. If the height of the balloon is increasing at a rate of , at what rate is the angle of inclination of your position to the balloon increasing after seconds?

Answer

Using right triangles we know that

$\tan(\theta(t))= \frac{h(t)}{300} \right$ .

Solving for $\theta(t)$ we get

$\theta(t)=\arctan\left ( \frac{h(t)}{300} \right )$ .

Taking the derivative, we need to remember to apply the chain rule to since the height depends on time,

$\frac{d\theta}{dt}=\frac{1}{1+\left(\frac{h}{300} \right )^2}\frac{1}{300}\frac{dh}{dt}$ .

We are asked to find $\frac{d\theta}{dt}|_{t=5}$ . We are given $\frac{dh}{dt}=50$ and since $\frac{dh}{dt}$ is constant, we know that the height of the balloon is given by $h(t)=\frac{dh}{dt}t$ .

Therefore, at we know that the height of the balloon is $h(5)=5\cdot 50=250$ .

Plugging these numbers into $\frac{d\theta}{dt}$ we find

$\frac{d\theta}{dt}|_{t=5} =\frac{1}{1+\left(\frac{250}{300} \right )^2}\frac{1}{300}50=0.0098$ radians.

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Question

Boat leaves a port at noon traveling . At the same time, boat leaves the port traveling east at . At what rate is the distance between the two boats changing at ?

Answer

This scenario describes a right triangle where the hypotenuse is the distance between the two boats. Let denote the distance boat is from the port, denote the distance boat is from the port, denote the distance between the two boats, and denote the time since they left the port. Applying the Pythagorean Theorem we have,

.

Implicitly differentiating this equation we get

$2D(t)\frac{dD}{dt}=2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}$ .

We need to find $\frac{dD}{dt}$ when .

We are given

$\frac{dx}{dt}=30, \frac{dy}{dt}=20$ which tells us

$x(3)=3\cdot 30=90, y(3)=3\cdot 20=60,$

$D(3)=\sqrt{90^2+60^2}=30\sqrt{13}$ .

Plugging this in we have

$2\cdot 30\sqrt{13}\frac{dD}{dt}|_{t=3}=2\cdot90\cdot 30+2\cdot60\cdot20$ .

Solving we get

$\frac{dD}{dt}|_{t=3}=10\sqrt{13}$ .

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Question

Determine the point on the function that is not changing:

Answer

In order to determine where the function is not changing, it is necessary to take the derivative and set the slope equal to zero. This will provide information on where the curve is not changing. Once we find the x value that gives the derivative a slope of zero, we can substitute the x-value back into the original function to obtain the point.

Substitute this value back to the original equation to solve for .

The point where the function is not changing is .

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