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Calculus 1 : Calculus

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Example Questions

Example Question #202 : How To Find Rate Of Change

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's volume is equal to an eigth of the rate of growth of its surface area?

Possible Answers:

$\sqrt{3}$ $\displaystyle \sqrt{3}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\frac{\sqrt{3}}{2}$ $\displaystyle \frac{\sqrt{3}}{2}$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

$\displaystyle 2$

Correct answer:

$\frac{\sqrt{3}}{2}$ $\displaystyle \frac{\sqrt{3}}{2}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the cube's volume is equal to an eigth of the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{8})12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\frac{1}{8})12s\frac{ds}{dt}$

$s=(\frac{1}{8})4$ $\displaystyle s=(\frac{1}{8})4$

$s=\frac{1}{2}$ $\displaystyle s=\frac{1}{2}$

The diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$ $\displaystyle d=\sqrt{3s^2}$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

$d=\frac{\sqrt{3}}{2}$ $\displaystyle d=\frac{\sqrt{3}}{2}$

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Example Question #203 : How To Find Rate Of Change

A cube is diminishing in size into nothingness. What is the area of one of the cube's sides at the time that the rate of shrinkage of the cube's volume is equal to thirteen times the rate of shrinkage of its surface area?

Possible Answers:

104 $\displaystyle 104$

$\frac{96}{169}$ $\displaystyle \frac{96}{169}$

16224 $\displaystyle 16224$

2704 $\displaystyle 2704$

$\frac{16}{169}$ $\displaystyle \frac{16}{169}$

Correct answer:

2704 $\displaystyle 2704$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's volume is equal to thirteen times the rate of shrinkage of its surface area:

$3s^2\frac{ds}{dt}=(13)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(13)12s\frac{ds}{dt}$

s=(13)4 $\displaystyle s=(13)4$

s=52 $\displaystyle s=52$

The area of one of a cube's is given by the equation:

a=s^2 $\displaystyle a=s^2$

a=52^2 $\displaystyle a=52^2$

a=2704 $\displaystyle a=2704$

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Example Question #204 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the surface area of the tetrahedron at the time the rate of growth of its volume is twice the rate of growth of its surface area?

Possible Answers:

$384\sqrt{3}$ $\displaystyle 384\sqrt{3}$

$84\sqrt{3}$ $\displaystyle 84\sqrt{3}$

$192\sqrt{3}$ $\displaystyle 192\sqrt{3}$

$48\sqrt{3}$ $\displaystyle 48\sqrt{3}$

$96\sqrt{3}$ $\displaystyle 96\sqrt{3}$

Correct answer:

$384\sqrt{3}$ $\displaystyle 384\sqrt{3}$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions in terms of the length of its sides:

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

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

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

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

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

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 twice the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

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

$s=8\sqrt{6}$ $\displaystyle s=8\sqrt{6}$

The surface area of the tetrahedron:

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

$A=\sqrt{3}(8\sqrt{6})^2$ $\displaystyle A=\sqrt{3}(8\sqrt{6})^2$

$A=384\sqrt{3}$ $\displaystyle A=384\sqrt{3}$

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Example Question #3121 : Calculus

A regular tetrahedron is growing in size. What is the height of the tetrahedron at the time the rate of growth of its volume is four times the rate of growth of one of its faces?

Possible Answers:

$\displaystyle 8$

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

$4\sqrt{3}$ $\displaystyle 4\sqrt{3}$

$8\sqrt{3}$ $\displaystyle 8\sqrt{3}$

$\displaystyle 4$

Correct answer:

$\displaystyle 8$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions in terms of the length of its sides:

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

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

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

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

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

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 four times the rate of growth of one of its faces, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(4)\frac{\sqrt{3}}{2}s\frac{ds}{dt}$ $\displaystyle \frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(4)\frac{\sqrt{3}}{2}s\frac{ds}{dt}$

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

$s=4\sqrt{6}$ $\displaystyle s=4\sqrt{6}$

The height of a tetrahedron is given by the equation:

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

$h=\frac{\sqrt{6}}{3}(4\sqrt{6})$ $\displaystyle h=\frac{\sqrt{6}}{3}(4\sqrt{6})$

h=8 $\displaystyle h=8$

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Example Question #2095 : Functions

A regular tetrahedron is growing in size. What is the area of a face of the tetrahedron at the time the rate of growth of its volume is a twelve times the rate of growth of its surface area?

Possible Answers:

$5126\sqrt{3}$ $\displaystyle 5126\sqrt{3}$

$1211\sqrt{3}$ $\displaystyle 1211\sqrt{3}$

$8920\sqrt{3}$ $\displaystyle 8920\sqrt{3}$

$13824\sqrt{3}$ $\displaystyle 13824\sqrt{3}$

$3456\sqrt{3}$ $\displaystyle 3456\sqrt{3}$

Correct answer:

$3456\sqrt{3}$ $\displaystyle 3456\sqrt{3}$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions in terms of the length of its sides:

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

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

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

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

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

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 twelve times the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(12)2\sqrt{3}s\frac{ds}{dt}$ $\displaystyle \frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(12)2\sqrt{3}s\frac{ds}{dt}$

$s^2=(12)4\sqrt{6}s$ $\displaystyle s^2=(12)4\sqrt{6}s$

$s=48\sqrt{6}$ $\displaystyle s=48\sqrt{6}$

The area of a face of a tetrahedron is given by:

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

Since a regular tetrahedron is composed of four identical faces, the area of one is the fourth of the total surface area.

$a=\frac{\sqrt{3}}{4}(48\sqrt{6})^2$ $\displaystyle a=\frac{\sqrt{3}}{4}(48\sqrt{6})^2$

$a=3456\sqrt{3}$ $\displaystyle a=3456\sqrt{3}$

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Example Question #212 : Rate Of Change

A cube is growing in size, such that it's becoming a threat to the planet. What is the volume of the cube at the time that the rate of growth of the cube's volume is equal to one hundred times the rate of growth of the area of one of its sides?

Possible Answers:

$\frac{800000}{27}$ $\displaystyle \frac{800000}{27}$

$\frac{80000}{27}$ $\displaystyle \frac{80000}{27}$

$\frac{8000000}{27}$ $\displaystyle \frac{8000000}{27}$

$\frac{8000}{27}$ $\displaystyle \frac{8000}{27}$

$\frac{800}{27}$ $\displaystyle \frac{800}{27}$

Correct answer:

$\frac{8000000}{27}$ $\displaystyle \frac{8000000}{27}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{da}{dt}=2s\frac{ds}{dt}$ $\displaystyle \frac{da}{dt}=2s\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to one hundred times the rate of growth of the area of one of its sides:

$3s^2\frac{ds}{dt}=(100)2s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(100)2s\frac{ds}{dt}$

$s=(100)\frac{2}{3}$ $\displaystyle s=(100)\frac{2}{3}$

$s=\frac{200}{3}$ $\displaystyle s=\frac{200}{3}$

The volume of this nightmarish cube is then:

$V=(\frac{200}{3})^3$ $\displaystyle V=(\frac{200}{3})^3$

$V=\frac{8000000}{27}$ $\displaystyle V=\frac{8000000}{27}$

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Example Question #211 : How To Find Rate Of Change

Use implicit differentiation to find the derrivative $\frac{dy}{dx}$ $\displaystyle \frac{dy}{dx}$.

find $\frac{dy}{dx}$ $\displaystyle \frac{dy}{dx}$ given the equation $y+x^2=\cos^2(x+y)$ $\displaystyle y+x^2=\cos^2(x+y)$.

Possible Answers:

$\frac{dy}{dx}=-\frac{\sin(x)\cos(x)-2y}{2\sin(x)\cos(x+y)+1}$ $\displaystyle \frac{dy}{dx}=-\frac{\sin(x)\cos(x)-2y}{2\sin(x)\cos(x+y)+1}$

$\frac{dy}{dx}=\frac{\cos(2x)\sin(2x)-y}{2\sin(x+y)\cos(x+y)}$ $\displaystyle \frac{dy}{dx}=\frac{\cos(2x)\sin(2x)-y}{2\sin(x+y)\cos(x+y)}$

$\frac{dy}{dx}=\frac{\cos(2x+2y)\sin(x+y)-x}{2-\cos(x+y)\sin(x+y)}$ $\displaystyle \frac{dy}{dx}=\frac{\cos(2x+2y)\sin(x+y)-x}{2-\cos(x+y)\sin(x+y)}$

$\frac{dy}{dx}=-\frac{\sin(x)\cos(x)-2y}{2\sin(x)\cos(x+y)}$ $\displaystyle \frac{dy}{dx}=-\frac{\sin(x)\cos(x)-2y}{2\sin(x)\cos(x+y)}$

$\frac{dy}{dx}=\frac{-2(\cos(x+y)\sin(x+y)+x)}{1+2\cos(x+y)\sin(x+y)}$ $\displaystyle \frac{dy}{dx}=\frac{-2(\cos(x+y)\sin(x+y)+x)}{1+2\cos(x+y)\sin(x+y)}$

Correct answer:

$\frac{dy}{dx}=\frac{-2(\cos(x+y)\sin(x+y)+x)}{1+2\cos(x+y)\sin(x+y)}$ $\displaystyle \frac{dy}{dx}=\frac{-2(\cos(x+y)\sin(x+y)+x)}{1+2\cos(x+y)\sin(x+y)}$

Explanation:

taking the derivative of both sides of the equation:

$\frac{d}{dx}[y+x^2]=\frac{d}{dx}[\cos^2(x+y)]$ $\displaystyle \frac{d}{dx}[y+x^2]=\frac{d}{dx}[\cos^2(x+y)]$

$\frac{dy}{dx}+2x=2\cos(x+y)*(-\sin(x+y))*(1+\frac{dy}{dx})$ $\displaystyle \frac{dy}{dx}+2x=2\cos(x+y)*(-\sin(x+y))*(1+\frac{dy}{dx})$

simplifying:

$\frac{dy}{dx}+2x=-2\cos(x+y)\sin(x+y)+-2\cos(x+y)\sin(x+y)\frac{dy}{dx}$ $\displaystyle \frac{dy}{dx}+2x=-2\cos(x+y)\sin(x+y)+-2\cos(x+y)\sin(x+y)\frac{dy}{dx}$

Solving for dy/dx

$\frac{dy}{dx}+2\cos(x+y)\sin(x+y)\frac{dy}{dx}=-2\cos(x+y)\sin(x+y)-2x$ $\displaystyle \frac{dy}{dx}+2\cos(x+y)\sin(x+y)\frac{dy}{dx}=-2\cos(x+y)\sin(x+y)-2x$

$\frac{dy}{dx}(1+2\cos(x+y)\sin(x+y))=-2\cos(x+y)\sin(x+y)-2x$ $\displaystyle \frac{dy}{dx}(1+2\cos(x+y)\sin(x+y))=-2\cos(x+y)\sin(x+y)-2x$

$\frac{dy}{dx}=\frac{-2\cos(x+y)\sin(x+y)-2x}{(1+2\cos(x+y)\sin(x+y))}$ $\displaystyle \frac{dy}{dx}=\frac{-2\cos(x+y)\sin(x+y)-2x}{(1+2\cos(x+y)\sin(x+y))}$

Which is our answer.

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Example Question #3121 : Calculus

We can interperet a derrivative as $\frac{dy}{dx}= \lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}$ $\displaystyle \frac{dy}{dx}= \lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}$ (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" ( $\Delta x$ $\displaystyle \Delta x$ and $\Delta y$ $\displaystyle \Delta y$) can be a good tool to use for aproximations. If we suppose that $\frac{dy}{dx}=f'(x)$ $\displaystyle \frac{dy}{dx}=f'(x)$, or equivalently $\displaystyle dy=f'(x)dx$. If we suppose a change in x (have a concrete value for $\Delta x$ $\displaystyle \Delta x$) we can find the change in $\displaystyle y$ with the afore mentioned relation.

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

Possible Answers:

0.3256 $\displaystyle 0.3256$

0.1754 $\displaystyle 0.1754$

0.0036 $\displaystyle 0.0036$

-0.4523 $\displaystyle -0.4523$

0.0175 $\displaystyle 0.0175$

Correct answer:

0.0175 $\displaystyle 0.0175$

Explanation:

We find the derivative of the function:

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

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

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

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

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

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

Letting dx =0.1 $\displaystyle dx =0.1$

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

Which is our answer.

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Example Question #211 : How To Find Rate Of Change

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's volume is equal to fifty times the rate of growth of its surface area?

Possible Answers:

$50\sqrt{3}$ $\displaystyle 50\sqrt{3}$

$40\sqrt{3}$ $\displaystyle 40\sqrt{3}$

$250\sqrt{3}$ $\displaystyle 250\sqrt{3}$

$100\sqrt{3}$ $\displaystyle 100\sqrt{3}$

$200\sqrt{3}$ $\displaystyle 200\sqrt{3}$

Correct answer:

$200\sqrt{3}$ $\displaystyle 200\sqrt{3}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to fifty times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(50)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(50)12s\frac{ds}{dt}$

s=(50)4 $\displaystyle s=(50)4$

s=200 $\displaystyle s=200$

The diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$ $\displaystyle d=\sqrt{3s^2}$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

$d=200\sqrt{3}$ $\displaystyle d=200\sqrt{3}$

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Example Question #211 : Rate Of Change

A spherical balloon is being filled with air. What is the surface area of the sphere at the instance the rate of growth of the volume is a twelth of the rate of growth of the surface area?

Possible Answers:

$\frac{\pi}{3}$ $\displaystyle \frac{\pi}{3}$

$\frac{\pi}{9}$ $\displaystyle \frac{\pi}{9}$

$\frac{\pi}{6}$ $\displaystyle \frac{\pi}{6}$

$3\pi$ $\displaystyle 3\pi$

$6\pi$ $\displaystyle 6\pi$

Correct answer:

$\frac{\pi}{9}$ $\displaystyle \frac{\pi}{9}$

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is a twelth of the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{12})8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\frac{1}{12})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{12})2$ $\displaystyle r =(\frac{1}{12})2$

$r=\frac{1}{6}$ $\displaystyle r=\frac{1}{6}$

Calculating the surface area:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$A=4\pi (\frac{1}{6})^2$ $\displaystyle A=4\pi (\frac{1}{6})^2$

$A=\frac{\pi}{9}$ $\displaystyle A=\frac{\pi}{9}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #202 : How To Find Rate Of Change

Example Question #203 : How To Find Rate Of Change

Example Question #204 : How To Find Rate Of Change

Example Question #3121 : Calculus

Example Question #2095 : Functions

Example Question #212 : Rate Of Change

Example Question #211 : How To Find Rate Of Change

Example Question #3121 : Calculus

Example Question #211 : How To Find Rate Of Change

Example Question #211 : Rate Of Change

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