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Calculus 1 : Calculus

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Example Questions

Example Question #2991 : Calculus

The legs of a right triangle are each growing at a rate of $0.6\frac{in}{s}$ . If one leg has a length of five inches and the other a length of twelve inches, what is the rate of growth of hypotenuse?

Possible Answers:

$0.47\frac{in}{s}$

$0.85\frac{in}{s}$

$0.6\frac{in}{s}$

$0.3\frac{in}{s}$

$0.79\frac{in}{s}$

Correct answer:

$0.79\frac{in}{s}$

Explanation:

Begin by finding the hypotenuse using the Pythagorean theorem:

c^2=a^2+b^2

c^2=5^2in^2+12^2in^2

c^2=169in^2

c=13in

Now, to relate the rates of growth, take the time derivative of the Pythagorean Theorem:

$2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$

$\frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}$

$\frac{dc}{dt}=\frac{5in(0.6\frac{in}{s})+12in(0.6\frac{in}{s})}{13in}$

$\frac{dc}{dt}=0.79\frac{in}{s}$

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Example Question #1962 : Functions

The radius of a circle with a square inscribed inside of it (such that each corner touches the edge of the circle) increases at a rate of $0.7\frac{in}{s}$ . What is the rate of growth of the difference in areas of the circle and square if the radius of the circle is $5\sqrt2$ ?

Possible Answers:

$7\sqrt2(2-\pi)\frac{in^2}{s}$

$7\sqrt2(\pi+2)\frac{in^2}{s}$

$14\sqrt2\frac{in^2}{s}$

$\pi7\sqrt2\frac{in^2}{s}$

$7\sqrt2(\pi-2)\frac{in^2}{s}$

Correct answer:

$7\sqrt2(\pi-2)\frac{in^2}{s}$

Explanation:

The areas of the circle and square can be written in terms of the radius of the circle as follows:

$A_{circle}=\pi r^2 = 50\pi in^2$

$A_{square}=(\frac{2r}{\sqrt2})^2=2r^2=50in^2$

The area of the square follows from the fact that the diameter of the circle forms the diagonal of the square, and each side is equal to the diagonal divided by the square root of two. This can be verified via the Pythagorean Theorem (s^2+s^2=d^2) .

Now, the rate of change of areas can be found by taking the derivative of each function with respect to time:

$\frac{dA_{circle}}{dt}=2\pi r \frac{dr}{dt}$

$\frac{dA_{square}}{dt}=4r \frac{dr}{dt}$

Therefore, the rate of growth of the difference in the two areas can be found by finding the difference of these two quantities:

$2r(\pi-2)\frac{dr}{dt}$

$2(5\sqrt2in)(\pi-2)(0.7\frac{in}{s})$

$7\sqrt2(\pi-2)\frac{in^2}{s}$

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Example Question #1963 : Functions

The sides of a square have lengths 5+t^2 . What is the rate of growth of the area at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

The area of a square is the square of its sides:

A=s^2

A(t)=(5+t^2)^2

At time t=2

A(2)=(5+4)^2

A=81

Now to find the rate of change of the area as related to the rate of change of the sides, take the derivative of the area equation with respect to time:

$\frac{dA(t)}{dt}=2s\frac{ds}{dt}$

Since s=5+t^2 , we will use the power rule which states,

$x^n\ dx=nx^{n-1}$

therefore

$\frac{ds}{dt}=2t$ , giving:

$\frac{dA(t)}{dt}=2(5+t^2)(2t)$

$\frac{dA(2)}{dt}=2(5+2^2)(2(2))$

$\frac{dA(2)}{dt}=72$

Note that the same result would be found if you took the derivative of the equation wherein the 5+t^2 was substituted in for :

A(t)=(5+t^2)^2

Use the power rule and chain rule to find the derivative of area with respect to time.

Chain Rule: $f(g(x)) dx=f'(g(x))\cdot g'(x)$

$\frac{dA(t)}{dt}=2(2t)(5+t^2)$

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Example Question #81 : Rate Of Change

The sides of a rectangle have lengths of 2+t^3 and 1+t^2 . What is the rate of growth of the area at time t=3 ?

Possible Answers:

444

290

145

162

212

Correct answer:

444

Explanation:

The area of the rectangle is given by the product of its sides:

A=wl

A(t)=(2+t^3)(1+t^2)

A(t)=t^5+t^3+2t^2+2

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

We will use the power rule which states,

$x^n\ dx=nx^{n-1}$

to find the following derivative.

$\frac{dA(t)}{dt}=5t^4+3t^2+4t$

$\frac{dA(3)}{dt}=5(81)+3(9)+4(3)$

$\frac{dA(3)}{dt}=405+27+12$

$\frac{dA(3)}{dt}=444$

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Example Question #82 : Rate Of Change

The radius of a circle at any time is given by the equation 3+t^4 . What is the rate of growth of the area of the circle when it has an area of $361\pi$ ?

Possible Answers:

$612\pi$

$608\pi$

$1024\pi$

$512\pi$

$1216\pi$

Correct answer:

$1216\pi$

Explanation:

The area of a circle is given by the equation:

$A=\pi r^2$

$A(t)=\pi(3+t^4)^2$

Find the time that satisfies the area given in the problem:

$361\pi=\pi(3+t^4)^2$

(3+t^4)^2=361

3+t^4=19

t^4=16

t=2

Now that the time is known, to find the rate of area growth, take the derivative of the area equation with respect to time:

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.

$\frac{dA(t)}{dt}=2\pi(4t^3)(3+t^4)$

$\frac{dA(2)}{dt}=2\pi(4(2)^3)(3+(2)^4)$

$\frac{dA(2)}{dt}=2\pi(32)(19)$

$\frac{dA(2)}{dt}=1216\pi$

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Example Question #81 : Rate Of Change

The two legs of a right triangle are growing at a rate of 2+t^2 and t^3 . What is the rate of growth of the hypotenuse at time t=2 ?

Possible Answers:

$2\sqrt10$

$4\sqrt10$

$2\sqrt5$

Correct answer:

Explanation:

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

c^2=a^2+b^2

c^2(t)=(2+t^2)^2+(t^3)^2

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

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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Example Question #1966 : Functions

The radius of a sphere at any time is determined by the expression 1+2t^2+t^3 . What is the rate of growth of the sphere's volume at time t=2 ?

Possible Answers:

$23120\pi$

$89632\pi$

$\frac{19652}{3}\pi$

$6000\pi$

$\frac{32000}{3}\pi$

Correct answer:

$23120\pi$

Explanation:

The volume of a sphere is given by the expression:

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

$V(t)=\frac{4}{3}\pi (1+2t^2+t^3)^3$

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

We will use the chain rule,

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

to find the following derivative.

$\frac{dV(t)}{dt}=4\pi (4t+3t^2)(1+2t^2+t^3)^2$

$\frac{dV(2)}{dt}=4\pi (20)(17)^2$

$\frac{dV(2)}{dt}=23120\pi$

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Example Question #85 : Rate Of Change

The volume of a cube is growing at a rate of 3+t^2 . What is the rate of growth of the side at time t=3 if the cube has an initial volume of ?

Possible Answers:

$\sqrt[3]{12}$

$\frac{4}{9}$

$\sqrt[3]6$

Correct answer:

$\frac{4}{9}$

Explanation:

We are given the rate of change of the cube's volume:

$\frac{dV(t)}{dt}=3+t^2$

The volume equation can be found by taking the integral of this function with respect to time:

We will use the following rule which states,

$\int x^n\ dx=\frac{x^{n+1}}{n+1}$ to find the following integral.

$V(t)=3t+\frac{1}{3}t^3+C$

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

$V(0)=3(0)+\frac{1}{3}(0^3)+C=9$

C=9

Giving the volume as a funciton of time:

$V(t)=3t+\frac{1}{3}t^3+9$

Now, find the volume at time t=3 :

$V(3)=3(3)+\frac{1}{3}3^3+9$

V(3)=9+9+9

V(3)=27

From this we can find the length of a side at this time:

V(t)=s(t)^3

s(3)^3=V(3)=27

s(3)=3

Now, let's return the volume equation V(t)=s(t)^3 ; we can relate rates of change by taking the derivative of each side with respect to time:

$\frac{dV(t)}{dt}=3s(t)^2\frac{ds(t)}{dt}$

At time t=3

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

$\frac{ds(3)}{dt}=\frac{3+t^2}{3s(3)^2}$

$\frac{ds(3)}{dt}=\frac{12}{27}$ (Recall that s(3)=3 )

$\frac{ds(3)}{dt}=\frac{4}{9}$

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Example Question #1967 : Functions

Find the rate of change of a line connected by the points (1, 7) and (4, 28) .

Possible Answers:

m=28

m=7

m=4

m=0

m=x

Correct answer:

m=7

Explanation:

To find the rate of change of a line connected by two points. We will use the following equation for slope.

$m=\frac{\Delta y}{\Delta x}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{28-7}{4-1}=\frac{21}{3}=7$

The rate of change of the line is

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Example Question #2998 : Calculus

$z=x^{3}-2y^{2}$ where , , and all change with respect time. When x=3 and y=4 , changes at rate of 5 units/s and changes at rate of 7 units/s . At what rate is changing?

Possible Answers:

$\frac{dz}{dt}=17$

$\frac{dz}{dt}=1$

$\frac{dz}{dt}=12$

$\frac{dz}{dt}=23$

$\frac{dz}{dt}=203$

Correct answer:

$\frac{dz}{dt}=23$

Explanation:

The rate at which changes is $\frac{dz}{dt}$ . The find $\frac{dz}{dt}$ , we will find the first derivative of the equation

$z=x^{3}-2y^{2}$

$\frac{d}{dt}(z)=\frac{d}{dt}\left [x^{3}-2y^{2} \right ]=3x^{2}\frac{dx}{dt}-4y\frac{dy}{dt}$

We are given the following parameters

x=3 , y=4 , $\frac{dx}{dt}=5 units/s$ , and $\frac{dy}{dt}=7 units/s$ .

Subsituting these values into our derivative equation

$\frac{dz}{dt}=3x^{2}\frac{dx}{dt}-4y\frac{dy}{dt}$

$\frac{dz}{dt}=3(3)^{2}(5)-4(4)(7)=135-112=23$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2991 : Calculus

Example Question #1962 : Functions

Example Question #1963 : Functions

Example Question #81 : Rate Of Change

Example Question #82 : Rate Of Change

Example Question #81 : Rate Of Change

Example Question #1966 : Functions

Example Question #85 : Rate Of Change

Example Question #1967 : Functions

Example Question #2998 : Calculus

All Calculus 1 Resources

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