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Calculus 3 : Calculus Review

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Integration

Consider the velocity function given by v(t) :

v(t)=5t^4-343t^6+51t^2

Find the position of a particle after seconds if its velocity can be modeled by v(t) and the graph of its position function passes through the point (2,2) .

Possible Answers:

-38289.79

3828979

-3828.979

Correct answer:

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

v(t)=5t^4-343t^6+51t^2

$p(t)=V(t)=\int 5t^4-343t^6+51t^2dt$

So we get:

p(t)=t^5-49t^7+17t^3+c

What we ultimately need is p(5), but first we need to find c: Use the point (2,2)

p(t)=t^5-49t^7+17t^3+c

$2=2^5-49\cdot 2^7+17\cdot 2^3+c$

c=6106

So our position function is:

p(t)=t^5-49t^7+17t^3+6106

$p(5)=5^5-49\cdot 5^7+17\cdot 5^3+6106=-3816769$

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Example Question #2 : Integration

The velocity equation of an object is given by the equation v(t) = (4t^2)(3t) . What is the position of the object at time t = 4 if the initial position of the object is ?

Possible Answers:

192

768

Correct answer:

768

Explanation:

The position of the object can be found by integrating the velocity equation and solving for x(4) . To integrate the velocity equation we first rewrite the equation.

v(t) = (4t^2)(3t) = 12t^3

To integrate this equation we must use the power rule where,

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Applying this to the velocity equation gives us,

$x(t) = \int (12t^3)dt = \frac{12t^{3+1}}{3+1} = \frac{12t^4}{4} = 3t^4 +C$ .

We must solve for the value of by using the initial position of the object.

x(0) = 3(0)^4 + C =0 + C= 0

Therefore, C = 0 and x(t) = 3t^4 .

x(4) = 3(4)^4 = 3(256) = 768

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Example Question #1 : Integration

The acceleration of an object is given by the equation a(t) = 4t+1 . What is the equation for the position of the object, if the object has an initial velocity of and an initial position of ?

Possible Answers:

x(t) = 2t^3 - t^2 + 10t

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 5t$

x(t) = 2t^3 + t^2 + 10t

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Correct answer:

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Explanation:

To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Therefore integrating the acceleration equation gives us

$v(t) = \int (4t+1)dt = \frac{4t^{1+1}}{1+1} +\frac{1t^{0+1}}{0+1} +C = 2t^2 + t +C$

We can solve for the value of by using the initial velocity of the object.

v(0) = 2(0)^2 +(0)+C = 10

Therefore C = 10 and v(t) = 2t^2 + t + 10

To find the position of the object we integrate the velocity equation.

$x(t) = \int (2t^2+t+10)dt = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} + \frac{10t^{0+1}}{0+1} +C = \frac{2t^3}{3}+\frac{t^2}{2}+10t +C$

We can solve for this new value of by using the object's initial position

$x(0) = \frac{2(0)^3}{3} + \frac{(0)^2}{2} + 10(0) +C = 0 + C = 0$

Therefore C = 0 and $x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

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Example Question #3 : Integration

The velocity of an object is given by the equation v(t) = 6t -3 . What is the position of the object at time t= 4 if the object has a position of and time t = 1 ?

Possible Answers:

Correct answer:

Explanation:

To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Using this rule we find that

$x(t) = \int (6t-3)dt = \frac{6t^{1+1}}{1+1} - \frac{3t^{0+1}}{0+1} +C = 3t^2 - 3t +C$

Using the position of the object at time t = 1 we can solve for

x(1) = 3(1)^2 - 3(1) + C = 3 - 3 + C =0+C= 6

Therefore C = 6 and x(t) = 3t^2 - 3t +6

We can now find the position at time t = 4 .

x(4) = 3(4)^2 - 3(4) +6 = 48-12 +6 = 42

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Example Question #5 : Integration

The velocity of an object is v(t) = 3t^2 . What is the position of the object if its initial position is 1000 ?

Possible Answers:

x(t) = 6t +1000

x(t) = t^3 +1000

$x(t) = \frac{t^4}{4} +1000$

x(t) = t^3 -1000

Correct answer:

x(t) = t^3 +1000

Explanation:

The position is the integral of the velocity. By integrating with the power rule we can find the object's position.

The power rule is where

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the position of the object is

$x(t) = \int (3t^2)dt = \frac{3t^{(2+1)}}{2+1} +C = t^3 +C$ .

We can solve for the constant using the object's initial position.

x(0) = (0)^3 + C = 0+C =1000

Therefore C= 1000 and x(t) = t^3 +1000 .

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Example Question #891 : Spatial Calculus

If the velocity function of a car is v(t)= 3cos(t) , what is the position when t=1 ?

Possible Answers:

$-\frac{9\pi}{2}$

3sin(1)

3cos(1)

$\frac{3\pi}{2}$

Correct answer:

3sin(1)

Explanation:

To find the position from the velocity function, take the integral of the velocity function.

$\int v(t)\:dt=\int 3cos(t)\:dt$

s(t)=3sin(t)

Substitute t=1 .

s(t=1)=3sin(1)

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Example Question #1 : Integration

If the velocity of a particle is $\small v(t)=t^3-3t^2$ , and its position at $\small \small t=2$ is $\small \small \small s(2)=4$ , what is its position at $\small \small t=4$ ?

Possible Answers:

$\small \small s(4)=4$

$\small s(4)=8$

$\small \small s(4)=0$

$\small \small s(4)=-4$

Correct answer:

$\small s(4)=8$

Explanation:

This problem can be done using the fundamental theorem of calculus. We hvae

$\small \small \int_2^4 v(t)dt=\int_2^4 s'(t)dt=s(4)-s(2)$

where $\small s(t)$ is the position at time $\small t$ . So we have

$\small \small \small s(4)=s(2)+\int_2^4 v(t)dt$

so we need to solve for $\small \small \small \small \int_2^4 v(t)dt$ :

$\small \small \small \small \small \small \int_2^4 v(t)dt=\int_2^4 (t^3-3t^2)dt=\frac{t^4}{4}-t^3|_2^4$

$\small =\left(\frac{4^4}{4}-4^3\right)-\left(\frac{2^4}{4}-2^3\right)=(4^3-4^3)-\left(\frac{16}{4}-8\right)=-(4-8)=4$

So then the position at $\small t=4$ is

$\small \small s(4)=s(2)+\int_2^4 v(t)dt=4+4=8$

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Example Question #11 : Integration

A particle's velocity in two dimensions is described by the functions:

v_x(t)=t^2+sin(t)

v_y(t)=tcos(t^2)+3

If the particle has an initial position of (0,0) , what will its position be at time $t=2\pi$ ?

Possible Answers:

$(\frac{8\pi^3-1}{3},6\pi-1)$

$(\frac{8\pi^3}{3}-1,6\pi)$

$(\frac{4\pi^3}{3},2\pi)$

$(\frac{8\pi^3}{3}+1,6\pi-1)$

$(\frac{8\pi^3}{3},6\pi)$

Correct answer:

$(\frac{8\pi^3}{3},6\pi)$

Explanation:

Position can be found by integrating velocity with respect to time:

$p(t)=\int v(t)dt$

For velocities:

v_x(t)=t^2+sin(t)

v_y(t)=tcos(t^2)+3

The position functions are:

$x(t)=\frac{t^3}{3}-cos(t)+C_x$

$y(t)=\frac{sin(t^2)}{2}+3t+C_y$

These constants of integration can be found by using the given initial conditions:

$x(0)=\frac{0^3}{3}-cos(0)+C_x=0$

C_x=1

$y(0)=\frac{sin(0^2)}{2}+3(0)+C_y=0$

C_y=0

Which gives the definite integrals:

$x(t)=\frac{t^3}{3}-cos(t)+1$

$y(t)=\frac{sin(t^2)}{2}+3t$

$x(2\pi)=\frac{(2\pi)^3}{3}-cos(2\pi)+1$

$x(2\pi)=\frac{8\pi^3}{3}$

$y(2\pi)=\frac{sin((2\pi)^2)}{2}+3(2\pi)$

$y(2\pi)=6\pi$

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Example Question #11 : Integration

Calculate antiderivative of $f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

Possible Answers:

$\ln(x^3+x^2+x)+C$

$\ln(x^3+x^2+x+20)+C$

$\ln(x^3+x^2+1)+C$

$\ln(x^3+x^2+x+1)+C$

Correct answer:

$\ln(x^3+x^2+x+1)+C$

Explanation:

We can calculate the antiderivative $\small f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

$\small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx$

by using $\small u$ -substitution. We set $\small u=x^3+x^2+x+1$ , so we get $\small du=3x^2+2x+1$ , so the integral becomes

$\small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u$

So we just plug $\small u=x^3+x^2+x+1$ to get

$\small \small \small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u=\ln (x^3+x^2+x+1)$

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Example Question #12 : Integration

Calculate antiderivative of $\small f(x)=ex$ .

Possible Answers:

$\small \small \frac{e}{2}x^2+C$

$\small 2e^x+C$

$\small e^x+C$

$\small \small e^2x+C$

Correct answer:

$\small \small \frac{e}{2}x^2+C$

Explanation:

We can calculate the antiderivative $\small \small f(x)=ex$

$\small \small \small \int ex dx$

by using the power rule for antiderivatives:

$\small \int x^a=\frac{x^{a+1}}{a+1}+C$

In this case $\small a=1$ , so we have

$\small \small \small \small \small \int ex dx=e\frac{x^2}{2}+C=\frac{ex^2}{2}+C$ .

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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Integration

Example Question #2 : Integration

Example Question #1 : Integration

Example Question #3 : Integration

Example Question #5 : Integration

Example Question #891 : Spatial Calculus

Example Question #1 : Integration

Example Question #11 : Integration

Example Question #11 : Integration

Example Question #12 : Integration

All Calculus 3 Resources

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