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Calculus 3 : Calculus 3

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

Example Questions

Example Question #102 : Derivatives

Find the derivative of the function $h(x)=\csc x\tan x$

Possible Answers:

$h'(x)=\csc x \cot x\cdot \tan x+\csc x\cdot \sec^2 x$

$h'(x)=\csc x \cot x\cdot \tan x-\csc x\cdot \sec^2 x$

$h'(x)=-\csc x \cot x\cdot \tan x+\csc x\cdot \sec x$

$h'(x)=-\csc x \cot x\cdot \tan x+\csc x\cdot \sec^2 x$

Correct answer:

$h'(x)=-\csc x \cot x\cdot \tan x+\csc x\cdot \sec^2 x$

Explanation:

To find the derivative of the function $h(x)=\csc x\tan x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with $f(x)=\csc x$ and $g(x)=\tan x$

so we get

$h'(x)=-\csc x \cot x\cdot \tan x+\csc x\cdot \sec^2 x$

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Example Question #103 : Derivatives

Find the derivative of the function $h(x)=e^{x}\cos x$

Possible Answers:

$h'(x)=e^x(\cos x-\sin x)$

$h'(x)=e^x(\cos x+\sin x)$

$h'(x)=e^x(-\cos x+\sin x)$

$h'(x)=-e^x(\cos x+\sin x)$

Correct answer:

$h'(x)=e^x(\cos x-\sin x)$

Explanation:

To find the derivative of the function $h(x)=e^{x}\cos x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=e^x and $g(x)=\cos x$

so we get

$h'(x)=e^x\cdot \cos x+e^x\cdot -\sin x=e^x(\cos x-\sin x)$

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Example Question #104 : Derivatives

Find the derivative of the function $h(x)=e^{\pi x}\sec x$

Possible Answers:

$h'(x)=e^{\pi x}(\pi\sec x+\sec x\tan x)$

$h'(x)=e^{\pi x}(\sec x+\sec^2 x\tan x)$

$h'(x)=e^{\pi x}(\sec^2 x+\sec x\tan x)$

$h'(x)=e^{\pi x}(\pi\sec x+\tan x)$

Correct answer:

$h'(x)=e^{\pi x}(\pi\sec x+\sec x\tan x)$

Explanation:

To find the derivative of the function $h(x)=e^{\pi x}\sec x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with $f(x)=e^{\pi x}$ and $g(x)=\sec x$

so we get

$h'(x)=\pi e^{\pi x}\cdot \sec x+e^{\pi x}\cdot \sec x\tan x=e^{\pi x}(\pi\sec x+\sec x\tan x)$

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Example Question #294 : Calculus 3

Find the derivative of the function $h(x)=(2+x^3)\sec x$

Possible Answers:

$h'(x)=3x^2\cdot \sec x+(2+x^3)\cdot \tan x$

$h'(x)=3x^2\cdot \sec x+(3x^2)\cdot \sec x\tan x$

$h'(x)=3x^2\cdot \sec^2 x+(2+x^3)\cdot \sec x\tan x$

$h'(x)=3x^2\cdot \sec x+(2+x^3)\cdot \sec x\tan x$

Correct answer:

$h'(x)=3x^2\cdot \sec x+(2+x^3)\cdot \sec x\tan x$

Explanation:

To find the derivative of the function $h(x)=(2+x^3)\sec x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=2+x^3 and $g(x)=\sec x$

so we get

$h'(x)=3x^2\cdot \sec x+(2+x^3)\cdot \sec x\tan x$

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Example Question #295 : Calculus 3

Find the derivative of the function $h(x)=x^5\ln x$

Possible Answers:

$h'(x)=5x^4\ln x+x^4$

$h'(x)=5x^4\ln x+5x^5$

$h'(x)=5x^4\ln x+5x^4$

$h'(x)=5x^4\ln x+x^5$

Correct answer:

$h'(x)=5x^4\ln x+x^4$

Explanation:

To find the derivative of the function $h(x)=x^5\ln x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=x^5 and $g(x)=\ln x$

so we get

$h'(x)=5x^4\cdot \ln x+x^5\cdot \frac{1}{x}=5x^4\ln x+x^4$

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Example Question #296 : Calculus 3

Find the derivative of the function $h(x)=x^2\ln x$

Possible Answers:

$h'(x)=x(2\ln x+1)$

h'(x)=2

h'(x)=2x

$h'(x)=x(2\ln x+2)$

Correct answer:

$h'(x)=x(2\ln x+1)$

Explanation:

To find the derivative of the function $h(x)=x^2\ln x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=x^2 and $g(x)=\ln x$

so we get

$h'(x)=2x\cdot \ln x+x^2\cdot \frac{1}{x}=2x\ln x+x=x(2\ln x+1)$

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Example Question #2 : How To Find Position

Function v(t) gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models the particle's postion as a function of time.

Possible Answers:

$\small \small h(t)=\frac{t^6}{5}+\frac{3t^5}{4}-\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}+\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{7}-\frac{7t^3}{27}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}$

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Correct answer:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Explanation:

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

We are starting with the following

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

We need to perform the following:

$\small \small \small \small\int v(t)dt=\int t^5+3t^4-\frac{7t^2}{3}dt$

Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.

$\small \small \int t^5+3t^4-\frac{7t^2}{3}dt=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Which makes our position function, h(t), the following:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

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Example Question #1 : Integration

Consider the velocity function modeled in meters per second by v(t).

v(t)=4t^3-5t^2+6

Find the position of a particle whose velocity is modeled by v(t) after seconds.

Possible Answers:

$8393.3\bar{3} m$

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

$93.3\bar{3}+c {m}$

$833.3\bar{3}+c m$

$393.3\bar{3}m$

Correct answer:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

Explanation:

Recall that velocity is the first derivative of position, so to find the position function we need to integrate v(t) .

$V(t)=\int 4t^3-5t^2+6dt$

Becomes,

$V(t)=t^4-\frac{5t^3}{3}+6t+c$

Then, we need to find V(10)

$V(10)=10^4-\frac{5\cdot10^3}{3}+6\cdot10+c=8393.3\bar{3}+c$

So our final answer is:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

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Example Question #1 : Integration

Find the position function given the velocity function:

v(t)=3t^5+4t

Possible Answers:

15t^4+4

$\frac{1}{2}t^6+\frac{1}{2}t^2$

t^6+t^2

$\frac{1}{2}t^6+2t^2$

3t^6+4t^2

Correct answer:

$\frac{1}{2}t^6+2t^2$

Explanation:

To find the position from the velocity function, integrate

v(t)=3t^5+4t by increasing the exponent of each t term and then dividing that term by the new exponent value.

$s(t)=\int v(t)dt= \frac{3}{6}t^6+2t^2=\frac{1}{2}t^6+2t^2$

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Example Question #2 : 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:

3828979

-3828.979

-38289.79

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #102 : Derivatives

Example Question #103 : Derivatives

Example Question #104 : Derivatives

Example Question #294 : Calculus 3

Example Question #295 : Calculus 3

Example Question #296 : Calculus 3

Example Question #2 : How To Find Position

Example Question #1 : Integration

Example Question #1 : Integration

Example Question #2 : Integration

All Calculus 3 Resources

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