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Calculus 1 : Calculus

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Example Questions

Example Question #101 : Acceleration

Find f''(x) of

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

Possible Answers:

$f''(x)=-\csc^2(x)$

$f''(x)=\csc(x)$

$f''(x)=\tan^2(x)$

$f''(x)=\sec^2(x)$

Correct answer:

$f''(x)=-\csc^2(x)$

Explanation:

In order to find f''(x) , we need to find f'(x) first using the definition of derivative for the natural log and sine which are,

$ln(u)\rightarrow \frac{1}{u}\cdot u'$ and $sin(x)\rightarrow cos(x)$ .

Therefore we get,

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

$f'(x)=\frac{1}{\sin(x)}\cdot \cos(x)=\cot(x)$

This was found by using the chain rule.

Remember that the chain rule is

$[f(g(x))]'=f'(g(x))\cdot g'(x)$

Now we can find f''(x) .

$f''(x)=-\csc^2(x)$

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Example Question #501 : Spatial Calculus

An apple rolls off of a table. It's position is given by the following: s(t)=-4t^2+8t+16 .

What is the acceleration of the apple?

Possible Answers:

0m/s^2

-16m/s^2

-8m/s^2

8m/s^2

Correct answer:

-8m/s^2

Explanation:

Take the second derivative of s(t) by using the Power Rule $x^n=nx^{n-1}$ twice .

Applying the power rule once will give us our velocity function,

s'(t)=-8t+8 .

Applying the power rule a second time results in our acceleration function,

s''(t)=-8 .

Therefore, the acceleration of the apple is -8m/s^2 .

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Example Question #502 : Spatial Calculus

A boulder rolls into a valley. Its position is given by s(t)=5t^2-10t+25 , where represents distance in yards and represents time in seconds.

What is the acceleration of the boulder at t=4 ?

Possible Answers:

10yd/s

5yd/s

20yd/s

2yd/s

Correct answer:

10yd/s

Explanation:

Take the second derivative by using the Power Rule $x^n=nx^{n-1}$ of s(t) ,

$\\s'(t)=v(t)\\ v(t)=5(2)t-10\\v(t)=10t-10$ .

Applying the power rule a second time we find the acceleration function,

s''(t)=10 .

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Example Question #503 : Spatial Calculus

Lola tosses an acorn into the lake. The position of the acorn is represented by s(t)=-3t^2+5t-4 , where represents distance in feet and represents time in seconds.

What is the acceleration of the acorn?

Possible Answers:

3ft/s

6ft/s

-3ft/s

-6ft/s

Correct answer:

-6ft/s

Explanation:

Take the second derivative by using the Power Rule $x^n=nx^{n-1}$ of s(t) ,

$\\s'(t)=-3(2)t+5\\s'(t)=-6t+5$ .

Applying the power rule again we find our acceleration function,

$\\s''(t)=a(t)\\s''(t)=-6$ .

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Example Question #504 : Spatial Calculus

Sian tosses an apple into the air. Its position is represented by s(t)=-4t^2+12t-12 , where represents distance in feet and represents time in seconds.

What is the acceleration of the apple?

Possible Answers:

-16ft/s

4ft/s

12ft/s

-8ft/s

Correct answer:

-8ft/s

Explanation:

Take the second derivative by using the Power Rule $x^n=nx^{n-1}$ of s(t) ,

s'(t)=-4(2)t+12 .

Applying the power rule again we get the acceleration function,

$\\s''(t)=a(t)\\a(t)=-8$ .

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Example Question #505 : Spatial Calculus

Piper throws a softball. Its position is represented by s(t)=-8t^2+20t-12 , where represents distance in meters and represents time in seconds.

What is the acceleration of the softball?

Possible Answers:

-16m/s

20m/s

-8m/s

4m/s

Correct answer:

-16m/s

Explanation:

Take the second derivative by using the Power Rule $x^n=nx^{n-1}$ of s(t) ,

s'(t)=-8(2)t+20 .

Applying the power rule a second time we find the acceleration function,

$\\s''(t)=a(t)\\a(t)=-16$ .

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Example Question #111 : Acceleration

Find the acceleration function

a(t)

s(t)=t^3+5t^2+2t+4 .

Possible Answers:

a(t)=6t+10

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

a(t)=6t

a(t)=16t

Correct answer:

a(t)=6t+10

Explanation:

In order to find the acceleration function from the position function we need to take the second derivative of the position function.

$a(t)=\frac{d^2}{dt^2}[s(t)]$

When taking the derivative, we will use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x} x^{n}=nx^{n-1}$

and by applying this rule to each term we get

$a(t)=\frac{d^2}{dt^2}[s(t)]$

$=\frac{d^2}{dt^2}[t^3+5t^2+2t+4]$

$=\frac{d}{dt}[3t^{3-1}+2\cdot 5t^{2-1}+2t^{1-1}+0]$

$=\frac{d}{dt}[3t^2+10t+2]$ .

Next we find the second derivative,
$a(t)=\frac{d}{dt}[3t^2+10t+2]$

$=2\cdot 3t^{2-1}+10t^{1-1}+0$

=6t+10 .

Hence,

a(t)=6t+10 .

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Example Question #507 : Spatial Calculus

Find a(4) if

v(t)=5t^2 +3 .

Possible Answers:

Correct answer:

Explanation:

In order to find the acceleration function from the velocity function we need to take the derivative of the velocity function.

$a(t)=\frac{d}{dt}[v(t)]$

When taking the derivative, we will use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x} x^{n}=nx^{n-1}$

and by applying this rule to each term we get

$\frac{d}{dt}[v(t)]=\frac{d}{dt}[5t^2+3]=2\cdot 5t^{2-1}+0$ .

As such,

a(t)=10t .

Finally, to solve for a(4) we set t=4 to get

a(4)=40 .

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Example Question #501 : Calculus

A particle is moving according to the following equation:

f(x)=2x^2+3x+1

What is the acceleration of the particle at the point (2,15) ?

Possible Answers:

Correct answer:

Explanation:

The position function of the particle is given as follows:

f(x)=2x^2+3x+1

The first derivative of the position function describes the velocity of the particle:

f'(x)=4x+3
This derivative was found by using the power rule $\frac{d}{dx}x^n=nx^{n-1}$ , the derivative of x=1, and the derivative of a constant equaling zero.

The second derivative of the function - the derivative of the first derivative of the position function - describes the acceleration of the particle:

f''(x)=4

The derivative was found by the derivative of x=1, and the derivative of the constant equaling zero.

Thus, the acceleration of the particle is 4.

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Example Question #509 : Spatial Calculus

A spaceship is traveling from Venus to Mercury according to the following equation:

s(t)=4t^4+28t^2+9

Find the acceleration of the spaceship at t=10 .

Possible Answers:

545

None of the other answers

16560

4856

Correct answer:

4856

Explanation:

The acceleration of the spaceship is given by the second derivative of the position equation.

The first derivative - which describes the velocity of the spaceship - is given by

v(t)=16t^3+56t .

This derivative was found using the power rule This derivative was found by using the power rule

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

The derivative of this function (the second derivative of the initial function)- the acceleration of the spaceship - is given by

a(t)=48t^2+56 .

This derivative was also found using the power rule, and the rule that the derivative of a constant (in this case, 56) equals zero.

At t=10 , the acceleration

a(10)=48(10)^2+56

a(10)=4856 .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #101 : Acceleration

Example Question #501 : Spatial Calculus

Example Question #502 : Spatial Calculus

Example Question #503 : Spatial Calculus

Example Question #504 : Spatial Calculus

Example Question #505 : Spatial Calculus

Example Question #111 : Acceleration

Example Question #507 : Spatial Calculus

Example Question #501 : Calculus

Example Question #509 : Spatial Calculus

All Calculus 1 Resources

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