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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #131 : Spatial Calculus

Margot's cat has a position defined by the equation $p(t)=t^{2}+2t-5$ . What is its velocity at time t=4 ?

Possible Answers:

Correct answer:

Explanation:

By definition, velocity is the first derivative of position, or v(t)=p'(t) .

Given a position

$p(t)=t^{2}+2t-5$ , we can use the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ to determine that v(t)=p'(t)=2t+2 .

Therefore, at t=4 ,

v(4)=p'(4)=2(4)+2=8+2=10 .

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Example Question #131 : Calculus

Find the velocity function v(t) if s(t)=3t^2+6t+3 .

Possible Answers:

v(t)=2t+6

v(t)=3t+6

v(t)=6t+6

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

Correct answer:

v(t)=6t+6

Explanation:

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

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

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

$v(t)=\frac{d}{dt}[s(t)]=\frac{d}{dt}(3t^2+6t)=2\cdot 3t^1+1\cdot 6t^0$ .

As such,

v(t)=6t+6 .

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Example Question #131 : How To Find Velocity

Find v(4)

if v(0)=5 and a(t)=t+2 .

Possible Answers:

Correct answer:

Explanation:

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

$v(t)=\int a(t) \, dt$

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule to each term we get,

$v(t)=\int a(t) \, dt=\int t+2 \, dt= \frac{1}{2}t^2+2t+c$ .

To find the value of the constant c we will use the initial condition given in the problem.

Setting the initial condition t=0 , we get

$v(0)=5=\frac{1}{2}0^2 +2(0)+c$

c=5 .

As such,

$v(t)=\frac{1}{2}t^2 +2t+5$

Finally, we set t=4 , which yields

v(4)=21 .

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Example Question #134 : Spatial Calculus

Find the derivative of the position function $(6x-7)^{3}$ .

Possible Answers:

$18(6x-7)^{2}$

$14(6x-7)^{2}$

$20(6x-7)^{2}$

$16(6x-7)^{2}$

$10(6x-7)^{2}$

Correct answer:

$18(6x-7)^{2}$

Explanation:

Looking over our problem, we have a composite function $f\circ g$ , where $f(x)=(g(x))^{3}$ and g(x)=6x-7 . We are trying to find the derivative of $f\circ g$ , or $(f\circ g)'$ , which by the Chain Rule can be defined as $(f\circ g)'=(f'\circ g)\times g'.$

Therefore:

$((6x-7)^{3})'$

$=3(6x-7)^{2}(6x-7)'$

$=3(6x-7)^{2}(6)$

$=18(6x-7)^{2}$

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Example Question #135 : Spatial Calculus

A given car has a position defined by the equation $p(t)=11t^{2}-20t+30$ . What is its velocity at time t=4 ?

Possible Answers:

Correct answer:

Explanation:

By definition, velocity is the first derivative of position, or v(t)=p'(t) .

Given $p(t)=11t^{2}-20t+30$

and using the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$

for all $n\neq0$ ,

we can deduce that

v(t)=p'(t)=22t-20 .

At time t=4 ,

v(4)=p'(4)=22(4)-20=88-20=68 .

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Example Question #131 : How To Find Velocity

Find the velocity of the falling water droplet at time t=2 if its position is given by the following equation:

f(x)=16t^2+48t+90

Possible Answers:

112

Correct answer:

Explanation:

The velocity of the falling water droplet is given by the first derivative of the position equation:

v(t)=32t+48 .

This derivative was found by using the power rule

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

At t=2 , the speed is given by v(2)=112 .

However, we were asked for the droplet's velocity, which is negative when the droplet is falling (moving in the negative direction). So, the final answer is .

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Example Question #137 : Spatial Calculus

Nick throws a ball across the room. Its position is represented by s(t)=-12t^2+36t+10 , where represents distance in feet and represents time in seconds.

What is the velocity of the ball at t=2 ?

Possible Answers:

$-12\frac{ft}{s}$

$12\frac{ft}{s}$

$-60\frac{ft}{s}$

$-36\frac{ft}{s}$

Correct answer:

$-12\frac{ft}{s}$

Explanation:

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

s'(t)= -24t+36

Then, plug in for t=2 :

s'(t)=-24(2)+36

s'(t)=-12

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Example Question #138 : Spatial Calculus

At t=5 , the velocity of a particle . Which of the following expresions describes the position of the particle as a function of time?

Possible Answers:

$s(t)=-\frac{1}{3}t^{3}+\frac{3}{2}t^{2}+\frac{5}{3}$

$s(t)=\frac{1}{3}t^{3}+\frac{3}{2}t^{2}+\frac{1}{3}$

$s(t)=\frac{1}{3}t^{3}-\frac{3}{2}t^{2}+\frac{1}{3}$

$s(t)=-\frac{1}{3}t^{3}-\frac{3}{2}t^{2}+\frac{4}{3}$

Correct answer:

$s(t)=\frac{1}{3}t^{3}-\frac{3}{2}t^{2}+\frac{1}{3}$

Explanation:

To find the equation describing the velocity of a particle given the position s(t), utilize the relationship

v(t)=s'(t)

to check each of the possible answers.

The first derivative of

$s(t)=\frac{1}{3}t^{3}-\frac{3}{2}t^{2}+\frac{1}{3}$

$s'(t)=t^{2}-3t$ .

Evaluating this at the given time of t=5:

$s'(5)=5^{2}-(3)(5)=25-15=10$ .

Therefore,

v(5)=10

which is the condition the question requires.

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Example Question #131 : Spatial Calculus

Find f'(x) , where $f(x)=ln(\tan(x))$ .

Possible Answers:

$\csc(x)\sec (x)$

$\sin(x)$

$\sin(x)\cos(x)$

$\cos(x)$

Correct answer:

$\csc(x)\sec (x)$

Explanation:

In order to find f'(x) , we need to remember the definition of the derivative for natural log.

$\\ g(x)=ln(h(x)) \\ \\ g'(x)=\frac{h'(x)}{h(x)}$

Now lets find f'(x) .

$f(x)=ln(\tan(x))$

$f'(x)=\frac{\sec^2(x)}{\tan(x)}=\csc(x)\sec(x)$

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Example Question #140 : Spatial Calculus

Find f'(x) , where

f(x)=ln(ln(x^2)) .

Possible Answers:

$f'(x)=\frac{ln(x^2)}{2}$

$f'(x)=\frac{2x}{ln(x^2)}$

$f'(x)=\frac{1}{ln(x^2)}$

$f'(x)=\frac{2}{xln(x^2)}$

$f'(x)=\frac{xln(x^2)}{2}$

Correct answer:

$f'(x)=\frac{2}{xln(x^2)}$

Explanation:

In order to find the derivative, we need to use the chain rule and the definition of the derivative of natural log.

Remember that the definition of the derivative of natural log is

g(x)=ln(h(x))

$g'(x)=\frac{h'(x)}{h(x)}$ .

Remember that the chain rule is

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

Lets apply these two rules to our problem.

f(x)=ln(ln(x^2))

$f'(x)=\frac{1}{ln(x^2)}\cdot \frac{1}{x^2}\cdot 2x=\frac{2}{xln(x^2)}$

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #131 : Spatial Calculus

Example Question #131 : Calculus

Example Question #131 : How To Find Velocity

Example Question #134 : Spatial Calculus

Example Question #135 : Spatial Calculus

Example Question #131 : How To Find Velocity

Example Question #137 : Spatial Calculus

Example Question #138 : Spatial Calculus

Example Question #131 : Spatial Calculus

Example Question #140 : Spatial Calculus

All Calculus 1 Resources

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