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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #111 : Velocity

$\\ f(x)=\frac{1}{x^{90}+ln(x)}$

Find f'(x)

Possible Answers:

${(x^{90}+ln(x))^2}$

$\frac{-90x^{89}+\frac{1}{x}}{x^{90}+ln(x)}$

$\frac{1}{(x^{90}+ln(x))^2}$

$\frac{-90x^{89}+\frac{1}{x}}{(x^{90}+ln(x))^2}$

$\frac{(x^{90}+ln(x))^2}{-90x^{89}+\frac{1}{x}}$

Correct answer:

$\frac{-90x^{89}+\frac{1}{x}}{(x^{90}+ln(x))^2}$

Explanation:

$\\ f(x)=\frac{1}{x^{90}+ln(x)}$

Rewrite f(x) as $f(x)=(x^{90}+ln(x))^{-1}$

Use the chain rule and power rule to obtain $\ f'(x)$ .

Remember that the chain rule is $[f(g(x))]'=f'(g(x))\cdot g'(x)$ and the power rule is,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying these rules we get,

$\\ \\ f'(x)=-1\cdot (x^{90}+ln(x))^{-2}\cdot \left(90x^{89}+\frac{1}{x}\right)$ .

This can be rewritten as

$\\ \\ f'(x)=\frac{-90x^{89}+\frac{1}{x}}{(x^{90}+ln(x))^2}$ .

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Example Question #111 : Spatial Calculus

A ball is thrown upwards at a speed of 10m/s from a 100 m building. (Assuming gravity is 9.8m/s^2 ).

After one second, in which direction is the ball travelling?

Possible Answers:

The ball is not moving (at a minimum).

The ball is not moving (at a maximum).

Down

The ball is moving horiztonally, not vertically.

Correct answer:

Explanation:

The velocity function is v=-9.8t+10 , which is derived from the general formula for velocity,

$v(t)=a \cdot t+v_{0}$ .

Where the acceleration due to gravity and initial velocity are given by $a=-9.8, v_{0}=10$ .

Plugging in t=1 we arrive at a velocity of 0.2 m/s .

So after one second the ball is travelling upward.

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Example Question #111 : Spatial Calculus

A ball is thrown upwards at a speed of 10 m/s from a 100m building. Assume gravity is 9.8m/s^2 .

What is the velocity of the ball after three seconds?

Possible Answers:

-13.2 m/s

-9.6 m/s

-19.6 m/s

-19.4 m/s

-4.4 m/s

Correct answer:

-19.4 m/s

Explanation:

We can integrate the constant acceleration of gravity by the general formula for integrating a constant, $\int a dt = at+c$ , where and are both constants.

This gives us a velocity function. Then plugging in our initial velocity, we arrive at an expression for the velocity of the ball over time: $v(t)=\int a(t)dt = \int -9.8 dt=-9.8t+v_0=-9.8t+10$ .

We then can plug in t=3 to get the velocity of the ball at seconds and arrive at our answer:

v(3)=-9.8(3)+10=-19.4 m/s .

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Example Question #114 : Velocity

A given frisbee has a position defined by the equation $p(t)=3t^{2}+5t-2$ . What is its velocity at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

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

Given

$p(t)=3t^{2}+5t-2$ and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then

v(t)=p'(t)=6t+5 .

Plugging in t=2 , we get

v(2)=6(2)+5=12+5=17 .

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Example Question #115 : Velocity

A given horse has a position defined by the equation $p(t)=5t^{2}-7t+12$ . What is its velocity at time t=3 ?

Possible Answers:

Correct answer:

Explanation:

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

Given

$p(t)=5t^{2}-7t+12$ and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then

v(t)=p'(t)=10t-7 .

Plugging in t=3 , we get

v(3)=10(3)-7=30-7=23 .

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Example Question #112 : Spatial Calculus

A given boat has a position defined by the equation $p(t)=10t^{2}-2t+19$ . 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

$p(t)=10t^{2}-2t+19$ and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then

v(t)=p'(t)=20t-2 .

Plugging in t=4 , we get

v(4)=20(4)-2=80-2=78 .

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Example Question #111 : Spatial Calculus

Given and , find . Round to the nearest hundredth. a(t) and v(t) represent the acceleration and velocity functions, respectively.

Possible Answers:

153.41

136.91

148.41

159.91

Not enough information

Correct answer:

159.91

Explanation:

In order to solve for velocity at five seconds, we need to integrate a(t) to get v(t) .

Recall that,

$\int e^t=e^t$ . The power rule, $\int x^n=\frac{x^{n+1}}{n+1}$ , will be used to integrate the second term in the function.

Thus resulting in,

$\small v(t)=\int a(t)dt=e^t+t^2/2+c$ .

To find $\small c$ , we'll use our initial condition

$\small v(0)=0$ . $\small \small \small \small 0=e^{0}+0^2/2+c=1+c \therefore 0=1+c \rightarrow c=-1$ .

Now we can find $\small v(5)$ by simply plugging in $\small t=5$ and $\small c=-1$ into our $\small v(t)$ .

e^5+5^2/2-1=v(5) =e^5+23/2 .

In decimal form, this equals $\small 159.91$ . No units were mentioned in the problem, so leaving the answer unitless is acceptable.

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Example Question #113 : Spatial Calculus

Given the equation of the position of a ball at time is S(t)=4.9t^2+6t-4 , find the velocity of the ball when t=4 .

Possible Answers:

44.2

None of the above.

46.2

45.2

Correct answer:

45.2

Explanation:

The velocity of an object is defined as the change in position of that object over the change in time. Since we are given the position equation in this problem, we simply take the derivative with respect to time of S(t) . To take the derivative of this equation, we must use the power rule,

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

We also must remember that the derivative of an constant is 0. After taking the derivative of the function by applying the power rule, the funciton becomes

v(t)=9.8t+6 .

The questions states we must find the velocity of the ball when t=4 , therefore we simply plug in 4 for t and find that the velocity of the ball at that time is 45.2 .

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Example Question #113 : Velocity

Boulder question

A boulder rolls off the side of a seaside cliff that is above the water. The boulder is traveling at directly to the right when it leaves the cliff.

What is the vertical velocity of the boulder when it is 200m above the water? Assume that gravity is the only force acting on the boulder, causing a downward acceleration of .

Possible Answers:

-117.1m/s

-70.0m/s

-62.6m/s

-132.8m/s

-99.0m/s

Correct answer:

-99.0m/s

Explanation:

The vertical acceleration is given by a(t)=-9.8 .

Initial vertical velocity is .

To integrate we do,

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

Integrating gives v(t)=-9.8t given our initial vertical velocity.

Integrating again gives a vertical position of y(t)= -4.9t^2+C .

Since we know the boulder starts at a height of 700 m , we determine that C=700 .

Solving 200=-4.9t^2+700 gives t=10.10

Evaluate the vertical velocity function when t=10.10

v(10.10)=-9.8(10.10)=-99.0m/s

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Example Question #120 : Velocity

Let $f(t) = 4sin(\pi t)$ represent the position of a swinging marble.

Find the marble's velocity when t=3 .

Possible Answers:

$-4\pi$

$12\pi$

$4\pi$

Correct answer:

$-4\pi$

Explanation:

First, find the function representing velocity by differentiating the position function.

$f(t) = 4sin(\pi t)$

Recall the derivative of sin(u) is $cos(u)\frac{du}{dx}$ .

Applying this rule we get,

$f'(t)=4\pi cos(\pi t)$ .

Then, evaluate f'(t) at t=3 to get the answer.

$f'(3)=4\pi cos(3\pi)$

$=4\pi (-1)$

$=-4\pi$

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #111 : Velocity

Example Question #111 : Spatial Calculus

Example Question #111 : Spatial Calculus

Example Question #114 : Velocity

Example Question #115 : Velocity

Example Question #112 : Spatial Calculus

Example Question #111 : Spatial Calculus

Example Question #113 : Spatial Calculus

Example Question #113 : Velocity

Example Question #120 : Velocity

All Calculus 1 Resources

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