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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #107 : How To Find Position

Given the two points in three-space:

A=(3,5,-10);B=(2,-4,1)

Find the midpoint

Possible Answers:

(-0.5,4.5,5.5)

(-2.5,0.5,-4.5)

(2.5,4.5,5.5)

(2.5,0.5,-4.5)

(0.5,-4.5,-5.5)

Correct answer:

(2.5,0.5,-4.5)

Explanation:

Given two points:

A=(x_1,y_1,z_1);B=(x_2,y_2,z_2)

The midpoint of $\overrightarrow{AB}$ is given as:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$

Using our given coordinates, the midpoint is found by simple application of this formula:

$(\frac{3+2}{2},\frac{5+(-4)}{2},\frac{-10+1}{2})=(\frac{5}{2},\frac{1}{2},\frac{-9}{2})=(2.5,0.5,-4.5)$

This is one of the answer choices.

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Example Question #108 : How To Find Position

You are given the velocity function v(t)=12t-6t^2 . When t = 0 , the position x(t) is also equal to . Find the position when t = 2 .

Possible Answers:

Correct answer:

Explanation:

To find the position function from the velocity function, take the integral of the velocity function, using the power rule for integration:

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

Applying this formula to the given equation, remembering to multiply by the constants in each term:

$\int12t-6t^2 = 6t^2 - 2t^3 + C = x(t)$

Find the value of by plugging t = 0 into x(t) .

$6\cdot0^2 - 2\cdot0^3 + C = 0$

C = 0

Therefore

x(t) = 6t^2 - 2t^3

Evaluate this at t=2 to find the position.

$x(2) = 6\cdot2^2 - 2\cdot2^3$

x(2) = 24 - 16

x(2) = 8

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Example Question #961 : Calculus

What is the position function if the velocity function is

v(t)=sec^2 (x) .

Possible Answers:

s(t)=sec(t)+c

s(t)=sin(t)+c

s(t)=tan(t)+c

s(t)=cos(t)+c

Correct answer:

s(t)=tan(t)+c

Explanation:

In order to find the position function from the velocity function we must take the integral of the velocity function since

$s(t)=\int v(t)\,dt$ .

When taking the integral, we use the trigonometric property which states

$\int sec^2(x)\,dx = tan(x)$ .

As such,

$\int v(t)\, dt=\int sec^2 (t)\,dt$

=tan(t)+c

Therefore the position function is

s(t)=tan(t)+c .

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Example Question #962 : Calculus

If the function b(s) models the acceleration of a ping pong ball being thrown across a table, find the function which models the ball's position if the velocity when s=3 is 108 and the position when s=5 is 45.

b(s)=-15s^2+18s-8

Possible Answers:

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s-378.75$

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s+45$

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s+473.25$

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s$

Correct answer:

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s-378.75$

Explanation:

b(s)=-15s^2+18s-8

We are asked to relate acceleration and position. This may seem confusing, but if we remember that acceleration is the first derivative of velocity, we know that we should start by integrating b(s).

$\int b(s)=\int -15s^2+18s-8ds=-5s^3+9s^2-8s+c$

Now we have the following:

B(s)=-5s^3+9s^2-8s+c

Which is almost our velocity function, but we need to use our given conditions to find c.

We are told that when s=3, the velocity is equal to 108. Let's plug those numbers in and solve for c!

108=-5(3)^3+9(3)^2-8(3)+c

108=-78+c

c=186

Which makes our velocity function:

B(s)=-5s^3+9s^2-8s+186

Now, from our velocity function we can integrate to get out position function. We know this because velocity is the first derivative of position. So...

$\int B(s)ds=\int -5s^3+9s^2-8s+186ds$

$\int -5s^3+9s^2-8s+186ds=\frac{-5s^4}{4}+3s^3-4s^2+186s+c$

We'll call our position function P(s), for position.

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s+c$

We are almost there, but we know that P(5)=45 , so we need to find "c"

$45=\frac{-5(5)^4}{4}+3(5)^3-4(5)^2+186(5)+c$

Simplify:

45=423.75+c

c=-378.75

Making our position function:

$P(s)=\frac{-5s^4}{4}+3s^3-4s^2+186s-378.75$

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Example Question #111 : How To Find Position

What is the position at t=2 if the velocity function is v(t) = 2t ? Assume that the initial position at t=0 is zero.

Possible Answers:

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

Explanation:

In order to determine the position function from the velocity function, we need to integrate the velocity function.

$s(t) = \int v(t)\: dt$

Integrate the velocity function.

$s(t)=\int v(t) \: dt= \int 2t\:dt = t^2 + C$

Since the initial condition is zero, substitution of s(t)=0 and t=0 will leave C=0 , which will eliminate the constant term. Rewrite the position function.

s(t)=t^2

Solve for s(t=2) .

s(t=2) = 2^2 =4

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Example Question #111 : How To Find Position

Given the velocity function below, find the position at t=2 .

v(t)=6x+1

Possible Answers:

Correct answer:

Explanation:

To solve, integrate to find position and then plug in . Thus,

p(t)=3t^2+t=3(2)^2+2=14

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Example Question #111 : Position

The height, in inches, of a ball thrown straight up in the air is modeled with the equation

h(t)=-16t^2+160t+25 .

What is the maximum height the ball will reach in feet?

Possible Answers:

Not enough information

5 feet

425 feet

42.5 feet

35.417 feet

Correct answer:

35.417 feet

Explanation:

To find the maximum height of the ball we will first need to find the moment which it has stoped moving. This is the same as finding when the velocity is equal to zero. The velocity at the maximum height will be , so we need to determine when that happens.

For this particular function we will need to apply the power rule which states, $(x^n)'=nx^{n-1}$ .

Applying the power rule to our position function we are able to find the velocity function.

$\\v(t)=h'(t)=(-16t^2+160t+25)'\\ \\v(t)=(2*-16)t^{2-1}+(1*160)t^{1-1}+0\\ \\ v(t)=-32t^1+160t^0\\ \\v(t)=-32t+160(1)\\ \\v(t)=-32t+160$

From here, set the velocity function equal to zero and solve for time.

0=-32t+160

-160=-32t

5=t

Now we can find the height of the ball at time t=5 .

h(5)=-16(5)^2+160(5)+25=-400+800+25=425

The original problem tells us that the calculations are in inches, so we need to convert to feet.

$\frac{425}{12}\approx35.417\ feet$ .

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Example Question #111 : How To Find Position

Let the acceleration due to gravity be $\pm32\ \frac{ft}{sec^2}$ depending on the direction of the object.

A bucket is accidentally pushed off of a window washer's gondola (platform) that is suspended 300 feet in the air. If the initial speed of the falling bucket is only $5\ \frac{ft}{sec}$ , how long does it take to reach the ground?

Possible Answers:

$32\ seconds$

Not enough information is given to determine the solution.

$300\ seconds$

$4.177\ seconds$

$5\ seconds$

Correct answer:

$4.177\ seconds$

Explanation:

We are given very little information, so let us determine what we know.

$a=-32\ \frac{ft}{sec^2}$

$h(0)=300\ ft$

$v(0)=-5\ \frac{ft}{sec}$

Note that both of these values are negative because the object is traveling downward. (This is personal preference. You can use down as the positive direction in this problem as well so long as the signs of the velocity and acceleration are the same.)

We know that we can integrate acceleration to derive velocity, so we can find v(t) .

Recall the rules for integration that apply to this specific problem,

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

applying this rule to the accelaration we are able to find the velocity function.

$v(t)=\int a(t)dt=\int -32dt=-32t+c$

We can now use the initial value for to determine .

v(t)=-32t+c .

Plugging in $v(0)=-5\ \frac{ft}{sec}$ , we find

-5=-32(0)+c

-5=c

v(t)=-32t-5 .

Again, we can integrate the velocity function to find the position function.

$\\h(t)=\int v(t)dt=\int (-32t-5)dt\\ h(t)=\frac{-32}{2}t^{1+1}-5t+d$

h(t)=-16t^2-5t+d .

We know that h(0)=300 , so we can use that to find .

300=-16(0)^2-5(0)+d

300=d

h(t)=-16t^2-5t+300

Now we simply need to determine when the height of the bucket is .

0=-16t^2-5t+300

Using the quadratic formula,

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a=-16, b=-5, c=300

$t=\frac{--5\pm \sqrt{(-5)^2-4(-16)(300)}}{2(-16)}$

we find that $t\approx -4.489, 4.177$

Because negative time does not make sense, our solution is $4.177\ seconds$ .

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Example Question #111 : How To Find Position

An arrow that is shot straight up in the air can be located using the general position function

$x(t)=-\frac{1}{2}gt^2+v_0t_+s_0$ .

The initial velocity of the arrow is $128\ \frac{ft}{sec}$ .

Let the acceleration due to gravity be $32\ \frac{ft}{sec^2}$ .

Find the maximum height of the arrow before it begins its descent.

Possible Answers:

$256\ feet$

$4\ feet$

$128\ feet$

$96\ feet$

There is not enough information given to determine a solution.

Correct answer:

$256\ feet$

Explanation:

Let us first determine what we are given.

$x(t)=-\frac{1}{2}gt^2+v_0t_+s_0$

$x_0=0\ feet$

$v_0=128\ \frac{ft}{sec}$

$a(t)=32\ \frac{ft}{sec^2}$ , (g=a)

Let's plug what we know into the general formula.

$x(t)=-\frac{1}{2}(32)t^2+(128)t+0$

x(t)=-16t^2+128t

The derivative of position will give us the velocity function. For this particular function we will need to use the power rule to find the derivative.

The power rule states, $(x^n)'=nx^{n-1}$ .

Applying the power rule to the position function we are able to find the following velocity function.

$\\v(t)=x'(t)=(-16t^2+128t)\\ \\v(t)=(2*-16)t^{2-1}+(1*128)t^{1-1}\\ \\ v(t)=-32t+128t^0=-32t+128(1)\\ \\v(t)=-32t+128$

When the arrow reaches its maximum height, the velocity will be . Therefore, set the velocity function that was found equal to zero and solve for time.

0=-32t+128

-128=-32t

t=4 seconds.

Now we can find the position of the arrow at t=4 seconds.

$x(4)=-16(4)^2+128(4)=256\ feet$ .

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Example Question #111 : How To Find Position

Find the position at t=1 given the following velocity equation.

v(t)=3t^2+2t+1

Possible Answers:

Correct answer:

Explanation:

To solve, simply integrate the following function and plug in t=1 .

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

p(1)=1^3+1^2+1=3

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #107 : How To Find Position

Given the two points in three-space:

Example Question #108 : How To Find Position

Example Question #961 : Calculus

Example Question #962 : Calculus

Example Question #111 : How To Find Position

Example Question #111 : How To Find Position

Example Question #111 : Position

Example Question #111 : How To Find Position

Example Question #111 : How To Find Position

Example Question #111 : How To Find Position

All Calculus 1 Resources

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