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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #1001 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=(cos(t^3))^{\frac{1}{3}}$ p(0)=4 . Approximate p(6) using Euler's Method and three steps.

Possible Answers:

7.982

6.411

4.405

5.331

6.808

Correct answer:

6.411

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of and divide by the number of steps to be used:

$\Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told $v(t)=(cos(t^3))^{\frac{1}{3}}$ p(0)=4

Knowing this, we may take the steps to estimate our function value at our desired value:

$\Delta t = \frac{6-0}{3}=2$

p_0=4;t_0=0

$p_1=4+(2)(cos(0^3))^{\frac{1}{3}}=6$

$p_2=6+(2)(cos(2^3))^{\frac{1}{3}}=4.948$

$p_3=4.948+(2)(cos(4^3))^{\frac{1}{3}}=6.411$

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Example Question #1002 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=1.08^{t^5}$ p(1.08)=0 . Approximate p(1.38) using Euler's Method and three steps.

Possible Answers:

0.218

0.499

0.361

0.558

0.120

Correct answer:

0.361

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of and divide by the number of steps to be used:

$\Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told $v(t)=1.08^{t^5}$ p(1.08)=0

Knowing this, we may take the steps to estimate our function value at our desired value:

$\Delta t = \frac{1.38-1.08}{3}=0.1$

p_0=0;t_0=1.08

$p_1=0+(0.1)1.08^{1.08^5}=0.112$

$p_2=0.112+(0.1)1.08^{1.18^5}=0.231$

$p_3=0.231+(0.1)1.08^{1.28^5}=0.361$

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Example Question #1003 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=\frac{t^4+7}{t^3}$ p(1)=4 . Approximate p(4) using Euler's Method and three steps.

Possible Answers:

18.134

24.768

22.001

20.916

16.395

Correct answer:

18.134

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of and divide by the number of steps to be used:

$\Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told $v(t)=\frac{t^4+7}{t^3}$ p(1)=4

Knowing this, we may take the steps to estimate our function value at our desired value:

$\Delta t = \frac{4-1}{3}=1$

p_0=4;t_0=1

$p_1=4+(1)(\frac{1^4+7}{1^3})=12$

$p_2=12+(1)(\frac{2^4+7}{2^3})=14.875$

$p_3=14.875+(1)(\frac{3^4+7}{3^3})=18.134$

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Example Question #1004 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=\frac{t+9}{t^2+2}$ p(1)=1 . Approximate p(2.5) using Euler's Method and three steps.

Possible Answers:

2.895

4.046

1.907

4.819

3.902

Correct answer:

4.819

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of and divide by the number of steps to be used:

$\Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told $v(t)=\frac{t+9}{t^2+2}$ p(1)=1

Knowing this, we may take the steps to estimate our function value at our desired value:

$\Delta t = \frac{2.5-1}{3}=0.5$

p_0=1;t_0=1

$p_1=1+(0.5)(\frac{1+9}{1^2+2})=2.667$

$p_2=2.667+(0.5)(\frac{1.5+9}{1.5^2+2})=3.902$

$p_3=3.902+(0.5)(\frac{2+9}{2^2+2})=4.819$

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Example Question #151 : How To Find Position

What is the position function if x(1)=2 when v(t)=3t-1 ?

Possible Answers:

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

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

$x(t)=\frac{3}{2}t^2-t-\frac{5}{2}$

$x(t)=t^2-t+\frac{5}{2}$

Correct answer:

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

Explanation:

To find the position function, you must integrate the velocity function. To integrate, add 1 to the exponent and then put that result on the denominator. $\int 3t-1 dt=\frac{3t^2}{2}-t+C$ . Then, plug in your initial values to find your C: $\frac{3}{2}(1^2)-1+C=2$ . Then, $C=\frac{3}{2}$ . Therefore, your answer is: $x(t)=\frac{3}{2}t^2-t+\frac{3}{2}$ .

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Example Question #151 : How To Find Position

Newton's Second Law of Motion tells us that a force exerted on an object will create an acceleration inversely proportional to the mass of the object being accelerated. This is expressed by the equation f=ma , where f is force, measured in Newtons, (1 N is the force required to accelerate 1 kilogram at 1 meter persecond per second), m is the mass of the object, measured in kilograms, and a is the acceleration, measured in meters per second per second.

A rocket-powered car is coasting at $25\,m/s$ when the driver starts up the engine again. The engine exerts a force of $30,000 \,N$ on the car, which weighs $1500\,kg$ . The driver shuts off the engine after 5 seconds. How far does the car travel in this time?

Possible Answers:

$125\,m$

$250\,m$

$625\,m$

$375\,m$

Correct answer:

$375\,m$

Explanation:

We'll need to find the acceleration, and then work backwards to find the position. We can rewrite Newton's Second Law as $a=\frac{f}{m}$ , which in this case gives us an acceleration of

$a(t)=\frac{30000 N}{1500 kg} = 20\,m/s^2$ .

The car's initial position and velocity are

$\begin{align*} p_0&=0\,m \\ v_0 &= 25\,m/s\end{align*}$

Initial and final time are given as

$\begin{align*} t_i &=0\,s \\ t_f &= 5\,s \end{align*}$

Acceleration is the rate of change in velocity, which is the rate of change in position, so to find the equation for position, we'll ned to integrate and use initial conditions twice.

$\int a(t)dt=\int20 dt$

To integrate, we use the following rule:

$\int a\,dt = ax+C$

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

$\int a(t)dt=\int 20dt =20t+C$

To solve for C, we'll use our initial condition, v(0)=25 .

$\\ v(0)=20(0)+C=25 \\ C=25 \\ v(t)=20t+25$

We'll need to integrate again to find position:

$\int v(t)=\int 20t+25 \, dt$

using the following rule in addition to the earlier one:

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

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

$\begin{align*} \int v(t)\,dt &= \int 20t+25\,dt \\ &=\frac{20t^2}{2}+25t+C \\ &=10t^2+25t+C \end{align*}$

To solve for C, we'll use our initial condition, p(0)=0 .

$\\ p(0)=10(0^2)+25(0)+C=0 \\ C=0 \\ p(t)=10t^2+25t$

which, evaluated at t=5 , gives us

$\begin{align*} p(4)&=10(5^2)+25(5) \\ &=10(25)+125 \\ &=375\,m\end{align*}$

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Example Question #1007 : Calculus

A car starts at rest and accelerates in a straight line at $15\,m/s^2$ for 20 seconds. How far does it travel?

Possible Answers:

$300\,m$

$3,000\,m$

$600\,m$

$6,000\,m$

Correct answer:

$3,000\,m$

Explanation:

The equation for position is $\frac{at^2}{2}$ , since we're starting from rest. Evaluating at t=20 gives us our answer, $3,000\,m$

We'll need to work backwards from the acceleration to find the position. We're given the car's acceleration,

$\begin{align*} a(t) &=15\,m/s^2\end{align*}$

The car begins at rest, so

$\begin{align*} p_0&=0\,m \\ v_0 &= 0\,m/s\end{align*}$

Initial and final time are given as

$\begin{align*} t_i &=0\,s \\ t_f &= 20\,s \end{align*}$

Acceleration is the rate of change in velocity, which is the rate of change in position, so to find the equation for position, we'll ned to integrate and use initial conditions twice.

$\int a(t)dt=\int15 dt$

To integrate, we use the following rule:

$\int a\,dt = ax+C$

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

$\int a(t)dt=\int 15dt =15t+C$

To solve for C, we'll use our initial condition, v(0)=0 .

$\\ v(0)=15(0)+C=0 \\ C=0 \\ v(t)=15t$

We'll need to integrate again to find position:

$\int v(t)=\int 15t \, dt$

using the following rule:

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

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

$\begin{align*} \int v(t)\,dt &= \int 15t\,dt \\ &=\frac{15t^2}{2}+C \\ &=7.5t^2+C \end{align*}$

To solve for C, we'll use our initial condition, p(0)=0 .

$\\ p(0)=7.5(0^2)+C=0 \\ C=0 \\ p(t)=7.5t^2$

which, evaluated at t=20 , gives us

$\begin{align*} p(4)&=7.5(20^2) \\ &=7.5(400) \\&=3000\,m\end{align*}$

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Example Question #154 : How To Find Position

The position of t= 1 is given by the following function:

$p(t)= (3x^2 + e^{2x})^2$

Find the velocity.

Possible Answers:

431.7

126.8

371.9

Answer not listed

23.4

Correct answer:

431.7

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t)=v(t)

In this case, the position function is: $p(t)= (3t^2 + e^{2t})^2$

Then take the derivative of the position function to get the velocity function: $v(t)= 2(3t^2 + e^{2t}) (6t+2e^{2t})$

Then, plug t= 1 into the velocity function: $v(1)= 2(3({\color{Blue} 1})^2 + e^{2({\color{Blue} 1})}) (6({\color{Blue}1 })+2e^{2({\color{Blue} 1})})$

Therefore, the answer is: 431.7

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Example Question #1008 : Calculus

The position of t= 0 is given by the following function:

$p(t)= \sin(t^2+3t^4)$

Find the velocity.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t)=v(t)

In this case, the position function is: $p(t)= \sin(t^2+3t^4)$

Then take the derivative of the position function to get the velocity function: $v(t)= \cos(t^2+3t^4) (2t+12t^3)$

Then, plug t= 0 into the velocity function: $v(t)= \cos({\color{Blue} 0}^2+3({\color{Blue} 0})^4) (2({\color{Blue} 0})+12({\color{Blue} 0})^3)$

Therefore, the answer is:

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Example Question #1008 : Calculus

The position of t= 0 is given by the following function:

$p(t)= \sin(t^2+3t^4)$

Find the velocity.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t)=v(t)

In this case, the position function is: $p(t)= \sin(t^2+3t^4)$

Then take the derivative of the position function to get the velocity function: $v(t)= \cos(t^2+3t^4) (2t+12t^3)$

Then, plug t= 0 into the velocity function: $v(t)= \cos({\color{Blue} 0}^2+3({\color{Blue} 0})^4) (2({\color{Blue} 0})+12({\color{Blue} 0})^3)$

Therefore, the answer is:

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1001 : Calculus

Example Question #1002 : Calculus

Example Question #1003 : Calculus

Example Question #1004 : Calculus

Example Question #151 : How To Find Position

Example Question #151 : How To Find Position

Example Question #1007 : Calculus

Example Question #154 : How To Find Position

Example Question #1008 : Calculus

Example Question #1008 : Calculus

All Calculus 1 Resources

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