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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #371 : How To Find Velocity

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=t^4+sin(t)$ p(0)=3 $\displaystyle p(0)=3$. Approximate $p(\pi)$ $\displaystyle p(\pi)$ using Euler's Method and three steps.

Possible Answers:

2.934 $\displaystyle 2.934$

5.166 $\displaystyle 5.166$

26.222 $\displaystyle 26.222$

33.288 $\displaystyle 33.288$

41.950 $\displaystyle 41.950$

Correct answer:

26.222 $\displaystyle 26.222$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=t^4+sin(t)$ p(0)=3 $\displaystyle p(0)=3$

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

$\Delta t = \frac{\pi-0}{3}=\frac{\pi}{3}$ $\displaystyle \Delta t = \frac{\pi-0}{3}=\frac{\pi}{3}$

$\displaystyle p_0=3;t_0=0$

$p_1=3+(\frac{\pi}{3})(0^4+sin(0))=3$ $\displaystyle p_1=3+(\frac{\pi}{3})(0^4+sin(0))=3$

$p_2=3+(\frac{\pi}{3})((\frac{\pi}{3})^4+sin(\frac{\pi}{3}))=5.166$ $\displaystyle p_2=3+(\frac{\pi}{3})((\frac{\pi}{3})^4+sin(\frac{\pi}{3}))=5.166$

$p_3=5.166+(\frac{\pi}{3})((\frac{2\pi}{3})^4+sin(\frac{2\pi}{3}))=26.222$ $\displaystyle p_3=5.166+(\frac{\pi}{3})((\frac{2\pi}{3})^4+sin(\frac{2\pi}{3}))=26.222$

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Example Question #372 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=ln(t^3)$ p(10)=0 $\displaystyle p(10)=0$. Approximate p(1000) $\displaystyle p(1000)$ using Euler's Method and three steps.

Possible Answers:

12340.9 $\displaystyle 12340.9$

8020.7 $\displaystyle 8020.7$

23017.1 $\displaystyle 23017.1$

5508.2 $\displaystyle 5508.2$

14492.5 $\displaystyle 14492.5$

Correct answer:

14492.5 $\displaystyle 14492.5$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=ln(t^3)$ p(10)=0 $\displaystyle p(10)=0$

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

$\Delta t = \frac{1000-10}{3}=330$ $\displaystyle \Delta t = \frac{1000-10}{3}=330$

$\displaystyle p_0=0;t_0=10$

$\displaystyle p_1=0+(330)ln(10^3)=2279.6$

$\displaystyle p_2=2279.6+(330)ln(340^3)=8050.3$

$\displaystyle p_3=8050.3+(330)ln(670^3)=14492.5$

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Example Question #372 : How To Find Velocity

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

Possible Answers:

1.096 $\displaystyle 1.096$

0.183 $\displaystyle 0.183$

0.734 $\displaystyle 0.734$

0.362 $\displaystyle 0.362$

$\displaystyle 0$

Correct answer:

0.362 $\displaystyle 0.362$

Explanation:

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

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

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

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

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

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

For this problem, we are told $v(t)=te^{t}cos(t)$ $\displaystyle v(t)=te^{t}cos(t)$ p(0)=0 $\displaystyle p(0)=0$

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

$\Delta t = \frac{1-0}{2}=0.5$ $\displaystyle \Delta t = \frac{1-0}{2}=0.5$

$\displaystyle p_0=0;t_0=0$

$p_1=0+(0.5)(0)e^{0}cos(0)=0$ $\displaystyle p_1=0+(0.5)(0)e^{0}cos(0)=0$

$p_2=0+(0.5)(0.5)e^{0.5}cos(0.5)=0.362$ $\displaystyle p_2=0+(0.5)(0.5)e^{0.5}cos(0.5)=0.362$

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Example Question #371 : Velocity

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=(t-7)^3$ p(6)=0 $\displaystyle p(6)=0$. Approximate p(7) $\displaystyle p(7)$ using Euler's Method and two steps

Possible Answers:

-0.063 $\displaystyle -0.063$

$\displaystyle 0$

0.5 $\displaystyle 0.5$

-0.563 $\displaystyle -0.563$

-0.5 $\displaystyle -0.5$

Correct answer:

-0.563 $\displaystyle -0.563$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=(t-7)^3$ p(6)=0 $\displaystyle p(6)=0$

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

$\Delta t = \frac{7-6}{2}=0.5$ $\displaystyle \Delta t = \frac{7-6}{2}=0.5$

$\displaystyle p_0=0;t_0=6$

$\displaystyle p_1=0+(0.5)(6-7)^3=-0.5$

$\displaystyle p_2=-0.5+(0.5)(6.5-7)^3=-0.563$

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Example Question #375 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=4t^2+3t+1$ p(0)=0 $\displaystyle p(0)=0$. Approximate p(2) $\displaystyle p(2)$ using Euler's Method and two steps.

Possible Answers:

$\displaystyle 23$

$\displaystyle 31$

$\displaystyle 34$

$\frac{56}{3}$ $\displaystyle \frac{56}{3}$

$\displaystyle 9$

Correct answer:

$\displaystyle 9$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=4t^2+3t+1$ p(0)=0 $\displaystyle p(0)=0$

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

$\Delta t = \frac{2-0}{2}=1$ $\displaystyle \Delta t = \frac{2-0}{2}=1$

$\displaystyle p_0=0;t_0=0$

$\displaystyle p_1=0+(1)(4(0)^2+3(0)+1)=1$

$\displaystyle p_2=1+(1)(4(1)^2+3(1)+1)=9$

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Example Question #376 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=4sin^2(t)$ and $p(\frac{\pi}{2})=1$ $\displaystyle p(\frac{\pi}{2})=1$. Approximate $p(\pi)$ $\displaystyle p(\pi)$ using Euler's Method and two steps.

Possible Answers:

4.142 $\displaystyle 4.142$

5.713 $\displaystyle 5.713$

2.089 $\displaystyle 2.089$

6.552 $\displaystyle 6.552$

1.726 $\displaystyle 1.726$

Correct answer:

5.713 $\displaystyle 5.713$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=4sin^2(t)$ and $p(\frac{\pi}{2})=1$ $\displaystyle p(\frac{\pi}{2})=1$

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

$\Delta t = \frac{\pi-\frac{\pi}{2}}{2}=\frac{\pi}{4}$ $\displaystyle \Delta t = \frac{\pi-\frac{\pi}{2}}{2}=\frac{\pi}{4}$

$p_0=1;t_0=\frac{\pi}{2}$ $\displaystyle p_0=1;t_0=\frac{\pi}{2}$

$p_1=1+(\frac{\pi}{4})(4sin^2(\frac{\pi}{2}))=4.142$ $\displaystyle p_1=1+(\frac{\pi}{4})(4sin^2(\frac{\pi}{2}))=4.142$

$p_2=4.142+(\frac{\pi}{4})(4sin^2(\frac{3\pi}{4}))=5.713$ $\displaystyle p_2=4.142+(\frac{\pi}{4})(4sin^2(\frac{3\pi}{4}))=5.713$

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Example Question #377 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=tan^2(t^2)$ and p(0)=4 $\displaystyle p(0)=4$. Approximate $p(0.5\pi)$ $\displaystyle p(0.5\pi)$ using Euler's Method and two steps.

Possible Answers:

4.126 $\displaystyle 4.126$

1.395 $\displaystyle 1.395$

4.395 $\displaystyle 4.395$

5.261 $\displaystyle 5.261$

1.261 $\displaystyle 1.261$

Correct answer:

4.395 $\displaystyle 4.395$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=tan^2(t^2)$ and p(0)=4 $\displaystyle p(0)=4$

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

$\Delta t = \frac{0.5\pi-0}{2}=0.25\pi$ $\displaystyle \Delta t = \frac{0.5\pi-0}{2}=0.25\pi$

$\displaystyle p_0=4;t_0=0$

$p_1=4+(0.25\pi)(tan^2(0^2))=4$ $\displaystyle p_1=4+(0.25\pi)(tan^2(0^2))=4$

$p_2=4+(0.25\pi)(tan^2((0.25\pi)^2))=4.395$ $\displaystyle p_2=4+(0.25\pi)(tan^2((0.25\pi)^2))=4.395$

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Example Question #378 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=cos^3(t)$ and p(0)=7 $\displaystyle p(0)=7$. Approximate $p(\pi)$ $\displaystyle p(\pi)$ using Euler's Method and two steps.

Possible Answers:

2.316 $\displaystyle 2.316$

1.571 $\displaystyle 1.571$

8.571 $\displaystyle 8.571$

9.316 $\displaystyle 9.316$

$\displaystyle 1$

Correct answer:

8.571 $\displaystyle 8.571$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=cos^3(t)$ and p(0)=7 $\displaystyle p(0)=7$

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

$\Delta t = \frac{\pi-0}{2}=\frac{\pi}{2}$ $\displaystyle \Delta t = \frac{\pi-0}{2}=\frac{\pi}{2}$

$\displaystyle p_0=7;t_0=0$

$p_1=7+(\frac{\pi}{2})cos^3(0)=8.571$ $\displaystyle p_1=7+(\frac{\pi}{2})cos^3(0)=8.571$

$p_2=8.571+(\frac{\pi}{2})cos^3(\frac{\pi}{2})=8.571$ $\displaystyle p_2=8.571+(\frac{\pi}{2})cos^3(\frac{\pi}{2})=8.571$

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Example Question #379 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $\displaystyle v(t)=ln(cos(t^3))$ and p(0)=1 $\displaystyle p(0)=1$. Approximate p(8) $\displaystyle p(8)$ using Euler's Method and two steps.

Possible Answers:

-0.75 $\displaystyle -0.75$

$\displaystyle 1$

4.75 $\displaystyle 4.75$

3.75 $\displaystyle 3.75$

-2.75 $\displaystyle -2.75$

Correct answer:

-2.75 $\displaystyle -2.75$

Explanation:

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

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

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

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

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

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

For this problem, we are told $\displaystyle v(t)=ln(cos(t^3))$ and p(0)=1 $\displaystyle p(0)=1$

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

$\Delta t = \frac{8-0}{2}=4$ $\displaystyle \Delta t = \frac{8-0}{2}=4$

$\displaystyle p_0=1;t_0=0$

$\displaystyle p_1=1+(4)ln(cos(0^3))=1$

$\displaystyle p_2=1+(4)ln(cos(4^3))=-2.75$

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Example Question #380 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=e^{\frac{t^2}{4}}$ $\displaystyle v(t)=e^{\frac{t^2}{4}}$ and p(1)=3 $\displaystyle p(1)=3$. Approximate p(3) $\displaystyle p(3)$ using Euler's Method and two steps.

Possible Answers:

8.432 $\displaystyle 8.432$

7.002 $\displaystyle 7.002$

4.284 $\displaystyle 4.284$

5.718 $\displaystyle 5.718$

$\displaystyle 3$

Correct answer:

7.002 $\displaystyle 7.002$

Explanation:

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

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

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

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

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

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

For this problem, we are told $v(t)=e^{\frac{t^2}{4}}$ $\displaystyle v(t)=e^{\frac{t^2}{4}}$ and p(1)=3 $\displaystyle p(1)=3$

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

$\Delta t = \frac{3-1}{2}=1$ $\displaystyle \Delta t = \frac{3-1}{2}=1$

$\displaystyle p_0=3;t_0=1$

$p_1=3+(1)e^{\frac{1^2}{4}}=4.284$ $\displaystyle p_1=3+(1)e^{\frac{1^2}{4}}=4.284$

$p_2=4.284+(1)e^{\frac{2^2}{4}}=7.002$ $\displaystyle p_2=4.284+(1)e^{\frac{2^2}{4}}=7.002$

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #371 : How To Find Velocity

Example Question #372 : Calculus

Example Question #372 : How To Find Velocity

Example Question #371 : Velocity

Example Question #375 : Calculus

Example Question #376 : Calculus

Example Question #377 : Calculus

Example Question #378 : Calculus

Example Question #379 : Calculus

Example Question #380 : Calculus

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