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Calculus 1 : Calculus

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Example Questions

Example Question #552 : Other Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=(cos(x))^{cos(x)}$ $\displaystyle f'(x)=(cos(x))^{cos(x)}$ and y(0)=4 $\displaystyle y(0)=4$, approximate y(1.5) $\displaystyle y(1.5)$ using Euler's Method and three steps.

Possible Answers:

4.783 $\displaystyle 4.783$

5.112 $\displaystyle 5.112$

4.292 $\displaystyle 4.292$

5.305 $\displaystyle 5.305$

5.496 $\displaystyle 5.496$

Correct answer:

5.305 $\displaystyle 5.305$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=(cos(x))^{cos(x)}$ $\displaystyle f'(x)=(cos(x))^{cos(x)}$ and y(0)=4 $\displaystyle y(0)=4$

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

$\Delta x = \frac{1.5-0}{3}=0.5$ $\displaystyle \Delta x = \frac{1.5-0}{3}=0.5$

$\displaystyle y_0=4,x_0=0$

$y_1=4+0.5(cos(0))^{cos(0)}=4.5$ $\displaystyle y_1=4+0.5(cos(0))^{cos(0)}=4.5$

$y_2=4.5+0.5(cos(0.5))^{cos(0.5)}=4.946$ $\displaystyle y_2=4.5+0.5(cos(0.5))^{cos(0.5)}=4.946$

$y_3=4.946+0.5(cos(1))^{cos(1)}=5.305$ $\displaystyle y_3=4.946+0.5(cos(1))^{cos(1)}=5.305$

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Example Question #1772 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $\displaystyle f'(x)=cot(x^2)$ and y(1)=5 $\displaystyle y(1)=5$, approximate y(1.6) $\displaystyle y(1.6)$ using Euler's Method and three steps.

Possible Answers:

5.276 $\displaystyle 5.276$

4.938 $\displaystyle 4.938$

5.191 $\displaystyle 5.191$

5.072 $\displaystyle 5.072$

5.409 $\displaystyle 5.409$

Correct answer:

5.072 $\displaystyle 5.072$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $\displaystyle f'(x)=cot(x^2)$ and y(1)=5 $\displaystyle y(1)=5$

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

$\Delta x = \frac{1.6-1}{3}=0.2$ $\displaystyle \Delta x = \frac{1.6-1}{3}=0.2$

$\displaystyle y_0=5,x_0=1$

$\displaystyle y_1=5+(0.2)cot(1^2)=5.128$

$\displaystyle y_2=5.128+(0.2)cot(1.2^2)=5.154$

$\displaystyle y_3=5.154+(0.2)cot(1.4^2)=5.072$

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Example Question #1773 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $\displaystyle f'(x)=tan(ln(ln(x)))$ and y(2)=0 $\displaystyle y(2)=0$, approximate y(5) $\displaystyle y(5)$ using Euler's Method and three steps.

Possible Answers:

-0.265 $\displaystyle -0.265$

0.049 $\displaystyle 0.049$

-0.517 $\displaystyle -0.517$

0.183 $\displaystyle 0.183$

-0.384 $\displaystyle -0.384$

Correct answer:

0.049 $\displaystyle 0.049$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $\displaystyle f'(x)=tan(ln(ln(x)))$ and y(2)=0 $\displaystyle y(2)=0$

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

$\Delta x = \frac{5-2}{3}=1$ $\displaystyle \Delta x = \frac{5-2}{3}=1$

$\displaystyle y_0=0;x_0=2$

$\displaystyle y_1=0+(1)tan(ln(ln(2)))=-0.384$

$\displaystyle y_2=-0.384+(1)tan(ln(ln(3)))=-0.290$

$\displaystyle y_3=-0.290+(1)tan(ln(ln(4)))=0.049$

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Example Question #1774 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $\displaystyle f'(x)=sec(x^4)$ and y(1)=10 $\displaystyle y(1)=10$, approximate y(2) $\displaystyle y(2)$ using Euler's Method and four steps.

Possible Answers:

10.615 $\displaystyle 10.615$

10.725 $\displaystyle 10.725$

10.515 $\displaystyle 10.515$

10.983 $\displaystyle 10.983$

10.422 $\displaystyle 10.422$

Correct answer:

10.615 $\displaystyle 10.615$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told$\displaystyle f'(x)=sec(x^4)$ and y(1)=10 $\displaystyle y(1)=10$

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

$\Delta x = \frac{2-1}{4}=0.25$ $\displaystyle \Delta x = \frac{2-1}{4}=0.25$

$\displaystyle y_0=10;x_0=1$

$\displaystyle y_1=10+(0.25)sec(1^4)=10.463$

$\displaystyle y_2=10.463+(0.25)sec(1.25^4)=10.136$

$\displaystyle y_3=10.136+(0.25)sec(1.5^4)=10.865$

$\displaystyle y_4=10.865+(0.25)sec(1.75^4)=10.615$

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Example Question #1775 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $\displaystyle f'(x)=csc(2^x)$ and y(1)=3 $\displaystyle y(1)=3$, approximate y(5) $\displaystyle y(5)$ using Euler's Method and four steps.

Possible Answers:

1.208 $\displaystyle 1.208$

3.790 $\displaystyle 3.790$

4.100 $\displaystyle 4.100$

2.779 $\displaystyle 2.779$

0.317 $\displaystyle 0.317$

Correct answer:

0.317 $\displaystyle 0.317$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $\displaystyle f'(x)=csc(2^x)$ and y(1)=3 $\displaystyle y(1)=3$

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

$\Delta x = \frac{5-1}{4}=1$ $\displaystyle \Delta x = \frac{5-1}{4}=1$

$\displaystyle y_0=3;x_0=1$

$\displaystyle y_1=3+(1)csc(2^1)=4.100$

$\displaystyle y_2=4.100+(1)csc(2^2)=2.779$

$\displaystyle y_3=2.779+(1)csc(2^3)=3.790$

$\displaystyle y_4=3.790+(1)csc(2^4)=0.317$

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Example Question #1776 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=11^{x^6}$ $\displaystyle f'(x)=11^{x^6}$ and y(1)=5 $\displaystyle y(1)=5$, approximate y(1.16) $\displaystyle y(1.16)$ using Euler's Method and four steps.

Possible Answers:

12.614 $\displaystyle 12.614$

12.025 $\displaystyle 12.025$

8.098 $\displaystyle 8.098$

15.622 $\displaystyle 15.622$

13.159 $\displaystyle 13.159$

Correct answer:

12.614 $\displaystyle 12.614$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=11^{x^6}$ $\displaystyle f'(x)=11^{x^6}$ and y(1)=5 $\displaystyle y(1)=5$

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

$\Delta x = \frac{1.16-1}{4}=0.04$ $\displaystyle \Delta x = \frac{1.16-1}{4}=0.04$

$\displaystyle y_0=5;x_0=1$

$y_1=5+(0.04)11^{1^6}=5.44$ $\displaystyle y_1=5+(0.04)11^{1^6}=5.44$

$y_2=5.44+(0.04)11^{1.04^6}=6.271$ $\displaystyle y_2=5.44+(0.04)11^{1.04^6}=6.271$

$y_3=6.271+(0.04)11^{1.08^6}=8.068$ $\displaystyle y_3=6.271+(0.04)11^{1.08^6}=8.068$

$y_4=8.068+(0.04)11^{1.12^6}=12.614$ $\displaystyle y_4=8.068+(0.04)11^{1.12^6}=12.614$

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Example Question #1777 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=sin(\pi cos(4\pi x))$ $\displaystyle f'(x)=sin(\pi cos(4\pi x))$ and y(1)=3 $\displaystyle y(1)=3$, approximate y(2) $\displaystyle y(2)$ using Euler's Method and five steps.

Possible Answers:

3.104 $\displaystyle 3.104$

3.377 $\displaystyle 3.377$

3.815 $\displaystyle 3.815$

3.904 $\displaystyle 3.904$

2.908 $\displaystyle 2.908$

Correct answer:

3.104 $\displaystyle 3.104$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=sin(\pi cos(4\pi x))$ $\displaystyle f'(x)=sin(\pi cos(4\pi x))$ and y(1)=3 $\displaystyle y(1)=3$

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

$\Delta x = \frac{2-1}{5}=0.2$ $\displaystyle \Delta x = \frac{2-1}{5}=0.2$

$\displaystyle y_0=3;x_0=1$

$y_1=3+(0.2)sin(\pi cos(4\pi (1)))=3$ $\displaystyle y_1=3+(0.2)sin(\pi cos(4\pi (1)))=3$

$y_2=3+(0.2)sin(\pi cos(4\pi (1.2)))=2.887$ $\displaystyle y_2=3+(0.2)sin(\pi cos(4\pi (1.2)))=2.887$

$y_3=2.887+(0.2)sin(\pi cos(4\pi (1.4)))=3.052$ $\displaystyle y_3=2.887+(0.2)sin(\pi cos(4\pi (1.4)))=3.052$

$y_4=3.052+(0.2)sin(\pi cos(4\pi (1.6)))=3.217$ $\displaystyle y_4=3.052+(0.2)sin(\pi cos(4\pi (1.6)))=3.217$

$y_5=3.217+(0.2)sin(\pi cos(4\pi (1.8)))=3.104$ $\displaystyle y_5=3.217+(0.2)sin(\pi cos(4\pi (1.8)))=3.104$

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Example Question #1778 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=e^{(\frac{1}{4})^x}$ $\displaystyle f'(x)=e^{(\frac{1}{4})^x}$ and $\displaystyle y(0.5)=0$, approximate y(1) $\displaystyle y(1)$ using Euler's Method and five steps.

Possible Answers:

0.738 $\displaystyle 0.738$

0.812 $\displaystyle 0.812$

0.611 $\displaystyle 0.611$

0.545 $\displaystyle 0.545$

0.689 $\displaystyle 0.689$

Correct answer:

0.738 $\displaystyle 0.738$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=e^{(\frac{1}{4})^x}$ $\displaystyle f'(x)=e^{(\frac{1}{4})^x}$ and $\displaystyle y(0.5)=0$

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

$\Delta x = \frac{1-0.5}{5}=0.1$ $\displaystyle \Delta x = \frac{1-0.5}{5}=0.1$

$\displaystyle y_0=0;x_0=0.5$

$y_1=0+(0.1)e^{(\frac{1}{4})^{0.5}}=0.165$ $\displaystyle y_1=0+(0.1)e^{(\frac{1}{4})^{0.5}}=0.165$

$y_2=0.165+(0.1)e^{(\frac{1}{4})^{0.6}}=0.320$ $\displaystyle y_2=0.165+(0.1)e^{(\frac{1}{4})^{0.6}}=0.320$

$y_3=0.320+(0.1)e^{(\frac{1}{4})^{0.7}}=0.466$ $\displaystyle y_3=0.320+(0.1)e^{(\frac{1}{4})^{0.7}}=0.466$

$y_4=0.466+(0.1)e^{(\frac{1}{4})^{0.8}}=0.605$ $\displaystyle y_4=0.466+(0.1)e^{(\frac{1}{4})^{0.8}}=0.605$

$y_5=0.605+(0.1)e^{(\frac{1}{4})^{0.9}}=0.738$ $\displaystyle y_5=0.605+(0.1)e^{(\frac{1}{4})^{0.9}}=0.738$

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Example Question #1771 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $\displaystyle f'(x)=ln(csc(x))$ and y(1)=0 $\displaystyle y(1)=0$, approximate y(1.75) $\displaystyle y(1.75)$ using Euler's Method and five steps.

Possible Answers:

0.040 $\displaystyle 0.040$

0.054 $\displaystyle 0.054$

0.051 $\displaystyle 0.051$

0.047 $\displaystyle 0.047$

0.033 $\displaystyle 0.033$

Correct answer:

0.047 $\displaystyle 0.047$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $\displaystyle f'(x)=ln(csc(x))$ and y(1)=0 $\displaystyle y(1)=0$

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

$\Delta x = \frac{1.75-1}{5}=0.15$ $\displaystyle \Delta x = \frac{1.75-1}{5}=0.15$

$\displaystyle y_0=0;x_0=1$

$\displaystyle y_1=0+(0.15)ln(csc(1))=0.026$

$\displaystyle y_2=0.026+(0.15)ln(csc(1.15))=0.040$

$\displaystyle y_3=0.040+(0.15)ln(csc(1.30))=0.046$

$\displaystyle y_4=0.046+(0.15)ln(csc(1.45))=0.047$

$\displaystyle y_5=0.047+(0.15)ln(csc(1.60))=0.047$

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Example Question #1772 : Calculus

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=1.4^{cos(\pi e^{x})}$ $\displaystyle f'(x)=1.4^{cos(\pi e^{x})}$ and y(0)=0 $\displaystyle y(0)=0$, approximate y(4) $\displaystyle y(4)$ using Euler's Method and five steps.

Possible Answers:

2.750 $\displaystyle 2.750$

5.598 $\displaystyle 5.598$

3.523 $\displaystyle 3.523$

4.285 $\displaystyle 4.285$

4.009 $\displaystyle 4.009$

Correct answer:

3.523 $\displaystyle 3.523$

Explanation:

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

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

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

$\Delta x = \frac{x_f-x_i}{Steps}$ $\displaystyle \Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=1.4^{cos(\pi e^{x})}$ $\displaystyle f'(x)=1.4^{cos(\pi e^{x})}$ and y(0)=0 $\displaystyle y(0)=0$

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

$\Delta x = \frac{4-0}{5}=0.8$ $\displaystyle \Delta x = \frac{4-0}{5}=0.8$

$\displaystyle y_0=0;x_0=0$

$y_1=0+(0.8)1.4^{cos(\pi e^{0})}=0.571$ $\displaystyle y_1=0+(0.8)1.4^{cos(\pi e^{0})}=0.571$

$y_2=0.571+(0.8)1.4^{cos(\pi e^{0.8})}=1.604$ $\displaystyle y_2=0.571+(0.8)1.4^{cos(\pi e^{0.8})}=1.604$

$y_3=1.604+(0.8)1.4^{cos(\pi e^{1.6})}=2.178$ $\displaystyle y_3=1.604+(0.8)1.4^{cos(\pi e^{1.6})}=2.178$

$y_4=2.178+(0.8)1.4^{cos(\pi e^{2.4})}=2.750$ $\displaystyle y_4=2.178+(0.8)1.4^{cos(\pi e^{2.4})}=2.750$

$y_5=2.750+(0.8)1.4^{cos(\pi e^{3.2})}=3.523$ $\displaystyle y_5=2.750+(0.8)1.4^{cos(\pi e^{3.2})}=3.523$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #552 : Other Differential Functions

Example Question #1772 : Calculus

Example Question #1773 : Calculus

Example Question #1774 : Calculus

Example Question #1775 : Calculus

Example Question #1776 : Calculus

Example Question #1777 : Calculus

Example Question #1778 : Calculus

Example Question #1771 : Calculus

Example Question #1772 : Calculus

All Calculus 1 Resources

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