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Calculus 1 : How to find differential functions

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Example Questions

Example Question #1777 : Calculus

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

Possible Answers:

3.104

3.377

3.815

3.904

2.908

Correct answer:

3.104

Explanation:

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

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

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

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

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

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

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

y_0=3;x_0=1

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

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

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

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

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

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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 $f'(x)=e^{(\frac{1}{4})^x}$ and y(0.5)=0 , approximate y(1) using Euler's Method and five steps.

Possible Answers:

0.738

0.812

0.611

0.545

0.689

Correct answer:

0.738

Explanation:

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

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

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

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

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

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

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

y_0=0;x_0=0.5

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

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

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

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

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

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Example Question #1779 : Calculus

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

Possible Answers:

0.040

0.051

0.054

0.033

0.047

Correct answer:

0.047

Explanation:

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

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

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

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

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

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

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

y_0=0;x_0=1

y_1=0+(0.15)ln(csc(1))=0.026

y_2=0.026+(0.15)ln(csc(1.15))=0.040

y_3=0.040+(0.15)ln(csc(1.30))=0.046

y_4=0.046+(0.15)ln(csc(1.45))=0.047

y_5=0.047+(0.15)ln(csc(1.60))=0.047

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Example Question #564 : How To Find Differential Functions

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

Possible Answers:

2.750

3.523

4.009

5.598

4.285

Correct answer:

3.523

Explanation:

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

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

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

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

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

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

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

y_0=0;x_0=0

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

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

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

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

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

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Example Question #1781 : Calculus

Find the divergence of the function $F(x,y)=e^{xy^2}\widehat{i}+2^{xy^2}\widehat{j}$ at (2,1.2)

Hint: $divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}$

Possible Answers:

50.15

52.87

54.18

46.75

48.90

Correct answer:

50.15

Explanation:

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

To visualize this, picture an open drain in a tub full of water; this drain may represent a 'sink,' and all of the velocities at each specific point in the tub represent the vector field. Close to the drain, the velocity will be greater than a spot farther away from the drain.

Vectorfield

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$F(x,y)=e^{xy^2}\widehat{i}+2^{xy^2}\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $(\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Derivative of an exponential:

d[e^u]=e^udu

d[a^u]=a^uduln(a)

Note that u may represent large functions, and not just individual variables!

$divF(x,y)=y^2e^{xy^2}+2xy(2^{xy^2})ln(2)$

At the point (2,1.2)

$divF(2,1.2)=1.2^2e^{2(1.2^2)}+2(2)(1.2)(2^{(2)(1.2^2)})ln(2)$

divF(2,1.2)=50.15

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Example Question #1781 : Calculus

Find the divergence of the function $F(x,y)=x^3y\widehat{i}+x^2y^2\widehat{j}$ at (-1,2)

Hint: $divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}$

Possible Answers:

Correct answer:

Explanation:

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

Vectorfield

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$F(x,y)=x^3y\widehat{i}+x^2y^2\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $(\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

divF(x,y)=3x^2y+2x^2y

At the point (-1,2)

divF(-1,2)=3(-1)^2(2)+2(-1)^2(2)

divF(-1,2)=10

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Example Question #1782 : Calculus

Find the divergence of the function $F(x,y)=e^{2y}\widehat{i}+e^{3x}\widehat{j}$ at (2,3)

Hint: $divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}$

Possible Answers:

5e^6

e^6

Correct answer:

Explanation:

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

Vectorfield

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$F(x,y)=e^{2y}\widehat{i}+e^{3x}\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $(\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

divF(x,y)=0+0

divF(x,y)=0

Since no variables were paired with their respective variable, the divergence goes to zero.

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Example Question #1784 : Calculus

Find the divergence of the function $F(x,y)=3^x\widehat{i}+2^{xy}\widehat{j}$ at (-2,1)

Hint: $divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}$

Possible Answers:

-0.225

-0.317

-0.592

-0.422

-0.100

Correct answer:

-0.225

Explanation:

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

Vectorfield

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$F(x,y)=3^x\widehat{i}+2^{xy}\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $(\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Derivative of an exponential:

d[a^u]=a^uduln(a)

Note that u may represent large functions, and not just individual variables!

$divF(x,y)=3^{x}ln(3)+2^{xy}xln(2)$

At the point (-2,1)

$divF(-2,1)=3^{-2}ln(3)+2^{(-2)(1)}(-2)ln(2)$

divF(-2,1)=-0.225

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Example Question #1785 : Calculus

Find the divergence of the function $F(x,y)=ytan(x)\widehat{i}+ln(xy)\widehat{j}$ at (3,2)

Hint: $divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}$

Possible Answers:

2.541

2.879

1.834

1.229

3.008

Correct answer:

2.541

Explanation:

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

Vectorfield

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$F(x,y)=ytan(x)\widehat{i}+ln(xy)\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $(\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

$divF(x,y)=\frac{y}{cos^2(x)}+\frac{1}{y}$

At the point (3,2)

$divF(3,2)=\frac{2}{cos^2(3)}+\frac{1}{2}$

divF(3,2)=2.541

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Example Question #561 : How To Find Differential Functions

Find the divergence of the function $F(x,y)=xsin(xy)\widehat{i}+xcos(xy)\widehat{j}$ at $(\pi,2)$

Hint: $divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}$

Possible Answers:

$2\pi-\pi^2$

$2\pi+\pi^2+1$

$2\pi$

$2\pi+\pi^2-1$

$2\pi-\pi^2+1$

Correct answer:

$2\pi$

Explanation:

Divergence can be viewed as a measure of the magnitude of a vector field's source or sink at a given point.

Vectorfield

What divergence can calculate is what this velocity is at a given point. Again, the magnitude of the vector field.

We're given the function

$F(x,y)=xsin(xy)\widehat{i}+xcos(xy)\widehat{j}$

What we will do is take the derivative of each vector element with respect to its variable $(\widehat{i}:x,\widehat{j}:y,\widehat{k}:z)$

Then sum the results together:

Trigonometric derivative:

d[sin(u)]=cos(u)du

d[cos(u)]=-sin(u)du

Product rule: d[uv]=udv+vdu

Note that u and v may represent large functions, and not just individual variables!

divF(x,y)=sin(xy)+xycos(xy)-x^2sin(xy)

At the point $(\pi,2)$

$divF(\pi,2)=sin(2\pi)+2\pi cos(2\pi)-\pi^2sin(2\pi)$

$divF(\pi,2)=2\pi$

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1777 : Calculus

Example Question #1778 : Calculus

Example Question #1779 : Calculus

Example Question #564 : How To Find Differential Functions

Example Question #1781 : Calculus

Example Question #1781 : Calculus

Example Question #1782 : Calculus

Example Question #1784 : Calculus

Example Question #1785 : Calculus

Example Question #561 : How To Find Differential Functions

All Calculus 1 Resources

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