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Calculus 1 : Other Differential Functions

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Example Questions

Example Question #571 : How To Find Differential Functions

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

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:

$\frac{5}{6}$

$\frac{6}{5}$

$\frac{5}{36}$

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.

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

$F(x,y)=\frac{x}{y}\widehat{i}+\frac{y}{x}\widehat{j}$

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:

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

At the point $(\frac{1}{2},\frac{1}{3})$

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

$divF(\frac{1}{2},\frac{1}{3})=5$

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Example Question #571 : Other Differential Functions

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

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:

$\frac{-35}{6}$

$\frac{-6}{5}$

$\frac{-5}{6}$

Correct answer:

$\frac{-35}{6}$

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

$F(x,y)=\frac{y}{x}\widehat{i}+\frac{x}{y}\widehat{j}$

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:

Quotient rule: $d[\frac{u}{v}]=\frac{vdu-udv}{v^2}$

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

$divF(x,y)=-\frac{y}{x^2}-\frac{x}{y^2}$

At the point $(\frac{1}{2},\frac{1}{3})$

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

$divF(\frac{1}{2},\frac{1}{3})=\frac{-35}{6}$

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Example Question #571 : Other Differential Functions

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

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:

ln(4)

ln(3)

Correct answer:

ln(3)

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

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

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:

$divF(x,y)=ln(y)x^{ln(y)-1}+ln(x)y^{ln(x)-1}$

At the point (3,1)

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

divF(3,1)=ln(3)

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Example Question #573 : Other Differential Functions

Find the divergence of the function $F(x,y)=\frac{tan(x)}{y}\widehat{i}+\frac{sin(y)}{x}\widehat{j}+\widehat{k}$ at $(\pi,\pi)$

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:

$\frac{1}{\pi}$

$\pi$

$\frac{2}{\pi}$

$-\pi$

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

$F(x,y)=\frac{tan(x)}{y}\widehat{i}+\frac{sin(y)}{x}\widehat{j}+\widehat{k}$

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

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

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

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

At the point $(\pi,\pi)$

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

$divF(\pi,\pi)=0$

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Example Question #571 : Other Differential Functions

Find the derivative of f(x)=4x^3+2x^2-3x+1 .

Possible Answers:

$f{}'(x)=12x^2+4x-3$

$f{}'(x)=12x^2+4x$

$f{}'(x)=12x^2+4x-6$

$f{}'(x)=12x^2-4x-3$

$f{}'(x)=x^2+4x-3$

Correct answer:

$f{}'(x)=12x^2+4x-3$

Explanation:

To find the derivative of a term, multiply by the exponent and then subtract one from the exponent. 4x^3 becomes 12x^2 , 2x^2 becomes , -3x becomes -, and the derivative of is because it is a constant. Put those all together to get your answer: $f{}'(x)=12x^2+4x-3$ .

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Example Question #575 : Other Differential Functions

Find the divergence of the function $F(x,y)=ycos(x)\widehat{i}+ysin(x)\widehat{j}$ at $(\frac{\pi}{2},3)$

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:

$\pi$

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

$F(x,y)=ycos(x)\widehat{i}+ysin(x)\widehat{j}$

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

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

divF(x,y)=-ysin(x)+sin(x)

At the point $(\frac{\pi}{2},3)$

$divF(\frac{\pi}{2},3)=-3sin(\frac{\pi}{2})+sin(\frac{\pi}{2})$

$divF(\frac{\pi}{2},3)=-2$

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Example Question #576 : Other Differential Functions

Find the divergence of the function $F(x,y)=xycos(xy)\widehat{i}+xysin(xy)\widehat{j}$ at $(1,\pi)$

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:

$2\pi$

$-2\pi$

$-\pi$

$\pi$

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

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

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 and may represent large functions, and not just individual variables!

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

At the point $(1,\pi)$

$divF(1,\pi)=\pi cos((1)(\pi))-(1)(\pi)^2sin((1)(\pi))+(1)sin((1)(\pi))+(1)^2(\pi)cos((1)(\pi))$

$divF(1,\pi)=-2\pi$

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Example Question #577 : Other Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=2^{1.2^{x}}$ and y(0)=4 , approximate y(3) using Euler's Method and three steps.

Possible Answers:

11.010

13.131

17.810

15.328

8.297

Correct answer:

11.010

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 $f'(x)=2^{1.2^{x}}$ and y(0)=4

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

$\Delta x = \frac{3-0}{3}=1$

y_0=4;x_0=0

$y_1=4+(1)2^{1.2^{0}}=6$

$y_2=6+(1)2^{1.2^{1}}=8.297$

$y_3=8.297+(1)2^{1.2^{2}}=11.010$

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Example Question #578 : Other Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=e^{x^{ln(x)}}$ and y(1)=2.5 , approximate y(7) using Euler's Method and three steps.

Possible Answers:

$1.235 \cdot 10^3$

$1.235 \cdot 10^6$

$1.235 \cdot 10^2$

$1.235 \cdot 10^5$

$1.235 \cdot 10^4$

Correct answer:

$1.235 \cdot 10^6$

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 $f'(x)=e^{x^{ln(x)}}$ and y(1)=2.5

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

$\Delta x = \frac{7-1}{3}=2$

y_0=2.5;x_0=1

$y_1=2.5+(2)e^{1^{ln(1)}}=7.937$

$y_2=7.937+(2)e^{3^{ln(3)}}=64.560$

$y_3=64.560+(2)e^{5^{ln(5)}}=1.235 \cdot 10^6$

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Example Question #579 : Other Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=(cos(x))^{x}$ and y(0)=3 , approximate y(0.9) using Euler's Method and three steps.

Possible Answers:

3.863

4.403

3.596

4.199

3.267

Correct answer:

3.863

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 $f'(x)=(cos(x))^{x}$ and y(0)=3

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

$\Delta x = \frac{0.9-0}{3}=0.3$

y_0=3;x_0=0

$y_1=3+(0.3)(cos(0))^{0}=3.3$

$y_2=3.3+(0.3)(cos(0.3))^{0.3}=3.596$

$y_3=3.596+(0.3)(cos(0.6))^{0.6}=3.863$

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Calculus 1 : Other Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #571 : How To Find Differential Functions

Example Question #571 : Other Differential Functions

Example Question #571 : Other Differential Functions

Example Question #573 : Other Differential Functions

Example Question #571 : Other Differential Functions

Example Question #575 : Other Differential Functions

Example Question #576 : Other Differential Functions

Example Question #577 : Other Differential Functions

Example Question #578 : Other Differential Functions

Example Question #579 : Other Differential Functions

All Calculus 1 Resources

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