Calculus 1 : Other Differential Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #668 : Functions

Find the divergence of the function \(\displaystyle F(x,y,z)=x^2y\widehat{i}+3yz^2\widehat{j}+xz\widehat{k}\) at \(\displaystyle (2,4,5)\)

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

Possible Answers:

\(\displaystyle 42\)

\(\displaystyle 118\)

\(\displaystyle 19\)

\(\displaystyle 72\)

\(\displaystyle 93\)

Correct answer:

\(\displaystyle 93\)

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.

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

 \(\displaystyle F(x,y,z)=x^2y\widehat{i}+3yz^2\widehat{j}+xz\widehat{k}\)

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

Then sum the results together:

 \(\displaystyle divF(x,y,z)=2xy+3z^2+x\)

At the point \(\displaystyle (2,4,5)\)

\(\displaystyle divF(2,4,5)=2(2)(4)+3(5)^2+2\)

\(\displaystyle divF(2,4,5)=93\)

Example Question #669 : Functions

Find the divergence of the function \(\displaystyle F(x,y,z)=4xyz\widehat{i}+2y^2z\widehat{j}+12x^2y\widehat{k}\) at \(\displaystyle (4,1,3)\)

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

Possible Answers:

\(\displaystyle 42\)

\(\displaystyle 24\)

\(\displaystyle 13\)

\(\displaystyle 33\)

\(\displaystyle 7\)

Correct answer:

\(\displaystyle 24\)

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.

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

 \(\displaystyle F(x,y,z)=4xyz\widehat{i}+2y^2z\widehat{j}+12x^2y\widehat{k}\)

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

Then sum the results together:

\(\displaystyle divF(x,y,z)=4yz+4yz+0=8yz\) 

At the point \(\displaystyle (4,1,3)\)

\(\displaystyle divF(4,1,3)=8(1)(3)=24\)

Example Question #670 : Functions

Find the divergence of the function \(\displaystyle F(x,y)=x^2e^y\widehat{i}+5ye^x\widehat{j}\) at \(\displaystyle (-1,2)\)

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

Possible Answers:

\(\displaystyle -15.88\)

\(\displaystyle 8.76\)

\(\displaystyle -12.94\)

\(\displaystyle -23.22\)

\(\displaystyle -5.12\)

Correct answer:

\(\displaystyle -12.94\)

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.

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

\(\displaystyle F(x,y)=x^2e^y\widehat{i}+5ye^x\widehat{j}\) 

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

Then sum the results together:

\(\displaystyle d(e^u)=e^udu\)

\(\displaystyle divF(x,y)=2xe^y+5e^x\)

At the point \(\displaystyle (-1,2)\)

\(\displaystyle divF(-1,2)=2(-1)e^2+5e^{-1}\)

\(\displaystyle divF(-1,2)=-12.94\)

Example Question #671 : Functions

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

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

Possible Answers:

\(\displaystyle \pi\)

\(\displaystyle 1\)

\(\displaystyle 1+\pi\)

\(\displaystyle 0\)

\(\displaystyle 1-\pi\)

Correct answer:

\(\displaystyle 1\)

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.

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

\(\displaystyle F(x,y)=xcos(y)\widehat{i}+xcos(y)\widehat{j}\) 

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

\(\displaystyle d(cos(u))=-sin(u)du\)

Then sum the results together:

\(\displaystyle divF(x,y)=cos(y)-xsin(y)\)

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

\(\displaystyle divF(\pi,\pi)=cos(\pi)-\pi sin(\pi)\)

\(\displaystyle divF(\pi,\pi)=1\)

Example Question #672 : Functions

Find the divergence of the function \(\displaystyle F(x,y)=x^2cos(y)\widehat{i}+x^2sin(y)\widehat{j}\) at \(\displaystyle (3,2\pi)\)

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

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle 9\)

\(\displaystyle 15\)

\(\displaystyle 6\)

\(\displaystyle -3\)

Correct answer:

\(\displaystyle 15\)

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.

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

\(\displaystyle F(x,y)=x^2cos(y)\widehat{i}+x^2sin(y)\widehat{j}\) 

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

Then sum the results together:

\(\displaystyle d(sin(u))=cos(u)du\)

\(\displaystyle divF(x,y)=2xcos(y)+x^2cos(y)\) 

At the point \(\displaystyle (3,2\pi)\)

\(\displaystyle divF(3,2\pi)=2(3)cos(2\pi)+3^2cos(2\pi)=6+9=15\)

Example Question #671 : Differential Functions

Find the divergence of the function \(\displaystyle F(x,y)=e^xln(y)\widehat{i}+cos(x)sin(y)\widehat{j}\) at \(\displaystyle (2,0.2)\)

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

Possible Answers:

\(\displaystyle 11.8\)

\(\displaystyle -6.6\)

\(\displaystyle -1.2\)

\(\displaystyle -12.3\)

\(\displaystyle 5.9\)

Correct answer:

\(\displaystyle -12.3\)

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.

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

\(\displaystyle F(x,y)=e^xln(y)\widehat{i}+cos(x)sin(y)\widehat{j}\) 

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

Then sum the results together:

\(\displaystyle d(e^u)=e^udu\)

\(\displaystyle d(sin(u))=cos(u)du\)

\(\displaystyle div F(x,y)=e^xln(y)+cos(x)cos(y)\)

At the point \(\displaystyle (2,0.2)\)

\(\displaystyle div F(2,0.2)=e^2ln(0.2)+cos(2)cos(0.2)=-12.3\)

Example Question #673 : Functions

Find the divergence of the function \(\displaystyle F(x,y)=18xy^2\widehat{i}+21x^2y\widehat{j}\) at \(\displaystyle (0.2,0.1)\)

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

Possible Answers:

\(\displaystyle 10.2\)

\(\displaystyle 1.02\)

\(\displaystyle 20.4\)

\(\displaystyle 3.12\)

Correct answer:

\(\displaystyle 1.02\)

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.

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

\(\displaystyle F(x,y)=18xy^2\widehat{i}+21x^2y\widehat{j}\) 

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

Then sum the results together:

\(\displaystyle divF(x,y)=18y^2+21x^2\)

At the point  \(\displaystyle (0.2,0.1)\)

\(\displaystyle divF(0.2,0.1)=18(0.1)^2+21(0.2)^2=1.02\)

Example Question #674 : Functions

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

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

Possible Answers:

\(\displaystyle 4.44\)

\(\displaystyle 1.62\)

\(\displaystyle 1.18\)

\(\displaystyle 3.95\)

\(\displaystyle 2.62\)

Correct answer:

\(\displaystyle 1.62\)

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.

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

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

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

Then sum the results together:

Derivative of an exponential: 

\(\displaystyle d[e^u]=e^udu\)

Trigonometric derivative: 

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

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

\(\displaystyle divF(x,y)=y^2cos(xy^2)+e^ycos(2x)\)

At the point \(\displaystyle (3,1)\)

\(\displaystyle divF(3,1)=1^2cos(3(1)^2)+e^1cos(2(3))=1.62\)

Example Question #675 : Functions

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

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

Possible Answers:

\(\displaystyle 16.2\)

\(\displaystyle 18\)

\(\displaystyle 14.2\)

\(\displaystyle 9\)

\(\displaystyle 3.7\)

Correct answer:

\(\displaystyle 18\)

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.

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

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

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

Then sum the results together:

\(\displaystyle divF(x,y)=yx^{y-1}+xy^{x-1}\)

At the point \(\displaystyle (2,3)\)

\(\displaystyle divF(2,3)=(3)(2^{3-1})+(2)(3^{2-1})=18\)

Example Question #676 : Functions

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

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

Possible Answers:

\(\displaystyle 15.4\)

\(\displaystyle 9\)

\(\displaystyle 12.2\)

\(\displaystyle 7.9\)

\(\displaystyle 18\)

Correct answer:

\(\displaystyle 15.4\)

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.

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

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

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

Then sum the results together:

Derivative of an exponential: 

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

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

\(\displaystyle divF(x,y)=y^xln(y)+x^yln(x)\)

At the point \(\displaystyle (2,3)\)

\(\displaystyle divF(2,3)=3^2ln(3)+2^3ln(2)=15.4\)

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