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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1671 : Calculus 3

Find the divergence of the vector $\left \langle \cos(x),-2\sin(y),x^2y^2z^2\right \rangle$

Possible Answers:

$\left \langle -\sin(x),2\cos(y),2x^2y^2z\right \rangle$

$\left \langle \sin(x),-2\cos(y),x^2y^2z\right \rangle$

$\left \langle \sin(x),\cos(y),x^2y^2\right \rangle$

$\left \langle -\sin(x),-2\cos(y),2x^2y^2z\right \rangle$

Correct answer:

$\left \langle -\sin(x),-2\cos(y),2x^2y^2z\right \rangle$

Explanation:

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=-\sin(x), \frac{\partial B}{\partial y}=-2\cos(x),\frac{\partial C}{\partial z}=2x^2y^2z$ . Adding all of these up, according to the definition, will produce the correct answer.

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Example Question #1672 : Calculus 3

Find the divergence of the vector $\left \langle 2x,6x^2y^3,z^4\right \rangle$

Possible Answers:

2+18xy^2+4z^3

2+6xy^3+z^5

2+18x^2y^2+4z^3

18x^2y^3+4z^3

Correct answer:

2+18x^2y^2+4z^3

Explanation:

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(2x)+\frac{\partial }{\partial y}(6x^2y^3)+\frac{\partial }{\partial z}(z^4)=2+18x^2y^2+4z^3$

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Example Question #1673 : Calculus 3

Find the divergence of the vector $\left \langle xz^4,6x^5y,x^4yz^2\right \rangle$

Possible Answers:

z^4+6x^5+x^4y

z^4+30x^6+8x^3yz

z^4+6x^5+2x^4yz

z^4+5x^4+2x^3yz^2

Correct answer:

z^4+6x^5+2x^4yz

Explanation:

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(xz^4)+\frac{\partial }{\partial y}(6x^5y)+\frac{\partial }{\partial z}(x^4yz^2)=z^4+6x^5+2x^4yz$

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Example Question #1674 : Calculus 3

Find the divergence of the vector $\left \langle 12x^2y,4y^2z,xy^5\right \rangle$

Possible Answers:

12x+8y

24xy+8yz

3xy+2yz+z

12x+4xyz

Correct answer:

24xy+8yz

Explanation:

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(12x^2y)+\frac{\partial }{\partial y}(4y^2z)+\frac{\partial }{\partial z}(xy^5)=24xy+8yz$

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Example Question #1675 : Calculus 3

Find the divergence of the force field $F = (x^2+y^2)\mathbf{i}+(x^2-y^2)\mathbf{j}+(z^2)\mathbf{k}$

Possible Answers:

2(x+y+z)

2(x+y-z)

x-y-z

2(x-y+z)

Correct answer:

2(x-y+z)

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(x^2+y^2) +\frac{\partial}{\partial y}(x^2-y^2) +\frac{\partial}{\partial z}(z^2)$

=2x-2y+2z

=2(x-y+2) .

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Example Question #23 : Line Integrals

Evaluate the divergence of the force field $F = \sin(xy)\mathbf{i}+\cos(xz)\mathbf{j}+e^{xy}\mathbf{k}$

Possible Answers:

$y\cos(xy)\mathbf{i}-x\sin(xz)\mathbf{j}$

None of the other answers

$y\cos(xy)$

$y\cos(xy)\mathbf{i}$

$y\cos(xy)-x\sin(xz)$

Correct answer:

$y\cos(xy)$

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(\sin(xy)) +\frac{\partial}{\partial y}(\cos(xz)) +\frac{\partial}{\partial z}(e^{xy})$

$=y\cos(xy)+0+0$

$=y\cos(xy)$ .

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Example Question #1677 : Calculus 3

Find the divergence of the force field $F=xyz\mathbf{i}+yz\mathbf{j}+z^3\mathbf{k}$ .

Possible Answers:

$\mathbf{0}$

$3z^2\mathbf{i}+z\mathbf{j}+yz\mathbf{k}$

(y+3z+1)z

None of the other answers

$yz\mathbf{i}+z\mathbf{j}+3z^2\mathbf{k}$

Correct answer:

(y+3z+1)z

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(xyz) +\frac{\partial}{\partial y}(yz) +\frac{\partial}{\partial z}(z^3)$

=yz+z+3z^2

=(y+3z+1)z .

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Example Question #24 : Line Integrals

Find the divergence of the force field $F=2x\mathbf{i}+2z\mathbf{j}+e^z\mathbf{k}$ .

Possible Answers:

None of the other answers

$\mathbf{0}$

2+e^z

$2\mathbf{i}+e^z\mathbf{k}$

Correct answer:

2+e^z

Explanation:

The correct formula for divergence is $div(F) = \bigtriangledown \cdot F$ .

Remember that the result of a divergence calculation must be a scalar, not a vector.

$\bigtriangledown \cdot F = \frac{\partial}{\partial x}(2x) +\frac{\partial}{\partial y}(2z) +\frac{\partial}{\partial z}(e^z)$

=2+0+e^z

=2+e^z .

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Example Question #21 : Divergence

Find the divergence of the vector $\left \langle \sin(x),\tan(y),2yz^2\right \rangle$

Possible Answers:

$\cos(x)-\sec^2(y)+4yz$

$\sin(x)+\sec^2(y)+4yz$

$-\cos(x)+\sec(y)+4yz$

$\cos(x)+\sec^2(y)+4yz$

Correct answer:

$\cos(x)+\sec^2(y)+4yz$

Explanation:

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(\tan(x))+\frac{\partial }{\partial z}(2yz^2)=\cos(x)+\sec^2(y)+4yz$

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Example Question #1680 : Calculus 3

Find $div \vec{F}$ , where F is given by

$\left \langle x^2z, z^2, ye^xz\right \rangle$

Possible Answers:

ye^x

2xz+ye^x

2xz

Correct answer:

2xz+ye^x

Explanation:

The divergence of a vector is given by

$\nabla \cdot \vec{F}$ , where $\nabla = \left \langle f_x, f_y, f_z\right \rangle$

Taking the partial respective partial derivatives of the x, y, and z components of our curve, we get

f_x=2xz

f_y=0

f_z=ye^x

The rules used to find the derivatives are as follows:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1671 : Calculus 3

Example Question #1672 : Calculus 3

Example Question #1673 : Calculus 3

Example Question #1674 : Calculus 3

Example Question #1675 : Calculus 3

Example Question #23 : Line Integrals

Example Question #1677 : Calculus 3

Example Question #24 : Line Integrals

Example Question #21 : Divergence

Example Question #1680 : Calculus 3

All Calculus 3 Resources

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