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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #883 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-ln(4x)sin(z^{2})sin(y^{2} - 4y))\widehat{i}+(2e^{(y^{4})}e^{(z^{3})})\widehat{j}+(\frac{(5ln(y))}{x})\widehat{k}\\&\text{at the point }(-1.43,1.74,-1.52)\end{align*}$

Possible Answers:

$-7.22\cdot10^{4}$

$1.20\cdot10^{4}$

-2407.26

6018.14

Correct answer:

$1.20\cdot10^{4}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-ln(4x)sin(z^{2})sin(y^{2} - 4y))\widehat{i}+(2e^{(y^{4})}e^{(z^{3})})\widehat{j}+(\frac{(5ln(y))}{x})\widehat{k}\\&\text{at the point }(-1.43,1.74,-1.52)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&divF(x,y,z)=(-\frac{(sin(z^{2})sin(y^{2} - 4y))}{x})+(8y^{3}e^{(y^{4})}e^{(z^{3})})+(0)\\&divF(-1.43,1.74,-1.52)=(0.367)+(1.204\cdot10^{4})+(0)=1.20\cdot10^{4}\end{align*}$

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Example Question #884 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(3^{(y^{2})}sin(4z + z^{4}))}{(5x^{2})})\widehat{i}+(\frac{(3x^{4}e^{(y)})}{z^{2}})\widehat{j}+(z^{3}e^{(x^{2})}e^{(y^{3})})\widehat{k}\\&\text{at the point }(-2.06,1.65,4.6)\end{align*}$

Possible Answers:

$6.58\cdot10^{4}$

$-7.90\cdot10^{4}$

$-1.18\cdot10^{6}$

$3.95\cdot10^{5}$

Correct answer:

$3.95\cdot10^{5}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(3^{(y^{2})}sin(4z + z^{4}))}{(5x^{2})})\widehat{i}+(\frac{(3x^{4}e^{(y)})}{z^{2}})\widehat{j}+(z^{3}e^{(x^{2})}e^{(y^{3})})\widehat{k}\\&\text{at the point }(-2.06,1.65,4.6)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(2\cdot 3^{(y^{2})}sin(4z + z^{4}))}{(5x^{3})})+(\frac{(3x^{4}e^{(y)})}{z^{2}})+(3z^{2}e^{(x^{2})}e^{(y^{3})})\\&divF(-2.06,1.65,4.6)=(-0.846)+(13.294)+(3.949\cdot10^{5})=3.95\cdot10^{5}\end{align*}$

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Example Question #885 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(7ln(4y)e^{(4z)}e^{(x)})}{2})\widehat{i}+(cos(y^{2})e^{(z^{2})})\widehat{j}+(\frac{(4y^{3}ln(z^{2}))}{x^{2}})\widehat{k}\\&\text{at the point }(-4.36,2.03,-2.88)\end{align*}$

Possible Answers:

$-2.70\cdot10^{4}$

2696.9

-3371.13

$1.35\cdot10^{4}$

Correct answer:

$1.35\cdot10^{4}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(7ln(4y)e^{(4z)}e^{(x)})}{2})\widehat{i}+(cos(y^{2})e^{(z^{2})})\widehat{j}+(\frac{(4y^{3}ln(z^{2}))}{x^{2}})\widehat{k}\\&\text{at the point }(-4.36,2.03,-2.88)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(7ln(4y)e^{(4z)}e^{(x)})}{2})+(-2ysin(y^{2})e^{(z^{2})})+(\frac{(8y^{3})}{(x^{2}z)})\\&divF(-4.36,2.03,-2.88)=(-9.301\cdot10^{-7})+(1.349\cdot10^{4})+(-1.222)=1.35\cdot10^{4}\end{align*}$

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Example Question #886 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(ztan(x^{3}))}{2})\widehat{i}+(\frac{(4y^{3}sin(z^{2}))}{x^{2}})\widehat{j}+(\frac{(4cos(y^{2})ln(z))}{x^{2}})\widehat{k}\\&\text{at the point }(-1.26,1.8,1.11)\end{align*}$

Possible Answers:

-72.16

36.08

6.01

-7.22

Correct answer:

36.08

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(ztan(x^{3}))}{2})\widehat{i}+(\frac{(4y^{3}sin(z^{2}))}{x^{2}})\widehat{j}+(\frac{(4cos(y^{2})ln(z))}{x^{2}})\widehat{k}\\&\text{at the point }(-1.26,1.8,1.11)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(\frac{(3x^{2}z\cdot (tan(x^{3})^{2} + 1))}{2})+(\frac{(12y^{2}sin(z^{2}))}{x^{2}})+(\frac{(4cos(y^{2}))}{(x^{2}z)})\\&divF(-1.26,1.8,1.11)=(15.239)+(23.098)+(-2.259)=36.08\end{align*}$

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Example Question #887 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(y^{2}sin(x^{3} - 3x)sin(z^{3} - 4z))}{5})\widehat{i}+(\frac{(2ln(y^{2})ln(z^{2}))}{x})\widehat{j}+(ln(z^{2})sin(y^{3})ln(x))\widehat{k}\\&\text{at the point }(4.17,3.22,1.88)\end{align*}$

Possible Answers:

25.43

-19.07

76.29

-305.15

Correct answer:

76.29

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(y^{2}sin(x^{3} - 3x)sin(z^{3} - 4z))}{5})\widehat{i}+(\frac{(2ln(y^{2})ln(z^{2}))}{x})\widehat{j}+(ln(z^{2})sin(y^{3})ln(x))\widehat{k}\\&\text{at the point }(4.17,3.22,1.88)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(\frac{(y^{2}cos(x^{3} - 3x)sin(z^{3} - 4z)\cdot (3x^{2} - 3))}{5})+(\frac{(4ln(z^{2}))}{(xy)})+(\frac{(2sin(y^{3})ln(x))}{z})\\&divF(4.17,3.22,1.88)=(74.511)+(0.376)+(1.399)=76.29\end{align*}$

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Example Question #888 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(2\cdot 2^{(4z)}xtan(y^{2} - 12y))\widehat{i}+(\frac{(2e^{(x^{2})}e^{(y^{2})})}{z})\widehat{j}+(ycos(z^{3})sin(x^{2}))\widehat{k}\\&\text{at the point }(3.84,1.07,4.63)\end{align*}$

Possible Answers:

$2.48\cdot10^{7}$

$8.26\cdot10^{6}$

$-3.30\cdot10^{7}$

$-1.38\cdot10^{6}$

Correct answer:

$8.26\cdot10^{6}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(2\cdot 2^{(4z)}xtan(y^{2} - 12y))\widehat{i}+(\frac{(2e^{(x^{2})}e^{(y^{2})})}{z})\widehat{j}+(ycos(z^{3})sin(x^{2}))\widehat{k}\\&\text{at the point }(3.84,1.07,4.63)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(2\cdot 2^{(4z)}tan(y^{2} - 12y))+(\frac{(4ye^{(x^{2})}e^{(y^{2})})}{z})+(-3yz^{2}sin(x^{2})sin(z^{3}))\\&divF(3.84,1.07,4.63)=(8.934\cdot10^{5})+(7.362\cdot10^{6})+(54.063)=8.26\cdot10^{6}\end{align*}$

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Example Question #889 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-cos(x^{2})e^{(4y)}tan(z^{2} - 12z))\widehat{i}+(2cos(y^{2})sin(x^{2})sin(z))\widehat{j}+(3e^{(z)}ln(x)ln(y))\widehat{k}\\&\text{at the point }(2.24,4.89,1.79)\end{align*}$

Possible Answers:

$4.31\cdot10^{9}$

$-1.73\cdot10^{8}$

$-8.63\cdot10^{8}$

$1.73\cdot10^{8}$

Correct answer:

$-8.63\cdot10^{8}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-cos(x^{2})e^{(4y)}tan(z^{2} - 12z))\widehat{i}+(2cos(y^{2})sin(x^{2})sin(z))\widehat{j}+(3e^{(z)}ln(x)ln(y))\widehat{k}\\&\text{at the point }(2.24,4.89,1.79)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(2xsin(x^{2})e^{(4y)}tan(z^{2} - 12z))+(-4ysin(x^{2})sin(y^{2})sin(z))+(3e^{(z)}ln(x)ln(y))\\&divF(2.24,4.89,1.79)=(-8.627\cdot10^{8})+(-17.105)+(23)=-8.63\cdot10^{8}\end{align*}$

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Example Question #890 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(5\cdot 2^{(3x)}e^{(3y)}cos(z^{4} - 3z))}{2})\widehat{i}+(3sin(y^{2})e^{(x^{2})}e^{(z)})\widehat{j}+(\frac{(3y^{4}sin(z^{3}))}{x^{2}})\widehat{k}\\&\text{at the point }(1.84,-1.34,2.83)\end{align*}$

Possible Answers:

-3362.8

5044.21

-280.23

840.7

Correct answer:

840.7

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(5\cdot 2^{(3x)}e^{(3y)}cos(z^{4} - 3z))}{2})\widehat{i}+(3sin(y^{2})e^{(x^{2})}e^{(z)})\widehat{j}+(\frac{(3y^{4}sin(z^{3}))}{x^{2}})\widehat{k}\\&\text{at the point }(1.84,-1.34,2.83)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-\frac{(15\cdot 2^{(3x)}e^{(3y)}ln(2)cos(z^{4} - 3z))}{2})+(6ycos(y^{2})e^{(x^{2})}e^{(z)})+(\frac{(9y^{4}z^{2}cos(z^{3}))}{x^{2}})\\&divF(1.84,-1.34,2.83)=(-2.675)+(897.008)+(-53.632)=840.7\end{align*}$

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Example Question #891 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(tan(y^{3})e^{(4x)})}{(3z)})\widehat{i}+(\frac{(4cos(y^{2})e^{(z^{2})})}{x^{3}})\widehat{j}+(\frac{(e^{(x^{2})}e^{(z)})}{y^{3}})\widehat{k}\\&\text{at the point }(-3.07,1.54,-0.67)\end{align*}$

Possible Answers:

434.21

-289.47

-8684.18

1736.84

Correct answer:

1736.84

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(tan(y^{3})e^{(4x)})}{(3z)})\widehat{i}+(\frac{(4cos(y^{2})e^{(z^{2})})}{x^{3}})\widehat{j}+(\frac{(e^{(x^{2})}e^{(z)})}{y^{3}})\widehat{k}\\&\text{at the point }(-3.07,1.54,-0.67)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(4tan(y^{3})e^{(4x)})}{(3z)})+(-\frac{(8ysin(y^{2})e^{(z^{2})})}{x^{3}})+(\frac{(e^{(x^{2})}e^{(z)})}{y^{3}})\\&divF(-3.07,1.54,-0.67)=(-5.177\cdot10^{-6})+(0.464)+(1736.37)=1736.84\end{align*}$

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Example Question #3260 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(tan(y^{3})cos(x^{3} - 4x))}{(3z)})\widehat{i}+(2y^{2}sin(x^{2})sin(z^{3}))\widehat{j}+(2xycos(z))\widehat{k}\\&\text{at the point }(3.04,-3.34,-2.37)\end{align*}$

Possible Answers:

2.14

-3.22

51.44

-12.86

Correct answer:

-12.86

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(tan(y^{3})cos(x^{3} - 4x))}{(3z)})\widehat{i}+(2y^{2}sin(x^{2})sin(z^{3}))\widehat{j}+(2xycos(z))\widehat{k}\\&\text{at the point }(3.04,-3.34,-2.37)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(tan(y^{3})sin(x^{3} - 4x)\cdot (3x^{2} - 4))}{(3z)})+(4ysin(x^{2})sin(z^{3}))+(-2xysin(z))\\&divF(3.04,-3.34,-2.37)=(-0.352)+(1.651)+(-14.16)=-12.86\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #883 : Partial Derivatives

Example Question #884 : Partial Derivatives

Example Question #885 : Partial Derivatives

Example Question #886 : Partial Derivatives

Example Question #887 : Partial Derivatives

Example Question #888 : Partial Derivatives

Example Question #889 : Partial Derivatives

Example Question #890 : Partial Derivatives

Example Question #891 : Partial Derivatives

Example Question #3260 : Calculus 3

All Calculus 3 Resources

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