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Calculus 3 : Calculus 3

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Example Questions

Example Question #3241 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(5e^{(y)}sin(x)sin(z))}{2})\widehat{i}+(2y^{2}tan(x^{2}))\widehat{j}+(5e^{(y)}sin(z))\widehat{k}\\&\text{at the point }(-1.27,-3.67,3.15)\end{align*}$

Possible Answers:

348.33

-69.67

174.16

-1741.65

Correct answer:

348.33

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(5e^{(y)}sin(x)sin(z))}{2})\widehat{i}+(2y^{2}tan(x^{2}))\widehat{j}+(5e^{(y)}sin(z))\widehat{k}\\&\text{at the point }(-1.27,-3.67,3.15)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(5e^{(y)}cos(x)sin(z))}{2})+(4ytan(x^{2}))+(5e^{(y)}cos(z))\\&divF(-1.27,-3.67,3.15)=(-0.000159)+(348.46)+(-0.13)=348.33\end{align*}$

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Example Question #874 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2sin(x^{2} - 4x)sin(z^{2} - 4z))}{(5y)})\widehat{i}+(4ye^{(z^{2})})\widehat{j}+(\frac{(5cos(z))}{y^{3}})\widehat{k}\\&\text{at the point }(-2.45,1.58,-4.29)\end{align*}$

Possible Answers:

$3.93\cdot10^{8}$

$1.97\cdot10^{8}$

$-6.56\cdot10^{7}$

$-1.97\cdot10^{9}$

Correct answer:

$3.93\cdot10^{8}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2sin(x^{2} - 4x)sin(z^{2} - 4z))}{(5y)})\widehat{i}+(4ye^{(z^{2})})\widehat{j}+(\frac{(5cos(z))}{y^{3}})\widehat{k}\\&\text{at the point }(-2.45,1.58,-4.29)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(2cos(x^{2} - 4x)sin(z^{2} - 4z)\cdot (2x - 4))}{(5y)})+(4e^{(z^{2})})+(-\frac{(5sin(z))}{y^{3}})\\&divF(-2.45,1.58,-4.29)=(1.9)+(3.93\cdot10^{8})+(-1.16)=3.93\cdot10^{8}\end{align*}$

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Example Question #875 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2cos(x^{2})tan(z^{2} - 12z))}{7})\widehat{i}+(\frac{x^{3}}{(yz^{2})})\widehat{j}+(3x^{3}y^{2}sin(z^{2}))\widehat{k}\\&\text{at the point }(-3.29,0.48,1.98)\end{align*}$

Possible Answers:

223.25

111.63

-223.25

-18.6

Correct answer:

111.63

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2cos(x^{2})tan(z^{2} - 12z))}{7})\widehat{i}+(\frac{x^{3}}{(yz^{2})})\widehat{j}+(3x^{3}y^{2}sin(z^{2}))\widehat{k}\\&\text{at the point }(-3.29,0.48,1.98)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-\frac{(4xsin(x^{2})tan(z^{2} - 12z))}{7})+(-\frac{x^{3}}{(y^{2}z^{2})})+(6x^{3}y^{2}zcos(z^{2}))\\&divF(-3.29,0.48,1.98)=(2.82)+(39.43)+(69.38)=111.63\end{align*}$

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Example Question #871 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(y^{2}sin(4x + x^{3})tan(z^{4} - 11z))\widehat{i}+(z^{4}ln(x)sin(y))\widehat{j}+(4xe^{(z^{4})}cos(y))\widehat{k}\\&\text{at the point }(0.43,-1.26,-1.7)\end{align*}$

Possible Answers:

$8.76\cdot10^{4}$

$2.19\cdot10^{4}$

$-4.38\cdot10^{4}$

$-1.31\cdot10^{5}$

Correct answer:

$-4.38\cdot10^{4}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(y^{2}sin(4x + x^{3})tan(z^{4} - 11z))\widehat{i}+(z^{4}ln(x)sin(y))\widehat{j}+(4xe^{(z^{4})}cos(y))\widehat{k}\\&\text{at the point }(0.43,-1.26,-1.7)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(y^{2}cos(4x + x^{3})tan(z^{4} - 11z)\cdot (3x^{2} + 4))+(z^{4}cos(y)ln(x))+(16xz^{3}e^{(z^{4})}cos(y))\\&divF(0.43,-1.26,-1.7)=(4.51)+(-2.16)+(-4.38\cdot10^{4})=-4.38\cdot10^{4}\end{align*}$

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Example Question #877 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(3ln(3y)sin(z^{2})e^{(x^{2})})\widehat{i}+(2sin(y^{2})ln(z))\widehat{j}+(\frac{(3ln(y^{2}))}{z})\widehat{k}\\&\text{at the point }(0.2,2.44,4.49)\end{align*}$

Possible Answers:

-32.01

Correct answer:

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(3ln(3y)sin(z^{2})e^{(x^{2})})\widehat{i}+(2sin(y^{2})ln(z))\widehat{j}+(\frac{(3ln(y^{2}))}{z})\widehat{k}\\&\text{at the point }(0.2,2.44,4.49)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(6xln(3y)sin(z^{2})e^{(x^{2})})+(4ycos(y^{2})ln(z))+(-\frac{(3ln(y^{2}))}{z^{2}})\\&divF(0.2,2.44,4.49)=(2.4)+(13.87)+(-0.27)=16\end{align*}$

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Example Question #878 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2tan(x^{2} - 12x)tan(z^{3} - 11z))}{(9y^{2})})\widehat{i}+(\frac{(5cos(y^{3})ln(x^{2}))}{z^{2}})\widehat{j}+(5xe^{(z^{3})}cos(y))\widehat{k}\\&\text{at the point }(3.13,1.5,-0.13)\end{align*}$

Possible Answers:

-353.11

2118.65

1059.33

-6355.96

Correct answer:

1059.33

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2tan(x^{2} - 12x)tan(z^{3} - 11z))}{(9y^{2})})\widehat{i}+(\frac{(5cos(y^{3})ln(x^{2}))}{z^{2}})\widehat{j}+(5xe^{(z^{3})}cos(y))\widehat{k}\\&\text{at the point }(3.13,1.5,-0.13)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(-\frac{(2tan(z^{3} - 11z)\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{(9y^{2})})+(-\frac{(15y^{2}ln(x^{2})sin(y^{3}))}{z^{2}})+(15xz^{2}e^{(z^{3})}cos(y))\\&divF(3.13,1.5,-0.13)=(5.177)+(1054.09)+(0.056)=1059.33\end{align*}$

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Example Question #879 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2\cdot 3^xe^{(4z)}ln(y))}{9})\widehat{i}+(4yz^{4}tan(x))\widehat{j}+(2x^{4}sin(y^{2})e^{(z^{3})})\widehat{k}\\&\text{at the point }(-4.23,4.28,2.65)\end{align*}$

Possible Answers:

$-8.25\cdot10^{11}$

$2.48\cdot10^{12}$

$-1.65\cdot10^{12}$

$1.65\cdot10^{11}$

Correct answer:

$-8.25\cdot10^{11}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2\cdot 3^xe^{(4z)}ln(y))}{9})\widehat{i}+(4yz^{4}tan(x))\widehat{j}+(2x^{4}sin(y^{2})e^{(z^{3})})\widehat{k}\\&\text{at the point }(-4.23,4.28,2.65)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(2\cdot 3^xe^{(4z)}ln(3)ln(y))}{9})+(4z^{4}tan(x))+(6x^{4}z^{2}sin(y^{2})e^{(z^{3})})\\&divF(-4.23,4.28,2.65)=(136.61)+(-376.705)+(-8.254\cdot10^{11})=-8.25\cdot10^{11}\end{align*}$

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Example Question #872 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(x^{2}y^{2}cos(z))}{4})\widehat{i}+(5ze^{(y^{3})})\widehat{j}+(\frac{(4tan(x^{2})ln(z))}{y^{3}})\widehat{k}\\&\text{at the point }(1.09,-3.26,-0.02)\end{align*}$

Possible Answers:

3.36

-120.79

-6.71

20.13

Correct answer:

20.13

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(x^{2}y^{2}cos(z))}{4})\widehat{i}+(5ze^{(y^{3})})\widehat{j}+(\frac{(4tan(x^{2})ln(z))}{y^{3}})\widehat{k}\\&\text{at the point }(1.09,-3.26,-0.02)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(\frac{(xy^{2}cos(z))}{2})+(15y^{2}ze^{(y^{3})})+(\frac{(4tan(x^{2}))}{(y^{3}z)})\\&divF(1.09,-3.26,-0.02)=(5.791)+(-2.864\cdot10^{-15})+(14.341)=20.13\end{align*}$

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Example Question #881 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2^{(3z)}ln(4y)ln(x))}{5})\widehat{i}+(cos(y^{2})e^{(z^{3})})\widehat{j}+(3zsin(y^{3})ln(x))\widehat{k}\\&\text{at the point }(2.98,2.96,-2.88)\end{align*}$

Possible Answers:

-0.59

9.42

-11.77

2.35

Correct answer:

2.35

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2^{(3z)}ln(4y)ln(x))}{5})\widehat{i}+(cos(y^{2})e^{(z^{3})})\widehat{j}+(3zsin(y^{3})ln(x))\widehat{k}\\&\text{at the point }(2.98,2.96,-2.88)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(2^{(3z)}ln(4y))}{(5x)})+(-2ysin(y^{2})e^{(z^{3})})+(3sin(y^{3})ln(x))\\&divF(2.98,2.96,-2.88)=(4.158\cdot10^{-4})+(-1.539\cdot10^{-10})+(2.354)=2.35\end{align*}$

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Example Question #882 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(e^{(4z)}e^{(y)}tan(12x + x^{2}))}{5})\widehat{i}+(2z^{4}cos(y^{2})ln(x))\widehat{j}+(5z^{3})\widehat{k}\\&\text{at the point }(3,-4.06,-2.57)\end{align*}$

Possible Answers:

2229.45

-445.89

111.47

-89.18

Correct answer:

-445.89

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(e^{(4z)}e^{(y)}tan(12x + x^{2}))}{5})\widehat{i}+(2z^{4}cos(y^{2})ln(x))\widehat{j}+(5z^{3})\widehat{k}\\&\text{at the point }(3,-4.06,-2.57)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(e^{(4z)}e^{(y)}\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1))}{5})+(-4yz^{4}sin(y^{2})ln(x))+(15z^{2})\\&divF(3,-4.06,-2.57)=(7.721\cdot10^{-6})+(-544.963)+(99.073)=-445.89\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3241 : Calculus 3

Example Question #874 : Partial Derivatives

Example Question #875 : Partial Derivatives

Example Question #871 : Partial Derivatives

Example Question #877 : Partial Derivatives

Example Question #878 : Partial Derivatives

Example Question #879 : Partial Derivatives

Example Question #872 : Partial Derivatives

Example Question #881 : Partial Derivatives

Example Question #882 : Partial Derivatives

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