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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #3229 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(3^{(4y)}\cdot 4^{(3x)}cos(z))}{3})\widehat{i}+(5zsin(x^{3})e^{(y^{2})})\widehat{j}+(\frac{(5x^{2}cos(y^{3}))}{z^{2}})\widehat{k}\\&\text{at the point }(2.35,0.68,2.68)\end{align*}$

Possible Answers:

$-4.326\cdot10^{5}$

$2.163\cdot10^{6}$

$1.082\cdot10^{5}$

$-2.596\cdot10^{6}$

Correct answer:

$-4.326\cdot10^{5}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(3^{(4y)}\cdot 4^{(3x)}cos(z))}{3})\widehat{i}+(5zsin(x^{3})e^{(y^{2})})\widehat{j}+(\frac{(5x^{2}cos(y^{3}))}{z^{2}})\widehat{k}\\&\text{at the point }(2.35,0.68,2.68)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&divF(x,y,z)=(3^{(4y)}\cdot 4^{(3x)}ln(4)cos(z))+(10yzsin(x^{3})e^{(y^{2})})+(-\frac{(10x^{2}cos(y^{3}))}{z^{3}})\\&divF(2.35,0.68,2.68)=(-4.33\cdot10^{5})+(11.57)+(-2.73)=-4.326\cdot10^{5}\end{align*}$

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Example Question #3230 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2\cdot 3^{(4x)}z^{2})}{(3y)})\widehat{i}+(5y^{2}sin(z^{2}))\widehat{j}+(5xe^{(z^{2})}ln(y))\widehat{k}\\&\text{at the point }(2.51,2.34,1.44)\end{align*}$

Possible Answers:

$-1.599\cdot10^{5}$

$2.665\cdot10^{4}$

$9.595\cdot10^{5}$

$-4.798\cdot10^{5}$

Correct answer:

$-1.599\cdot10^{5}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2\cdot 3^{(4x)}z^{2})}{(3y)})\widehat{i}+(5y^{2}sin(z^{2}))\widehat{j}+(5xe^{(z^{2})}ln(y))\widehat{k}\\&\text{at the point }(2.51,2.34,1.44)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&divF(x,y,z)=(-\frac{(8\cdot 3^{(4x)}z^{2}ln(3))}{(3y)})+(10ysin(z^{2}))+(10xze^{(z^{2})}ln(y))\\&divF(2.51,2.34,1.44)=(-1.6\cdot10^{5})+(20.5)+(244.39)=-1.599\cdot10^{5}\end{align*}$

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Example Question #861 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(4^{(3y)}\cdot 4^xz^{2})}{2})\widehat{i}+(3y^{3}z^{3})\widehat{j}+(2x^{2}zcos(y^{3}))\widehat{k}\\&\text{at the point }(2.47,1.75,1.18)\end{align*}$

Possible Answers:

$4.296\cdot10^{4}$

-7159.35

$8.591\cdot10^{4}$

$-8.591\cdot10^{4}$

Correct answer:

$4.296\cdot10^{4}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(4^{(3y)}\cdot 4^xz^{2})}{2})\widehat{i}+(3y^{3}z^{3})\widehat{j}+(2x^{2}zcos(y^{3}))\widehat{k}\\&\text{at the point }(2.47,1.75,1.18)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&divF(x,y,z)=(\frac{(4^{(3y)}\cdot 4^xz^{2}ln(4))}{2})+(9y^{2}z^{3})+(2x^{2}cos(y^{3}))\\&divF(2.47,1.75,1.18)=(4.29\cdot10^{4})+(45.29)+(7.36)=4.296\cdot10^{4}\end{align*}$

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Example Question #3232 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2\cdot 3^{(4y)}sin(3z + z^{4}))}{(9x)})\widehat{i}+(\frac{(cos(y^{2})e^{(z^{2})})}{x^{2}})\widehat{j}+(3e^{(z)}sin(x)sin(y))\widehat{k}\\&\text{at the point }(2.62,0.44,1.22)\end{align*}$

Possible Answers:

1.96

-0.65

-7.83

11.75

Correct answer:

1.96

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2\cdot 3^{(4y)}sin(3z + z^{4}))}{(9x)})\widehat{i}+(\frac{(cos(y^{2})e^{(z^{2})})}{x^{2}})\widehat{j}+(3e^{(z)}sin(x)sin(y))\widehat{k}\\&\text{at the point }(2.62,0.44,1.22)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(2\cdot 3^{(4y)}sin(3z + z^{4}))}{(9x^{2})})+(-\frac{(2ysin(y^{2})e^{(z^{2})})}{x^{2}})+(3e^{(z)}sin(x)sin(y))\\&divF(2.62,0.44,1.22)=(-0.09)+(-0.11)+(2.16)=1.96\end{align*}$

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Example Question #3233 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2ln(4z)ln(y))}{(9x)})\widehat{i}+(\frac{(2cos(z^{2})ln(x))}{y})\widehat{j}+(ln(x^{2})sin(y^{2})e^{(z)})\widehat{k}\\&\text{at the point }(0.59,0.83,1.57)\end{align*}$

Possible Answers:

-18.55

1.16

27.82

-4.64

Correct answer:

-4.64

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2ln(4z)ln(y))}{(9x)})\widehat{i}+(\frac{(2cos(z^{2})ln(x))}{y})\widehat{j}+(ln(x^{2})sin(y^{2})e^{(z)})\widehat{k}\\&\text{at the point }(0.59,0.83,1.57)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(2ln(4z)ln(y))}{(9x^{2})})+(-\frac{(2cos(z^{2})ln(x))}{y^{2}})+(ln(x^{2})sin(y^{2})e^{(z)})\\&divF(0.59,0.83,1.57)=(-0.22)+(-1.19)+(-3.22)=-4.64\end{align*}$

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Example Question #3234 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(tan(x^{4})tan(y^{4})e^{(z^{2})})}{2})\widehat{i}+(\frac{(e^{(x)}sin(y))}{z^{3}})\widehat{j}+(\frac{x^{4}}{(yz^{2})})\widehat{k}\\&\text{at the point }(0.23,0.33,2.81)\end{align*}$

Possible Answers:

-0.72

0.18

-0.24

2.89

Correct answer:

-0.72

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(tan(x^{4})tan(y^{4})e^{(z^{2})})}{2})\widehat{i}+(\frac{(e^{(x)}sin(y))}{z^{3}})\widehat{j}+(\frac{x^{4}}{(yz^{2})})\widehat{k}\\&\text{at the point }(0.23,0.33,2.81)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-2x^{3}tan(y^{4})e^{(z^{2})}\cdot (tan(x^{4})^{2} + 1))+(\frac{(e^{(x)}cos(y))}{z^{3}})+(-\frac{(2x^{4})}{(yz^{3})})\\&divF(0.23,0.33,2.81)=(-0.78)+(0.05)+(-0.000764)=-0.72\end{align*}$

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Example Question #862 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(e^{(3x)}sin(y)sin(z))}{5})\widehat{i}+(4e^{(x^{2})}e^{(y^{3})}sin(z))\widehat{j}+(4y^{2}z^{3})\widehat{k}\\&\text{at the point }(3,1.83,0.99)\end{align*}$

Possible Answers:

$-4.996\cdot10^{8}$

$2.498\cdot10^{7}$

$1.249\cdot10^{8}$

$-2.498\cdot10^{7}$

Correct answer:

$1.249\cdot10^{8}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(e^{(3x)}sin(y)sin(z))}{5})\widehat{i}+(4e^{(x^{2})}e^{(y^{3})}sin(z))\widehat{j}+(4y^{2}z^{3})\widehat{k}\\&\text{at the point }(3,1.83,0.99)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(3e^{(3x)}sin(y)sin(z))}{5})+(12y^{2}e^{(x^{2})}e^{(y^{3})}sin(z))+(12y^{2}z^{2})\\&divF(3,1.83,0.99)=(3928.85)+(1.25\cdot10^{8})+(39.39)=1.249\cdot10^{8}\end{align*}$

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Example Question #863 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2^{(x^{2})}\cdot 4^{(4y)}ln(3z))}{5})\widehat{i}+(3x^{4}sin(z^{2})cos(y))\widehat{j}+(\frac{(2xsin(y^{2}))}{z})\widehat{k}\\&\text{at the point }(2.02,2.08,2.63)\end{align*}$

Possible Answers:

$-3.997\cdot10^{5}$

$-1.999\cdot10^{6}$

$3.331\cdot10^{5}$

$1.999\cdot10^{6}$

Correct answer:

$1.999\cdot10^{6}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2^{(x^{2})}\cdot 4^{(4y)}ln(3z))}{5})\widehat{i}+(3x^{4}sin(z^{2})cos(y))\widehat{j}+(\frac{(2xsin(y^{2}))}{z})\widehat{k}\\&\text{at the point }(2.02,2.08,2.63)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(2\cdot 2^{(x^{2})}\cdot 4^{(4y)}xln(3z)ln(2))}{5})+(-3x^{4}sin(z^{2})sin(y))+(-\frac{(2xsin(y^{2}))}{z^{2}})\\&divF(2.02,2.08,2.63)=(2\cdot10^{6})+(-25.82)+(0.54)=1.999\cdot10^{6}\end{align*}$

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Example Question #3237 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(sin(x^{2})tan(z^{3})e^{(3y)})\widehat{i}+(\frac{cos(y)}{z})\widehat{j}+(\frac{(4x^{4}z)}{y^{3}})\widehat{k}\\&\text{at the point }(-3.53,1.49,-4.42)\end{align*}$

Possible Answers:

$1.45\cdot10^{4}$

$-1.45\cdot10^{4}$

-3618.05

4824.06

Correct answer:

$1.45\cdot10^{4}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(sin(x^{2})tan(z^{3})e^{(3y)})\widehat{i}+(\frac{cos(y)}{z})\widehat{j}+(\frac{(4x^{4}z)}{y^{3}})\widehat{k}\\&\text{at the point }(-3.53,1.49,-4.42)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(2xcos(x^{2})tan(z^{3})e^{(3y)})+(-\frac{sin(y)}{z})+(\frac{(4x^{4})}{y^{3}})\\&divF(-3.53,1.49,-4.42)=(1.43\cdot10^{4})+(0.23)+(187.76)=1.45\cdot10^{4}\end{align*}$

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Example Question #3238 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2^{(x^{2})}ln(3y)e^{(z)})}{2})\widehat{i}+(\frac{(4cos(y^{3})e^{(x^{2})})}{z^{3}})\widehat{j}+(2ln(y^{2})sin(z^{3})e^{(x)})\widehat{k}\\&\text{at the point }(1.98,4.46,0.95)\end{align*}$

Possible Answers:

-1874.46

$2.81\cdot10^{4}$

1874.46

-9372.28

Correct answer:

-9372.28

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2^{(x^{2})}ln(3y)e^{(z)})}{2})\widehat{i}+(\frac{(4cos(y^{3})e^{(x^{2})})}{z^{3}})\widehat{j}+(2ln(y^{2})sin(z^{3})e^{(x)})\widehat{k}\\&\text{at the point }(1.98,4.46,0.95)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(2^{(x^{2})}xln(3y)e^{(z)}ln(2))+(-\frac{(12y^{2}sin(y^{3})e^{(x^{2})})}{z^{3}})+(6z^{2}cos(z^{3})ln(y^{2})e^{(x)})\\&divF(1.98,4.46,0.95)=(139.37)+(-9588.4)+(76.75)=-9372.28\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3229 : Calculus 3

Example Question #3230 : Calculus 3

Example Question #861 : Partial Derivatives

Example Question #3232 : Calculus 3

Example Question #3233 : Calculus 3

Example Question #3234 : Calculus 3

Example Question #862 : Partial Derivatives

Example Question #863 : Partial Derivatives

Example Question #3237 : Calculus 3

Example Question #3238 : Calculus 3

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