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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #811 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(2z^{2}cos(y^{4} - 4y)sin(x))\widehat{i}+(2z^{4}e^{(y^{3})}ln(x))\widehat{j}+(4ln(y^{2})cos(z))\widehat{k}\\&\text{at the point }(1.13,2.56,0.68)\end{align*}$

Possible Answers:

$1.1918\cdot10^{8}$

$-3.97267\cdot10^{6}$

$1.98634\cdot10^{7}$

$-9.93168\cdot10^{7}$

Correct answer:

$1.98634\cdot10^{7}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(2z^{2}cos(y^{4} - 4y)sin(x))\widehat{i}+(2z^{4}e^{(y^{3})}ln(x))\widehat{j}+(4ln(y^{2})cos(z))\widehat{k}\\&\text{at the point }(1.13,2.56,0.68)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(2z^{2}cos(y^{4} - 4y)cos(x))+(6y^{2}z^{4}e^{(y^{3})}ln(x))+(-4ln(y^{2})sin(z))\\&divF(1.13,2.56,0.68)=(0.11)+(1.98634\cdot10^{7})+(-4.73)=1.98634\cdot10^{7}\end{align*}$

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Example Question #812 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2x^{2}ln(4y)tan(z^{4}))}{3})\widehat{i}+(\frac{4}{(y^{3}z^{3})})\widehat{j}+(y^{2}sin(z^{3})e^{(x^{2})})\widehat{k}\\&\text{at the point }(1.57,1.71,1)\end{align*}$

Possible Answers:

60.61

15.15

-15.15

-60.61

Correct answer:

60.61

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2x^{2}ln(4y)tan(z^{4}))}{3})\widehat{i}+(\frac{4}{(y^{3}z^{3})})\widehat{j}+(y^{2}sin(z^{3})e^{(x^{2})})\widehat{k}\\&\text{at the point }(1.57,1.71,1)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(4xln(4y)tan(z^{4}))}{3})+(-\frac{12}{(y^{4}z^{3})})+(3y^{2}z^{2}cos(z^{3})e^{(x^{2})})\\&divF(1.57,1.71,1)=(6.27)+(-1.4)+(55.75)=60.61\end{align*}$

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Example Question #813 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2\cdot 2^{(4x)}\cdot 3^{(z^{2})}ln(y))}{3})\widehat{i}+(\frac{(5sin(x^{3})e^{(y^{3})})}{z^{2}})\widehat{j}+(\frac{(5y^{3}ln(z^{2}))}{x^{2}})\widehat{k}\\&\text{at the point }(1.81,1.4,0.9)\end{align*}$

Possible Answers:

829.91

-414.95

207.48

-1244.86

Correct answer:

-414.95

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2\cdot 2^{(4x)}\cdot 3^{(z^{2})}ln(y))}{3})\widehat{i}+(\frac{(5sin(x^{3})e^{(y^{3})})}{z^{2}})\widehat{j}+(\frac{(5y^{3}ln(z^{2}))}{x^{2}})\widehat{k}\\&\text{at the point }(1.81,1.4,0.9)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(8\cdot 2^{(4x)}\cdot 3^{(z^{2})}ln(2)ln(y))}{3})+(\frac{(15y^{2}sin(x^{3})e^{(y^{3})})}{z^{2}})+(\frac{(10y^{3})}{(x^{2}z)})\\&divF(1.81,1.4,0.9)=(-228.91)+(-195.35)+(9.31)=-414.95\end{align*}$

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Example Question #814 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(e^{(y^{2})}sin(3z + z^{2}))}{x})\widehat{i}+(\frac{z^{2}}{y^{3}})\widehat{j}+(5ycos(z^{3})ln(x^{2}))\widehat{k}\\&\text{at the point }(0.42,1.89,2.58)\end{align*}$

Possible Answers:

87.03

2610.93

-522.19

-1566.56

Correct answer:

-522.19

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(e^{(y^{2})}sin(3z + z^{2}))}{x})\widehat{i}+(\frac{z^{2}}{y^{3}})\widehat{j}+(5ycos(z^{3})ln(x^{2}))\widehat{k}\\&\text{at the point }(0.42,1.89,2.58)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(e^{(y^{2})}sin(3z + z^{2}))}{x^{2}})+(-\frac{(3z^{2})}{y^{4}})+(-15yz^{2}ln(x^{2})sin(z^{3}))\\&divF(0.42,1.89,2.58)=(-195.02)+(-1.56)+(-325.6)=-522.19\end{align*}$

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Example Question #815 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(2cos(3z + z^{3})tan(y^{4} - 12y)cos(x))\widehat{i}+(4x^{3}y^{2}z)\widehat{j}+(ln(z^{2})sin(y))\widehat{k}\\&\text{at the point }(1.47,1.16,0.19)\end{align*}$

Possible Answers:

-86.59

14.43

-4.81

57.73

Correct answer:

14.43

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(2cos(3z + z^{3})tan(y^{4} - 12y)cos(x))\widehat{i}+(4x^{3}y^{2}z)\widehat{j}+(ln(z^{2})sin(y))\widehat{k}\\&\text{at the point }(1.47,1.16,0.19)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-2cos(3z + z^{3})tan(y^{4} - 12y)sin(x))+(8x^{3}yz)+(\frac{(2sin(y))}{z})\\&divF(1.47,1.16,0.19)=(-0.82)+(5.6)+(9.65)=14.43\end{align*}$

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Example Question #816 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(4\cdot 2^{(x^{2})}\cdot 4^{(3y)}ln(4z))\widehat{i}+(\frac{(5ln(x^{2})e^{(y^{2})})}{z^{2}})\widehat{j}+(4x^{3}cos(y^{2})e^{(z^{2})})\widehat{k}\\&\text{at the point }(1.13,2.31,0.51)\end{align*}$

Possible Answers:

33094.4

165472

Correct answer:

165472

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(4\cdot 2^{(x^{2})}\cdot 4^{(3y)}ln(4z))\widehat{i}+(\frac{(5ln(x^{2})e^{(y^{2})})}{z^{2}})\widehat{j}+(4x^{3}cos(y^{2})e^{(z^{2})})\widehat{k}\\&\text{at the point }(1.13,2.31,0.51)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&divF(x,y,z)=(8\cdot 2^{(x^{2})}\cdot 4^{(3y)}xln(4z)ln(2))+(\frac{(10yln(x^{2})e^{(y^{2})})}{z^{2}})+(8x^{3}zcos(y^{2})e^{(z^{2})})\\&divF(1.13,2.31,0.51)=(160958)+(4508.94)+(4.46)=165472\end{align*}$

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Example Question #817 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(z^{2}e^{(x^{2})}tan(y^{4} - 12y))}{2})\widehat{i}+(\frac{(5z^{3})}{y^{3}})\widehat{j}+(\frac{cos(z^{3})}{y^{2}})\widehat{k}\\&\text{at the point }(1.44,0.92,1.55)\end{align*}$

Possible Answers:

-647.27

215.76

21.58

-107.88

Correct answer:

-107.88

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(z^{2}e^{(x^{2})}tan(y^{4} - 12y))}{2})\widehat{i}+(\frac{(5z^{3})}{y^{3}})\widehat{j}+(\frac{cos(z^{3})}{y^{2}})\widehat{k}\\&\text{at the point }(1.44,0.92,1.55)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(xz^{2}e^{(x^{2})}tan(y^{4} - 12y))+(-\frac{(15z^{3})}{y^{4}})+(-\frac{(3z^{2}sin(z^{3}))}{y^{2}})\\&divF(1.44,0.92,1.55)=(-34.59)+(-77.97)+(4.68)=-107.88\end{align*}$

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Example Question #818 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(cos(y^{2})ln(3x)ln(3z))}{3})\widehat{i}+(3zsin(x^{2}))\widehat{j}+(\frac{(4ln(z))}{y})\widehat{k}\\&\text{at the point }(1.21,1.39,2.13)\end{align*}$

Possible Answers:

1.53

-6.13

-0.38

3.06

Correct answer:

1.53

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(cos(y^{2})ln(3x)ln(3z))}{3})\widehat{i}+(3zsin(x^{2}))\widehat{j}+(\frac{(4ln(z))}{y})\widehat{k}\\&\text{at the point }(1.21,1.39,2.13)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(cos(y^{2})ln(3z))}{(3x)})+(0)+(\frac{4}{(yz)})\\&divF(1.21,1.39,2.13)=(0.18)+(0)+(1.35)=1.53\end{align*}$

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Example Question #818 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(e^{(x^{2})}e^{(z)})}{(4y)})\widehat{i}+(2sin(z))\widehat{j}+(xy^{2}sin(z^{3}))\widehat{k}\\&\text{at the point }(0.65,1.04,2.25)\end{align*}$

Possible Answers:

1.24

-0.41

0.1

-0.83

Correct answer:

-0.41

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(e^{(x^{2})}e^{(z)})}{(4y)})\widehat{i}+(2sin(z))\widehat{j}+(xy^{2}sin(z^{3}))\widehat{k}\\&\text{at the point }(0.65,1.04,2.25)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-\frac{(xe^{(x^{2})}e^{(z)})}{(2y)})+(0)+(3xy^{2}z^{2}cos(z^{3}))\\&divF(0.65,1.04,2.25)=(-4.52)+(0)+(4.11)=-0.41\end{align*}$

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Example Question #819 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2tan(y^{3})cos(z^{2} - 4z)cos(x))}{9})\widehat{i}+(\frac{(2x^{3}cos(z))}{y^{3}})\widehat{j}+(\frac{(5cos(z^{2})e^{(x)})}{y^{3}})\widehat{k}\\&\text{at the point }(1.08,0.35,1.63)\end{align*}$

Possible Answers:

491.82

122.95

-491.82

-122.95

Correct answer:

-491.82

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2tan(y^{3})cos(z^{2} - 4z)cos(x))}{9})\widehat{i}+(\frac{(2x^{3}cos(z))}{y^{3}})\widehat{j}+(\frac{(5cos(z^{2})e^{(x)})}{y^{3}})\widehat{k}\\&\text{at the point }(1.08,0.35,1.63)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(2tan(y^{3})cos(z^{2} - 4z)sin(x))}{9})+(-\frac{(6x^{3}cos(z))}{y^{4}})+(-\frac{(10zsin(z^{2})e^{(x)})}{y^{3}})\\&divF(1.08,0.35,1.63)=(-0.01)+(29.8)+(-521.61)=-491.82\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #811 : Partial Derivatives

Example Question #812 : Partial Derivatives

Example Question #813 : Partial Derivatives

Example Question #814 : Partial Derivatives

Example Question #815 : Partial Derivatives

Example Question #816 : Partial Derivatives

Example Question #817 : Partial Derivatives

Example Question #818 : Partial Derivatives

Example Question #818 : Partial Derivatives

Example Question #819 : Partial Derivatives

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