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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #851 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(4\cdot 3^{(z^{2})}y^{2}e^{(x^{2})})\widehat{i}+(\frac{(5ln(x))}{(y^{3}z)})\widehat{j}+(2xz)\widehat{k}\\&\text{at the point }(0.83,2.86,1.05)\end{align*}$

Possible Answers:

-121.62

364.87

1459.49

-1824.36

Correct answer:

364.87

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(4\cdot 3^{(z^{2})}y^{2}e^{(x^{2})})\widehat{i}+(\frac{(5ln(x))}{(y^{3}z)})\widehat{j}+(2xz)\widehat{k}\\&\text{at the point }(0.83,2.86,1.05)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(8\cdot 3^{(z^{2})}xy^{2}e^{(x^{2})})+(-\frac{(15ln(x))}{(y^{4}z)})+(2x)\\&divF(0.83,2.86,1.05)=(363.17)+(0.04)+(1.66)=364.87\end{align*}$

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Example Question #3212 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(tan(y^{3})e^{(x)}tan(z^{4} - 12z))\widehat{i}+(2sin(x^{2})e^{(y^{2})}ln(z))\widehat{j}+(y^{4}e^{(z^{2})})\widehat{k}\\&\text{at the point }(2.06,1.91,0.43)\end{align*}$

Possible Answers:

249.27

-747.81

1495.62

-124.63

Correct answer:

249.27

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(tan(y^{3})e^{(x)}tan(z^{4} - 12z))\widehat{i}+(2sin(x^{2})e^{(y^{2})}ln(z))\widehat{j}+(y^{4}e^{(z^{2})})\widehat{k}\\&\text{at the point }(2.06,1.91,0.43)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(tan(y^{3})e^{(x)}tan(z^{4} - 12z))+(4ysin(x^{2})e^{(y^{2})}ln(z))+(2y^{4}ze^{(z^{2})})\\&divF(2.06,1.91,0.43)=(14.6)+(220.9)+(13.77)=249.27\end{align*}$

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Example Question #851 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(3ln(y^{2})cos(z)ln(x))}{2})\widehat{i}+(4x^{4}cos(z^{3})cos(y))\widehat{j}+(\frac{(5sin(y^{3})sin(z))}{x^{2}})\widehat{k}\\&\text{at the point }(1.86,1.68,0.68)\end{align*}$

Possible Answers:

22.86

-91.46

91.46

-45.73

Correct answer:

-45.73

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(3ln(y^{2})cos(z)ln(x))}{2})\widehat{i}+(4x^{4}cos(z^{3})cos(y))\widehat{j}+(\frac{(5sin(y^{3})sin(z))}{x^{2}})\widehat{k}\\&\text{at the point }(1.86,1.68,0.68)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(3ln(y^{2})cos(z))}{(2x)})+(-4x^{4}cos(z^{3})sin(y))+(\frac{(5sin(y^{3})cos(z))}{x^{2}})\\&divF(1.86,1.68,0.68)=(0.65)+(-45.26)+(-1.12)=-45.73\end{align*}$

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Example Question #852 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2^ye^{(4x)}cos(z^{4} - 3z))}{5})\widehat{i}+(5cos(y^{3})e^{(z^{3})})\widehat{j}+(3x^{2}z^{2}cos(y^{2}))\widehat{k}\\&\text{at the point }(1.87,0.15,2.99)\end{align*}$

Possible Answers:

$-2.32\cdot10^{9}$

$2.32\cdot10^{9}$

$4.63\cdot10^{8}$

$-4.63\cdot10^{8}$

Correct answer:

$-4.63\cdot10^{8}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2^ye^{(4x)}cos(z^{4} - 3z))}{5})\widehat{i}+(5cos(y^{3})e^{(z^{3})})\widehat{j}+(3x^{2}z^{2}cos(y^{2}))\widehat{k}\\&\text{at the point }(1.87,0.15,2.99)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(4\cdot 2^ye^{(4x)}cos(z^{4} - 3z))}{5})+(-15y^{2}sin(y^{3})e^{(z^{3})})+(6x^{2}zcos(y^{2}))\\&divF(1.87,0.15,2.99)=(-418.93)+(-4.63\cdot10^{8})+(62.72)=-4.63\cdot10^{8}\end{align*}$

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Example Question #855 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(4^{(x^{2})}y^{2}cos(3z + z^{3}))}{2})\widehat{i}+(5x^{2}cos(y^{2})cos(z^{3}))\widehat{j}+(2cos(y^{3})tan(x^{2})sin(z^{3}))\widehat{k}\\&\text{at the point }(2.61,1.68,0.07)\end{align*}$

Possible Answers:

$-1.26\cdot10^{5}$

$-2.1\cdot10^{4}$

$5.05\cdot10^{5}$

$1.26\cdot10^{5}$

Correct answer:

$-1.26\cdot10^{5}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(4^{(x^{2})}y^{2}cos(3z + z^{3}))}{2})\widehat{i}+(5x^{2}cos(y^{2})cos(z^{3}))\widehat{j}+(2cos(y^{3})tan(x^{2})sin(z^{3}))\widehat{k}\\&\text{at the point }(2.61,1.68,0.07)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-4^{(x^{2})}xy^{2}ln(4)cos(3z + z^{3}))+(-10x^{2}ycos(z^{3})sin(y^{2}))+(6z^{2}cos(y^{3})cos(z^{3})tan(x^{2}))\\&divF(2.61,1.68,0.07)=(-1.26\cdot10^{5})+(-35.91)+(0.000502)=-1.26\cdot10^{5}\end{align*}$

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Example Question #3222 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2tan(z^{2} - 12z))}{(x^{2}y)})\widehat{i}+(\frac{(4sin(x^{3})e^{(y^{3})})}{z})\widehat{j}+(\frac{(3x^{2}cos(y))}{z^{3}})\widehat{k}\\&\text{at the point }(0.2,0.97,0.3)\end{align*}$

Possible Answers:

-670.08

111.68

1116.8

-223.36

Correct answer:

-223.36

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2tan(z^{2} - 12z))}{(x^{2}y)})\widehat{i}+(\frac{(4sin(x^{3})e^{(y^{3})})}{z})\widehat{j}+(\frac{(3x^{2}cos(y))}{z^{3}})\widehat{k}\\&\text{at the point }(0.2,0.97,0.3)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(4tan(z^{2} - 12z))}{(x^{3}y)})+(\frac{(12y^{2}sin(x^{3})e^{(y^{3})})}{z})+(-\frac{(9x^{2}cos(y))}{z^{4}})\\&divF(0.2,0.97,0.3)=(-198.99)+(0.75)+(-25.12)=-223.36\end{align*}$

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Example Question #853 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(4\cdot 4^{(z^{2})}tan(x^{4} - 11x)ln(y))\widehat{i}+(\frac{(3ytan(x))}{z})\widehat{j}+(4x^{3}y^{4}z^{4})\widehat{k}\\&\text{at the point }(1.62,2.24,0.53)\end{align*}$

Possible Answers:

$4.406\cdot10^{4}$

-7344.02

7344.02

-2448.01

Correct answer:

7344.02

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(4\cdot 4^{(z^{2})}tan(x^{4} - 11x)ln(y))\widehat{i}+(\frac{(3ytan(x))}{z})\widehat{j}+(4x^{3}y^{4}z^{4})\widehat{k}\\&\text{at the point }(1.62,2.24,0.53)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&divF(x,y,z)=(4\cdot 4^{(z^{2})}ln(y)\cdot (tan(x^{4} - 11x)^{2} + 1)\cdot (4x^{3} - 11))+(\frac{(3tan(x))}{z})+(16x^{3}y^{4}z^{3})\\&divF(1.62,2.24,0.53)=(7204)+(-114.95)+(254.97)=7344.02\end{align*}$

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Example Question #852 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(2ln(x^{2})sin(y^{2})tan(12z + z^{3}))\widehat{i}+(\frac{(4x^{2}ln(z))}{y^{3}})\widehat{j}+(5x^{3}z^{4}cos(y^{2}))\widehat{k}\\&\text{at the point }(0.74,0.66,2.93)\end{align*}$

Possible Answers:

-29.85

-746.16

149.23

49.74

Correct answer:

149.23

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(2ln(x^{2})sin(y^{2})tan(12z + z^{3}))\widehat{i}+(\frac{(4x^{2}ln(z))}{y^{3}})\widehat{j}+(5x^{3}z^{4}cos(y^{2}))\widehat{k}\\&\text{at the point }(0.74,0.66,2.93)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(\frac{(4sin(y^{2})tan(12z + z^{3}))}{x})+(-\frac{(12x^{2}ln(z))}{y^{4}})+(20x^{3}z^{3}cos(y^{2}))\\&divF(0.74,0.66,2.93)=(1.64)+(-37.23)+(184.82)=149.23\end{align*}$

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Example Question #859 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(3\cdot 3^xtan(y^{3})tan(11z + z^{2}))}{2})\widehat{i}+(\frac{(5cos(y^{2}))}{z^{3}})\widehat{j}+(2e^{(y^{4})}e^{(z)}sin(x))\widehat{k}\\&\text{at the point }(1.25,1.56,0.61)\end{align*}$

Possible Answers:

-6270.87

313.54

1254.17

-250.83

Correct answer:

1254.17

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(3\cdot 3^xtan(y^{3})tan(11z + z^{2}))}{2})\widehat{i}+(\frac{(5cos(y^{2}))}{z^{3}})\widehat{j}+(2e^{(y^{4})}e^{(z)}sin(x))\widehat{k}\\&\text{at the point }(1.25,1.56,0.61)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(-\frac{(3\cdot 3^xtan(y^{3})ln(3)tan(11z + z^{2}))}{2})+(-\frac{(10ysin(y^{2}))}{z^{3}})+(2e^{(y^{4})}e^{(z)}sin(x))\\&divF(1.25,1.56,0.61)=(-5.13)+(-44.69)+(1304)=1254.17\end{align*}$

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Example Question #860 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(2ln(4z)tan(x^{3})sin(y))\widehat{i}+(\frac{(2y^{3}tan(x))}{z^{3}})\widehat{j}+(3ln(y^{2})e^{(z^{3})}sin(x))\widehat{k}\\&\text{at the point }(1.34,0.1,0.04)\end{align*}$

Possible Answers:

3985.99

$-1.594\cdot10^{4}$

-797.2

$2.392\cdot10^{4}$

Correct answer:

3985.99

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(2ln(4z)tan(x^{3})sin(y))\widehat{i}+(\frac{(2y^{3}tan(x))}{z^{3}})\widehat{j}+(3ln(y^{2})e^{(z^{3})}sin(x))\widehat{k}\\&\text{at the point }(1.34,0.1,0.04)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(6x^{2}ln(4z)sin(y)\cdot (tan(x^{3})^{2} + 1))+(\frac{(6y^{2}tan(x))}{z^{3}})+(9z^{2}ln(y^{2})e^{(z^{3})}sin(x))\\&divF(1.34,0.1,0.04)=(-3.58)+(3989.64)+(-0.06)=3985.99\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #851 : Partial Derivatives

Example Question #3212 : Calculus 3

Example Question #851 : Partial Derivatives

Example Question #852 : Partial Derivatives

Example Question #855 : Partial Derivatives

Example Question #3222 : Calculus 3

Example Question #853 : Partial Derivatives

Example Question #852 : Partial Derivatives

Example Question #859 : Partial Derivatives

Example Question #860 : Partial Derivatives

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