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Calculus 3 : Partial Derivatives

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

Example Questions

Example Question #831 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(7x^{2}z^{2}ln(y))}{2})\widehat{i}+(4sin(z^{2})e^{(y^{2})})\widehat{j}+(3e^{(x)}e^{(y)})\widehat{k}\\&\text{at the point }(0.08,0.18,0.43)\end{align*}$

Possible Answers:

0.48

-0.05

0.1

-0.29

Correct answer:

0.1

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(7x^{2}z^{2}ln(y))}{2})\widehat{i}+(4sin(z^{2})e^{(y^{2})})\widehat{j}+(3e^{(x)}e^{(y)})\widehat{k}\\&\text{at the point }(0.08,0.18,0.43)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(7xz^{2}ln(y))+(8ysin(z^{2})e^{(y^{2})})+(0)\\&divF(0.08,0.18,0.43)=(-0.18)+(0.27)+(0)=0.1\end{align*}$

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Example Question #832 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2\cdot 4^{(3x)}y^{2}sin(z^{2}))}{3})\widehat{i}+(2ln(x^{2})cos(y)cos(z))\widehat{j}+(\frac{(4ln(x^{2})e^{(y^{3})})}{z})\widehat{k}\\&\text{at the point }(2.39,0.57,0.09)\end{align*}$

Possible Answers:

-2377.61

594.4

-1188.8

7132.82

Correct answer:

-1188.8

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2\cdot 4^{(3x)}y^{2}sin(z^{2}))}{3})\widehat{i}+(2ln(x^{2})cos(y)cos(z))\widehat{j}+(\frac{(4ln(x^{2})e^{(y^{3})})}{z})\widehat{k}\\&\text{at the point }(2.39,0.57,0.09)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-2\cdot 4^{(3x)}y^{2}sin(z^{2})ln(4))+(-2ln(x^{2})cos(z)sin(y))+(-\frac{(4ln(x^{2})e^{(y^{3})})}{z^{2}})\\&divF(2.39,0.57,0.09)=(-151.32)+(-1.87)+(-1035.61)=-1188.8\end{align*}$

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Example Question #3201 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-ln(4x)ln(3z)e^{(3y)})\widehat{i}+(\frac{(4cos(z))}{(xy^{2})})\widehat{j}+(\frac{(5e^{(x)}ln(z))}{y^{2}})\widehat{k}\\&\text{at the point }(2.72,1.36,0.22)\end{align*}$

Possible Answers:

-64.81

194.42

-583.27

972.12

Correct answer:

194.42

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-ln(4x)ln(3z)e^{(3y)})\widehat{i}+(\frac{(4cos(z))}{(xy^{2})})\widehat{j}+(\frac{(5e^{(x)}ln(z))}{y^{2}})\widehat{k}\\&\text{at the point }(2.72,1.36,0.22)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(ln(3z)e^{(3y)})}{x})+(-\frac{(8cos(z))}{(xy^{3})})+(\frac{(5e^{(x)})}{(y^{2}z)})\\&divF(2.72,1.36,0.22)=(9.04)+(-1.14)+(186.53)=194.42\end{align*}$

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Example Question #3202 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(tan(y^{3})sin(z^{2})sin(3x + x^{4}))\widehat{i}+(4sin(z^{2})e^{(x)}e^{(y)})\widehat{j}+(4ze^{(y^{3})}tan(x))\widehat{k}\\&\text{at the point }(1.65,2.36,0.59)\end{align*}$

Possible Answers:

$-2.57562\cdot10^{7}$

$5.15124\cdot10^{6}$

$-6.43905\cdot10^{6}$

$5.15124\cdot10^{7}$

Correct answer:

$-2.57562\cdot10^{7}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(tan(y^{3})sin(z^{2})sin(3x + x^{4}))\widehat{i}+(4sin(z^{2})e^{(x)}e^{(y)})\widehat{j}+(4ze^{(y^{3})}tan(x))\widehat{k}\\&\text{at the point }(1.65,2.36,0.59)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(tan(y^{3})sin(z^{2})cos(3x + x^{4})\cdot (4x^{3} + 3))+(4sin(z^{2})e^{(x)}e^{(y)})+(4e^{(y^{3})}tan(x))\\&divF(1.65,2.36,0.59)=(4.57)+(75.25)+(-2.57563\cdot10^{7})=-2.57562\cdot10^{7}\end{align*}$

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Example Question #3201 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(5x^{2}cos(y^{2})tan(12z + z^{4}))}{2})\widehat{i}+(5y^{3}cos(z^{3})ln(x^{2}))\widehat{j}+(2y^{2}tan(x))\widehat{k}\\&\text{at the point }(2.24,1.44,1.97)\end{align*}$

Possible Answers:

-94.19

56.51

18.84

-3.14

Correct answer:

18.84

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(5x^{2}cos(y^{2})tan(12z + z^{4}))}{2})\widehat{i}+(5y^{3}cos(z^{3})ln(x^{2}))\widehat{j}+(2y^{2}tan(x))\widehat{k}\\&\text{at the point }(2.24,1.44,1.97)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&divF(x,y,z)=(-5xcos(y^{2})tan(12z + z^{4}))+(15y^{2}cos(z^{3})ln(x^{2}))+(0)\\&divF(2.24,1.44,1.97)=(8.45)+(10.39)+(0)=18.84\end{align*}$

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Example Question #3202 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2^{(x^{2})}sin(y^{3} - 4y)sin(z))}{2})\widehat{i}+(5y^{4}ln(x^{2}))\widehat{j}+(\frac{(4y^{2}cos(z^{2}))}{x^{3}})\widehat{k}\\&\text{at the point }(1.78,0.98,2.97)\end{align*}$

Possible Answers:

-9.86

98.59

19.72

-59.15

Correct answer:

19.72

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2^{(x^{2})}sin(y^{3} - 4y)sin(z))}{2})\widehat{i}+(5y^{4}ln(x^{2}))\widehat{j}+(\frac{(4y^{2}cos(z^{2}))}{x^{3}})\widehat{k}\\&\text{at the point }(1.78,0.98,2.97)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-2^{(x^{2})}xln(2)sin(y^{3} - 4y)sin(z))+(20y^{3}ln(x^{2}))+(-\frac{(8y^{2}zsin(z^{2}))}{x^{3}})\\&divF(1.78,0.98,2.97)=(0.31)+(21.71)+(-2.3)=19.72\end{align*}$

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Example Question #3205 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(4^{(y^{2})}z^{2}ln(4x))}{3})\widehat{i}+(\frac{(3z)}{y})\widehat{j}+(x^{3}sin(z^{3})e^{(y^{4})})\widehat{k}\\&\text{at the point }(0.75,1.32,2.01)\end{align*}$

Possible Answers:

154.84

-10.32

12.9

-51.61

Correct answer:

-51.61

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(4^{(y^{2})}z^{2}ln(4x))}{3})\widehat{i}+(\frac{(3z)}{y})\widehat{j}+(x^{3}sin(z^{3})e^{(y^{4})})\widehat{k}\\&\text{at the point }(0.75,1.32,2.01)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-\frac{(4^{(y^{2})}z^{2})}{(3x)})+(-\frac{(3z)}{y^{2}})+(3x^{3}z^{2}cos(z^{3})e^{(y^{4})})\\&divF(0.75,1.32,2.01)=(-20.1)+(-3.46)+(-28.05)=-51.61\end{align*}$

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Example Question #3206 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(y^{2}e^{(3x)})}{(2z)})\widehat{i}+(5y^{3}e^{(z^{3})})\widehat{j}+(5y^{2}z^{2}e^{(x^{2})})\widehat{k}\\&\text{at the point }(2.34,2.11,1.48)\end{align*}$

Possible Answers:

-67472.6

22490.9

5622.72

-3748.48

Correct answer:

22490.9

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(y^{2}e^{(3x)})}{(2z)})\widehat{i}+(5y^{3}e^{(z^{3})})\widehat{j}+(5y^{2}z^{2}e^{(x^{2})})\widehat{k}\\&\text{at the point }(2.34,2.11,1.48)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(\frac{(3y^{2}e^{(3x)})}{(2z)})+(15y^{2}e^{(z^{3})})+(10y^{2}ze^{(x^{2})})\\&divF(2.34,2.11,1.48)=(5048.26)+(1708.24)+(15734.4)=22490.9\end{align*}$

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Example Question #3203 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(5ln(x^{2})e^{(4y)}e^{(z)})}{2})\widehat{i}+(3zln(y^{2}))\widehat{j}+(5y^{2}ln(z)tan(x))\widehat{k}\\&\text{at the point }(1.02,0.82,0.52)\end{align*}$

Possible Answers:

-204.79

40.96

-51.2

409.58

Correct answer:

-204.79

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(5ln(x^{2})e^{(4y)}e^{(z)})}{2})\widehat{i}+(3zln(y^{2}))\widehat{j}+(5y^{2}ln(z)tan(x))\widehat{k}\\&\text{at the point }(1.02,0.82,0.52)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(5e^{(4y)}e^{(z)})}{x})+(\frac{(6z)}{y})+(\frac{(5y^{2}tan(x))}{z})\\&divF(1.02,0.82,0.52)=(-219.12)+(3.8)+(10.53)=-204.79\end{align*}$

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Example Question #3204 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(e^{(y^{2})}ln(z)sin(x))}{4})\widehat{i}+(\frac{(4cos(z^{3}))}{(x^{3}y)})\widehat{j}+(3ln(z^{2})sin(x^{3})e^{(y)})\widehat{k}\\&\text{at the point }(1.51,0.83,0.36)\end{align*}$

Possible Answers:

-2.17

2.17

51.99

Correct answer:

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(e^{(y^{2})}ln(z)sin(x))}{4})\widehat{i}+(\frac{(4cos(z^{3}))}{(x^{3}y)})\widehat{j}+(3ln(z^{2})sin(x^{3})e^{(y)})\widehat{k}\\&\text{at the point }(1.51,0.83,0.36)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(e^{(y^{2})}cos(x)ln(z))}{4})+(-\frac{(4cos(z^{3}))}{(x^{3}y^{2})})+(\frac{(6sin(x^{3})e^{(y)})}{z})\\&divF(1.51,0.83,0.36)=(0.03)+(-1.68)+(-11.34)=-13\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #831 : Partial Derivatives

Example Question #832 : Partial Derivatives

Example Question #3201 : Calculus 3

Example Question #3202 : Calculus 3

Example Question #3201 : Calculus 3

Example Question #3202 : Calculus 3

Example Question #3205 : Calculus 3

Example Question #3206 : Calculus 3

Example Question #3203 : Calculus 3

Example Question #3204 : Calculus 3

All Calculus 3 Resources

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