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Calculus 3 : Calculus 3

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

Example Questions

Example Question #843 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(3^{(y^{2})}\cdot 4^zcos(x))}{5})\widehat{i}+(\frac{(5cos(y)tan(x))}{z^{2}})\widehat{j}+(2cos(z^{3})e^{(x^{2})}sin(y))\widehat{k}\\&\text{at the point }(2.36,0.75,1.67)\end{align*}$

Possible Answers:

-995.41

-2986.22

2986.22

1493.11

Correct answer:

2986.22

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(3^{(y^{2})}\cdot 4^zcos(x))}{5})\widehat{i}+(\frac{(5cos(y)tan(x))}{z^{2}})\widehat{j}+(2cos(z^{3})e^{(x^{2})}sin(y))\widehat{k}\\&\text{at the point }(2.36,0.75,1.67)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(3^{(y^{2})}\cdot 4^zsin(x))}{5})+(-\frac{(5tan(x)sin(y))}{z^{2}})+(-6z^{2}sin(z^{3})e^{(x^{2})}sin(y))\\&divF(2.36,0.75,1.67)=(-2.65)+(1.21)+(2987.65)=2986.22\end{align*}$

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Example Question #844 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2e^{(x)}cos(y)sin(z))}{7})\widehat{i}+(4ln(x^{2})e^{(y^{3})}e^{(z^{4})})\widehat{j}+(cos(z^{2})sin(x^{3})e^{(y)})\widehat{k}\\&\text{at the point }(2.87,2.85,2.9)\end{align*}$

Possible Answers:

$-1.21124\cdot10^{43}$

$-6.05621\cdot10^{42}$

$4.84497\cdot10^{43}$

$1.21124\cdot10^{43}$

Correct answer:

$1.21124\cdot10^{43}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2e^{(x)}cos(y)sin(z))}{7})\widehat{i}+(4ln(x^{2})e^{(y^{3})}e^{(z^{4})})\widehat{j}+(cos(z^{2})sin(x^{3})e^{(y)})\widehat{k}\\&\text{at the point }(2.87,2.85,2.9)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(2e^{(x)}cos(y)sin(z))}{7})+(12y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{4})})+(-2zsin(x^{3})sin(z^{2})e^{(y)})\\&divF(2.87,2.85,2.9)=(-1.15)+(1.21124\cdot10^{43})+(84.91)=1.21124\cdot10^{43}\end{align*}$

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Example Question #845 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(3^{(3x)}ln(z^{2})e^{(y^{2})})}{5})\widehat{i}+(z^{4}cos(y))\widehat{j}+(5z^{2}cos(y^{2}))\widehat{k}\\&\text{at the point }(0.23,0.89,0.71)\end{align*}$

Possible Answers:

15.97

-0.89

-2.66

2.66

Correct answer:

2.66

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(3^{(3x)}ln(z^{2})e^{(y^{2})})}{5})\widehat{i}+(z^{4}cos(y))\widehat{j}+(5z^{2}cos(y^{2}))\widehat{k}\\&\text{at the point }(0.23,0.89,0.71)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(3\cdot 3^{(3x)}ln(z^{2})e^{(y^{2})}ln(3))}{5})+(-z^{4}sin(y))+(10zcos(y^{2}))\\&divF(0.23,0.89,0.71)=(-2.13)+(-0.2)+(4.99)=2.66\end{align*}$

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Example Question #846 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(7z^{2}ln(x^{2})cos(3y + y^{2}))}{2})\widehat{i}+(y^{3}z^{2}sin(x))\widehat{j}+(5x^{4}y^{4}cos(z))\widehat{k}\\&\text{at the point }(0.21,0.62,2.18)\end{align*}$

Possible Answers:

-32.56

-97.68

293.03

24.42

Correct answer:

-97.68

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(7z^{2}ln(x^{2})cos(3y + y^{2}))}{2})\widehat{i}+(y^{3}z^{2}sin(x))\widehat{j}+(5x^{4}y^{4}cos(z))\widehat{k}\\&\text{at the point }(0.21,0.62,2.18)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(7z^{2}cos(3y + y^{2}))}{x})+(3y^{2}z^{2}sin(x))+(-5x^{4}y^{4}sin(z))\\&divF(0.21,0.62,2.18)=(-98.82)+(1.14)+(-0.00118)=-97.68\end{align*}$

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Example Question #847 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-2sin(x^{3} - 3x)sin(3y + y^{3})sin(z))\widehat{i}+(4cos(z^{2})cos(y))\widehat{j}+(\frac{(4z^{4}e^{(y^{2})})}{x^{2}})\widehat{k}\\&\text{at the point }(1.2,0.98,1.67)\end{align*}$

Possible Answers:

-45.93

-551.2

137.8

413.4

Correct answer:

137.8

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-2sin(x^{3} - 3x)sin(3y + y^{3})sin(z))\widehat{i}+(4cos(z^{2})cos(y))\widehat{j}+(\frac{(4z^{4}e^{(y^{2})})}{x^{2}})\widehat{k}\\&\text{at the point }(1.2,0.98,1.67)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-2cos(x^{3} - 3x)sin(3y + y^{3})sin(z)\cdot (3x^{2} - 3))+(-4cos(z^{2})sin(y))+(\frac{(16z^{3}e^{(y^{2})})}{x^{2}})\\&divF(1.2,0.98,1.67)=(-0.53)+(3.12)+(135.21)=137.8\end{align*}$

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Example Question #848 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-2\cdot 2^{(3x)}tan(y^{4})tan(z^{3}))\widehat{i}+(\frac{(3z^{2}tan(x))}{y^{2}})\widehat{j}+(3ze^{(y^{2})})\widehat{k}\\&\text{at the point }(2.91,1.31,2.36)\end{align*}$

Possible Answers:

748.68

-62.39

-1247.81

249.56

Correct answer:

249.56

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-2\cdot 2^{(3x)}tan(y^{4})tan(z^{3}))\widehat{i}+(\frac{(3z^{2}tan(x))}{y^{2}})\widehat{j}+(3ze^{(y^{2})})\widehat{k}\\&\text{at the point }(2.91,1.31,2.36)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&divF(x,y,z)=(-6\cdot 2^{(3x)}tan(y^{4})tan(z^{3})ln(2))+(-\frac{(6z^{2}tan(x))}{y^{3}})+(3e^{(y^{2})})\\&divF(2.91,1.31,2.36)=(229.37)+(3.51)+(16.69)=249.56\end{align*}$

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Example Question #849 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(z^{2}tan(x^{2} - 12x)tan(11y + y^{4}))\widehat{i}+(2xyz^{4})\widehat{j}+(3x^{2}cos(y^{2})e^{(z)})\widehat{k}\\&\text{at the point }(0.32,2.19,1.95)\end{align*}$

Possible Answers:

68.46

-2.85

-45.64

11.41

Correct answer:

11.41

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(z^{2}tan(x^{2} - 12x)tan(11y + y^{4}))\widehat{i}+(2xyz^{4})\widehat{j}+(3x^{2}cos(y^{2})e^{(z)})\widehat{k}\\&\text{at the point }(0.32,2.19,1.95)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&divF(x,y,z)=(z^{2}tan(11y + y^{4})\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))+(2xz^{4})+(3x^{2}cos(y^{2})e^{(z)})\\&divF(0.32,2.19,1.95)=(1.98)+(9.25)+(0.18)=11.41\end{align*}$

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Example Question #850 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(9sin(x^{2})sin(z^{2} - 4z))}{(2y^{2})})\widehat{i}+(\frac{(x^{2}ln(z))}{y^{3}})\widehat{j}+(5ln(y^{2})sin(z^{3})ln(x))\widehat{k}\\&\text{at the point }(0.85,0.74,0.86)\end{align*}$

Possible Answers:

5.02

-2.51

0.42

-0.84

Correct answer:

-2.51

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(9sin(x^{2})sin(z^{2} - 4z))}{(2y^{2})})\widehat{i}+(\frac{(x^{2}ln(z))}{y^{3}})\widehat{j}+(5ln(y^{2})sin(z^{3})ln(x))\widehat{k}\\&\text{at the point }(0.85,0.74,0.86)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(9xcos(x^{2})sin(z^{2} - 4z))}{y^{2}})+(-\frac{(3x^{2}ln(z))}{y^{4}})+(15z^{2}cos(z^{3})ln(y^{2})ln(x))\\&divF(0.85,0.74,0.86)=(-4.48)+(1.09)+(0.87)=-2.51\end{align*}$

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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:

1459.49

-121.62

364.87

-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 #852 : Partial Derivatives

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

Possible Answers:

1495.62

249.27

-124.63

-747.81

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #843 : Partial Derivatives

Example Question #844 : Partial Derivatives

Example Question #845 : Partial Derivatives

Example Question #846 : Partial Derivatives

Example Question #847 : Partial Derivatives

Example Question #848 : Partial Derivatives

Example Question #849 : Partial Derivatives

Example Question #850 : Partial Derivatives

Example Question #851 : Partial Derivatives

Example Question #852 : Partial Derivatives

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