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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 #3231 : Calculus 3

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

Possible Answers:

$8.591\cdot10^{4}$

$4.296\cdot10^{4}$

$-8.591\cdot10^{4}$

-7159.35

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 #864 : Partial Derivatives

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

Possible Answers:

-7.83

1.96

-0.65

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 #865 : Partial Derivatives

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

Possible Answers:

-18.55

-4.64

27.82

1.16

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 #866 : Partial Derivatives

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

Possible Answers:

2.89

-0.24

-0.72

0.18

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 #867 : 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:

$1.249\cdot10^{8}$

$2.498\cdot10^{7}$

$-4.996\cdot10^{8}$

$-2.498\cdot10^{7}$

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 #868 : 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:

$1.999\cdot10^{6}$

$3.331\cdot10^{5}$

$-3.997\cdot10^{5}$

$-1.999\cdot10^{6}$

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 #861 : Partial Derivatives

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

Possible Answers:

$1.45\cdot10^{4}$

$-1.45\cdot10^{4}$

4824.06

-3618.05

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 #870 : Partial Derivatives

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

Possible Answers:

1874.46

$2.81\cdot10^{4}$

-9372.28

-1874.46

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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Example Question #871 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-2\cdot 3^yz^{2}cos(x))\widehat{i}+(\frac{(3y^{3}tan(x))}{z})\widehat{j}+(2z^{2}e^{(x)}cos(y))\widehat{k}\\&\text{at the point }(2.86,3.5,4.68)\end{align*}$

Possible Answers:

256.34

85.45

-1538.03

-51.27

Correct answer:

256.34

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-2\cdot 3^yz^{2}cos(x))\widehat{i}+(\frac{(3y^{3}tan(x))}{z})\widehat{j}+(2z^{2}e^{(x)}cos(y))\widehat{k}\\&\text{at the point }(2.86,3.5,4.68)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(2\cdot 3^yz^{2}sin(x))+(\frac{(9y^{2}tan(x))}{z})+(4ze^{(x)}cos(y))\\&divF(2.86,3.5,4.68)=(569.26)+(-6.81)+(-306.11)=256.34\end{align*}$

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Example Question #3232 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2tan(z^{4})sin(y^{4} - 3y)ln(x))}{7})\widehat{i}+(\frac{(4x^{4}ln(z))}{y})\widehat{j}+(\frac{(3ln(z^{2}))}{(x^{3}y)})\widehat{k}\\&\text{at the point }(1.94,-3.17,1.8)\end{align*}$

Possible Answers:

-6.64

-3.32

0.66

3.32

Correct answer:

-3.32

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2tan(z^{4})sin(y^{4} - 3y)ln(x))}{7})\widehat{i}+(\frac{(4x^{4}ln(z))}{y})\widehat{j}+(\frac{(3ln(z^{2}))}{(x^{3}y)})\widehat{k}\\&\text{at the point }(1.94,-3.17,1.8)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&divF(x,y,z)=(-\frac{(2tan(z^{4})sin(y^{4} - 3y))}{(7x)})+(-\frac{(4x^{4}ln(z))}{y^{2}})+(\frac{6}{(x^{3}yz)})\\&divF(1.94,-3.17,1.8)=(0.14)+(-3.31)+(-0.14)=-3.32\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3231 : Calculus 3

Example Question #864 : Partial Derivatives

Example Question #865 : Partial Derivatives

Example Question #866 : Partial Derivatives

Example Question #867 : Partial Derivatives

Example Question #868 : Partial Derivatives

Example Question #861 : Partial Derivatives

Example Question #870 : Partial Derivatives

Example Question #871 : Partial Derivatives

Example Question #3232 : Calculus 3

All Calculus 3 Resources

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