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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 #823 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-3^{(4z)}\cdot 4^{(y^{2})}sin(x^{3} - 3x))\widehat{i}+(\frac{(3x^{2}ln(z))}{y^{2}})\widehat{j}+(x^{2}y^{2}sin(z^{3}))\widehat{k}\\&\text{at the point }(1.63,2.37,2.61)\end{align*}$

Possible Answers:

$1.94322\cdot10^{8}$

$-5.82965\cdot10^{9}$

$5.82965\cdot10^{9}$

$-9.71609\cdot10^{8}$

Correct answer:

$-9.71609\cdot10^{8}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-3^{(4z)}\cdot 4^{(y^{2})}sin(x^{3} - 3x))\widehat{i}+(\frac{(3x^{2}ln(z))}{y^{2}})\widehat{j}+(x^{2}y^{2}sin(z^{3}))\widehat{k}\\&\text{at the point }(1.63,2.37,2.61)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-3^{(4z)}\cdot 4^{(y^{2})}cos(x^{3} - 3x)\cdot (3x^{2} - 3))+(-\frac{(6x^{2}ln(z))}{y^{3}})+(3x^{2}y^{2}z^{2}cos(z^{3}))\\&divF(1.63,2.37,2.61)=(-9.71609\cdot10^{8})+(-1.15)+(146.44)=-9.71609\cdot10^{8}\end{align*}$

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Example Question #824 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(2sin(y^{2}))}{(7xz^{2})})\widehat{i}+(sin(x^{3})cos(y))\widehat{j}+(\frac{(2cos(y)ln(x))}{z})\widehat{k}\\&\text{at the point }(1.15,1.19,1.74)\end{align*}$

Possible Answers:

1.03

-1.03

0.52

-0.26

Correct answer:

-1.03

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(2sin(y^{2}))}{(7xz^{2})})\widehat{i}+(sin(x^{3})cos(y))\widehat{j}+(\frac{(2cos(y)ln(x))}{z})\widehat{k}\\&\text{at the point }(1.15,1.19,1.74)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(2sin(y^{2}))}{(7x^{2}z^{2})})+(-sin(x^{3})sin(y))+(-\frac{(2cos(y)ln(x))}{z^{2}})\\&divF(1.15,1.19,1.74)=(-0.07)+(-0.93)+(-0.03)=-1.03\end{align*}$

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Example Question #825 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(tan(y^{4})e^{(3x)}cos(3z + z^{2}))}{4})\widehat{i}+(2x^{2}sin(z^{2})e^{(y)})\widehat{j}+(\frac{(5y^{2}ln(x^{2}))}{z^{2}})\widehat{k}\\&\text{at the point }(0.17,1.92,1.28)\end{align*}$

Possible Answers:

61.27

-10.21

245.08

-245.08

Correct answer:

61.27

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(tan(y^{4})e^{(3x)}cos(3z + z^{2}))}{4})\widehat{i}+(2x^{2}sin(z^{2})e^{(y)})\widehat{j}+(\frac{(5y^{2}ln(x^{2}))}{z^{2}})\widehat{k}\\&\text{at the point }(0.17,1.92,1.28)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&divF(x,y,z)=(-\frac{(3tan(y^{4})e^{(3x)}cos(3z + z^{2}))}{4})+(2x^{2}sin(z^{2})e^{(y)})+(-\frac{(10y^{2}ln(x^{2}))}{z^{3}})\\&divF(0.17,1.92,1.28)=(-1.42)+(0.39)+(62.3)=61.27\end{align*}$

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Example Question #826 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(3\cdot 3^{(3y)}tan(x^{3})e^{(z)})}{2})\widehat{i}+(cos(y))\widehat{j}+(2y^{3}z^{3}sin(x^{2}))\widehat{k}\\&\text{at the point }(1.38,0.71,0.09)\end{align*}$

Possible Answers:

-383.03

63.84

-25.54

127.68

Correct answer:

127.68

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(3\cdot 3^{(3y)}tan(x^{3})e^{(z)})}{2})\widehat{i}+(cos(y))\widehat{j}+(2y^{3}z^{3}sin(x^{2}))\widehat{k}\\&\text{at the point }(1.38,0.71,0.09)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(9\cdot 3^{(3y)}x^{2}e^{(z)}\cdot (tan(x^{3})^{2} + 1))}{2})+(-sin(y))+(6y^{3}z^{2}sin(x^{2}))\\&divF(1.38,0.71,0.09)=(128.31)+(-0.65)+(0.02)=127.68\end{align*}$

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Example Question #827 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-3^xe^{(3y)}e^{(3z)})\widehat{i}+(5sin(x^{3})ln(z))\widehat{j}+(5xze^{(y^{2})})\widehat{k}\\&\text{at the point }(2.51,2.58,1.58)\end{align*}$

Possible Answers:

$-2.72668\cdot10^{7}$

$-4.54446\cdot10^{6}$

$1.13611\cdot10^{6}$

$1.81778\cdot10^{7}$

Correct answer:

$-4.54446\cdot10^{6}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-3^xe^{(3y)}e^{(3z)})\widehat{i}+(5sin(x^{3})ln(z))\widehat{j}+(5xze^{(y^{2})})\widehat{k}\\&\text{at the point }(2.51,2.58,1.58)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&divF(x,y,z)=(-3^xe^{(3y)}e^{(3z)}ln(3))+(0)+(5xe^{(y^{2})})\\&divF(2.51,2.58,1.58)=(-4.55422\cdot10^{6})+(0)+(9760.71)=-4.54446\cdot10^{6}\end{align*}$

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Example Question #828 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(sin(z^{2})e^{(3y)}cos(4x + x^{4}))}{2})\widehat{i}+(\frac{(5x^{3}z^{4})}{y})\widehat{j}+(3z^{3}sin(y^{3}))\widehat{k}\\&\text{at the point }(0.48,0.46,0.42)\end{align*}$

Possible Answers:

-0.3

5.99

1.5

-7.49

Correct answer:

1.5

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(sin(z^{2})e^{(3y)}cos(4x + x^{4}))}{2})\widehat{i}+(\frac{(5x^{3}z^{4})}{y})\widehat{j}+(3z^{3}sin(y^{3}))\widehat{k}\\&\text{at the point }(0.48,0.46,0.42)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(sin(z^{2})e^{(3y)}sin(4x + x^{4})\cdot (4x^{3} + 4))}{2})+(-\frac{(5x^{3}z^{4})}{y^{2}})+(9z^{2}sin(y^{3}))\\&divF(0.48,0.46,0.42)=(1.43)+(-0.08)+(0.15)=1.5\end{align*}$

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Example Question #829 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-3ln(3y)ln(z^{2})sin(x))\widehat{i}+(3ln(z)sin(y))\widehat{j}+(e^{(y^{2})}sin(z))\widehat{k}\\&\text{at the point }(1.96,0.92,1.83)\end{align*}$

Possible Answers:

-11.39

1.9

-0.63

0.38

Correct answer:

1.9

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-3ln(3y)ln(z^{2})sin(x))\widehat{i}+(3ln(z)sin(y))\widehat{j}+(e^{(y^{2})}sin(z))\widehat{k}\\&\text{at the point }(1.96,0.92,1.83)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(-3ln(3y)ln(z^{2})cos(x))+(3cos(y)ln(z))+(e^{(y^{2})}cos(z))\\&divF(1.96,0.92,1.83)=(1.4)+(1.1)+(-0.6)=1.9\end{align*}$

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Example Question #830 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(-\frac{(2e^{(x^{2})}cos(z^{4} - 3z)ln(y))}{9})\widehat{i}+(3cos(z^{3})e^{(y^{3})}e^{(x)})\widehat{j}+(2cos(z^{3})sin(x)sin(y))\widehat{k}\\&\text{at the point }(1.4,2.86,1.07)\end{align*}$

Possible Answers:

$2.43585\cdot10^{11}$

$-4.87169\cdot10^{11}$

$-7.30754\cdot10^{12}$

$1.46151\cdot10^{12}$

Correct answer:

$1.46151\cdot10^{12}$

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(-\frac{(2e^{(x^{2})}cos(z^{4} - 3z)ln(y))}{9})\widehat{i}+(3cos(z^{3})e^{(y^{3})}e^{(x)})\widehat{j}+(2cos(z^{3})sin(x)sin(y))\widehat{k}\\&\text{at the point }(1.4,2.86,1.07)\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(4xe^{(x^{2})}cos(z^{4} - 3z)ln(y))}{9})+(9y^{2}cos(z^{3})e^{(y^{3})}e^{(x)})+(-6z^{2}sin(z^{3})sin(x)sin(y))\\&divF(1.4,2.86,1.07)=(1.5)+(1.46151\cdot10^{12})+(-1.77)=1.46151\cdot10^{12}\end{align*}$

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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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #823 : Partial Derivatives

Example Question #824 : Partial Derivatives

Example Question #825 : Partial Derivatives

Example Question #826 : Partial Derivatives

Example Question #827 : Partial Derivatives

Example Question #828 : Partial Derivatives

Example Question #829 : Partial Derivatives

Example Question #830 : Partial Derivatives

Example Question #831 : Partial Derivatives

Example Question #832 : Partial Derivatives

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