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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #1333 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}y^{4}cos(z^{3}) - 2^xe^{(y^{2})}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(18,-2,0).\end{align*}$

Possible Answers:

${3.92x^{3}y^{4}cos(z^{3}) + 0.0484x^{4}y^{3}cos(z^{3}) - 0.679\cdot 2^xe^{(y^{2})}cos(z) - 0.0242\cdot 2^xye^{(y^{2})}cos(z)}$

${3.96x^{3}y^{4}cos(z^{3}) - 0.44x^{4}y^{3}cos(z^{3}) - 0.686\cdot 2^xe^{(y^{2})}cos(z) + 0.22\cdot 2^xye^{(y^{2})}cos(z)}$

${72.4x^{3}y^{4}cos(z^{3}) + 72.4x^{4}y^{3}cos(z^{3}) - 12.6\cdot 2^xe^{(y^{2})}cos(z) + 18.1\cdot 2^xe^{(y^{2})}sin(z) - 36.2\cdot 2^xye^{(y^{2})}cos(z) - 54.3x^{4}y^{4}z^{2}sin(z^{3})}$

${25.1\cdot 2^xye^{(y^{2})}sin(z) - 869x^{3}y^{3}z^{2}sin(z^{3})}$

Correct answer:

${3.96x^{3}y^{4}cos(z^{3}) - 0.44x^{4}y^{3}cos(z^{3}) - 0.686\cdot 2^xe^{(y^{2})}cos(z) + 0.22\cdot 2^xye^{(y^{2})}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(-2)^2+(0)^2}=18.11\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{18.11}=0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{18.11}=-0.11\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.11}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.99)(4x^{3}y^{4}cos(z^{3}) - 0.693\cdot 2^xe^{(y^{2})}cos(z))+(-0.11)(4x^{4}y^{3}cos(z^{3}) - 2\cdot 2^xye^{(y^{2})}cos(z))+(0)(2^xe^{(y^{2})}sin(z) - 3x^{4}y^{4}z^{2}sin(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=3.96x^{3}y^{4}cos(z^{3}) - 0.44x^{4}y^{3}cos(z^{3}) - 0.686\cdot 2^xe^{(y^{2})}cos(z) + 0.22\cdot 2^xye^{(y^{2})}cos(z)\end{align*}$

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Example Question #1334 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(2^{(z^{4})}x)}{y}- 3zcos(y)\\&\text{In the direction of }\overrightarrow{v}=(-20,-3,0).\end{align*}$

Possible Answers:

${-\frac{(56.1\cdot 2^{(z^{4})}z^{3})}{y^{2}}}$

${\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}}$

${\frac{(20.2\cdot 2^{(z^{4})})}{y}- 60.7cos(y) + 60.7zsin(y) -\frac{ (20.2\cdot 2^{(z^{4})}x)}{y^{2}}+\frac{ (56.1\cdot 2^{(z^{4})}xz^{3})}{y}}$

${\frac{(3\cdot 2^{(z^{4})}x)}{y^{2}}- 9zsin(y) -\frac{ (20\cdot 2^{(z^{4})})}{y}}$

Correct answer:

${\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-3)^2+(0)^2}=20.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{20.22}=-0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{20.22}=-0.15\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.22}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{2^{(z^{4})}}{y})+(-0.15)(3zsin(y) -\frac{ (1\cdot 2^{(z^{4})}x)}{y^{2}})+(0)(\frac{(2.77\cdot 2^{(z^{4})}xz^{3})}{y}- 3cos(y))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}\end{align*}$

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Example Question #3701 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(ln(x^{3})ln(y^{4}))}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-20,-5,0).\end{align*}$

Possible Answers:

${\frac{(20ln(x^{3}))}{(yz^{2})}+\frac{ (60ln(y^{4}))}{(xz^{2})}}$

${\frac{495.0}{(xyz^{3})}}$

${\frac{(41.2ln(x^{3})ln(y^{4}))}{z^{3}}-\frac{ (61.9ln(y^{4}))}{(xz^{2})}-\frac{ (82.5ln(x^{3}))}{(yz^{2})}}$

${\frac{(0.96ln(x^{3}))}{(yz^{2})}+\frac{ (2.91ln(y^{4}))}{(xz^{2})}}$

Correct answer:

${\frac{(0.96ln(x^{3}))}{(yz^{2})}+\frac{ (2.91ln(y^{4}))}{(xz^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-5)^2+(0)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{20.62}=-0.97\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{20.62}=-0.24\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.97)(-\frac{(3ln(y^{4}))}{(xz^{2})})+(-0.24)(-\frac{(4ln(x^{3}))}{(yz^{2})})+(0)(\frac{(2ln(x^{3})ln(y^{4}))}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.96ln(x^{3}))}{(yz^{2})}+\frac{ (2.91ln(y^{4}))}{(xz^{2})}\end{align*}$

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Example Question #1336 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2z^{3}tan(x^{2}) + ln(x^{4})e^{(y^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-17,1,0).\end{align*}$

Possible Answers:

${4y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 68xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (68e^{(y^{4})}sin(z))}{x}}$

${102z^{2}tan(x^{2}) +\frac{ (68.1e^{(y^{4})}sin(z))}{x}+ 17ln(x^{4})e^{(y^{4})}cos(z) + 68.1xz^{3}\cdot (tan(x^{2})^{2} + 1) + 68.1y^{3}ln(x^{4})e^{(y^{4})}sin(z)}$

${\frac{(272y^{3}e^{(y^{4})}cos(z))}{x}}$

${0.24y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 4xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (4e^{(y^{4})}sin(z))}{x}}$

Correct answer:

${0.24y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 4xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (4e^{(y^{4})}sin(z))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-17)^2+(1)^2+(0)^2}=17.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-17}{17.03}=-1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{17.03}=0.06\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-1)(\frac{(4e^{(y^{4})}sin(z))}{x}+ 4xz^{3}\cdot (tan(x^{2})^{2} + 1))+(0.06)(4y^{3}ln(x^{4})e^{(y^{4})}sin(z))+(0)(6z^{2}tan(x^{2}) + ln(x^{4})e^{(y^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.24y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 4xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (4e^{(y^{4})}sin(z))}{x}\end{align*}$

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Example Question #1337 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^xyln(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,3,-10).\end{align*}$

Possible Answers:

${0.29\cdot 2^xln(z^{2}) -\frac{ (1.92\cdot 2^xy)}{z}}$

${10.4\cdot 2^xln(z^{2}) + 7.24\cdot 2^xyln(z^{2}) +\frac{ (20.9\cdot 2^xy)}{z}}$

${0.0841\cdot 2^xln(z^{2}) +\frac{ (1.84\cdot 2^xy)}{z}}$

${\frac{(14.5\cdot 2^x)}{z}}$

Correct answer:

${0.29\cdot 2^xln(z^{2}) -\frac{ (1.92\cdot 2^xy)}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(3)^2+(-10)^2}=10.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.44}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{3}{10.44}=0.29\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{10.44}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.693\cdot 2^xyln(z^{2}))+(0.29)(2^xln(z^{2}))+(-0.96)(\frac{(2\cdot 2^xy)}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.29\cdot 2^xln(z^{2}) -\frac{ (1.92\cdot 2^xy)}{z}\end{align*}$

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Example Question #1331 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(4cos(z^{3})ln(y))}{x}- z^{4}sin(y^{4})e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(20,-5,0).\end{align*}$

Possible Answers:

${0.96y^{3}z^{4}cos(y^{4})e^{(x)} -\frac{ (3.88cos(z^{3})ln(y))}{x^{2}}- 0.97z^{4}sin(y^{4})e^{(x)} -\frac{ (0.96cos(z^{3}))}{(xy)}}$

${\frac{(247z^{2}sin(z^{3}))}{(x^{2}y)}- 330y^{3}z^{3}cos(y^{4})e^{(x)}}$

${\frac{(82.5cos(z^{3}))}{(xy)}-\frac{ (82.5cos(z^{3})ln(y))}{x^{2}}- 82.5z^{3}sin(y^{4})e^{(x)} - 20.6z^{4}sin(y^{4})e^{(x)} - 82.5y^{3}z^{4}cos(y^{4})e^{(x)} -\frac{ (247z^{2}sin(z^{3})ln(y))}{x}}$

${\frac{(0.23cos(z^{3}))}{(xy)}-\frac{ (3.76cos(z^{3})ln(y))}{x^{2}}- 0.941z^{4}sin(y^{4})e^{(x)} - 0.23y^{3}z^{4}cos(y^{4})e^{(x)}}$

Correct answer:

${0.96y^{3}z^{4}cos(y^{4})e^{(x)} -\frac{ (3.88cos(z^{3})ln(y))}{x^{2}}- 0.97z^{4}sin(y^{4})e^{(x)} -\frac{ (0.96cos(z^{3}))}{(xy)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(-5)^2+(0)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{20.62}=0.97\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{20.62}=-0.24\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.97)(-\frac{ (4cos(z^{3})ln(y))}{x^{2}}- z^{4}sin(y^{4})e^{(x)})+(-0.24)(\frac{(4cos(z^{3}))}{(xy)}- 4y^{3}z^{4}cos(y^{4})e^{(x)})+(0)(- 4z^{3}sin(y^{4})e^{(x)} -\frac{ (12z^{2}sin(z^{3})ln(y))}{x})\\&D_{\overrightarrow{u}}(x,y,z)=0.96y^{3}z^{4}cos(y^{4})e^{(x)} -\frac{ (3.88cos(z^{3})ln(y))}{x^{2}}- 0.97z^{4}sin(y^{4})e^{(x)} -\frac{ (0.96cos(z^{3}))}{(xy)}\end{align*}$

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Example Question #272 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4z^{2}e^{(y)} - x^{3}cos(y^{4})sin(z^{3})\\&\text{In the direction of }\overrightarrow{v}=(-18,12,0).\end{align*}$

Possible Answers:

${2.2z^{2}e^{(y)} + 2.49x^{2}cos(y^{4})sin(z^{3}) + 2.2x^{3}y^{3}sin(y^{4})sin(z^{3})}$

${48z^{2}e^{(y)} + 54x^{2}cos(y^{4})sin(z^{3}) + 48x^{3}y^{3}sin(y^{4})sin(z^{3})}$

${86.5z^{2}e^{(y)} + 173ze^{(y)} - 64.9x^{2}cos(y^{4})sin(z^{3}) - 64.9x^{3}z^{2}cos(y^{4})cos(z^{3}) + 86.5x^{3}y^{3}sin(y^{4})sin(z^{3})}$

${779x^{2}y^{3}z^{2}cos(z^{3})sin(y^{4})}$

Correct answer:

${2.2z^{2}e^{(y)} + 2.49x^{2}cos(y^{4})sin(z^{3}) + 2.2x^{3}y^{3}sin(y^{4})sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(12)^2+(0)^2}=21.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{21.63}=-0.83\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{21.63}=0.55\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{21.63}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.83)(-3x^{2}cos(y^{4})sin(z^{3}))+(0.55)(4z^{2}e^{(y)} + 4x^{3}y^{3}sin(y^{4})sin(z^{3}))+(0)(8ze^{(y)} - 3x^{3}z^{2}cos(y^{4})cos(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=2.2z^{2}e^{(y)} + 2.49x^{2}cos(y^{4})sin(z^{3}) + 2.2x^{3}y^{3}sin(y^{4})sin(z^{3})\end{align*}$

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Example Question #273 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2tan(y^{2})cos(z) - sin(x^{2})e^{(z^{4})}\\&\text{In the direction of }\overrightarrow{v}=(0,-4,8).\end{align*}$

Possible Answers:

${0.81ycos(z)\cdot (tan(y^{2})^{2} + 1) - 1.58tan(y^{2})sin(z) - 3.17z^{3}sin(x^{2})e^{(z^{4})}}$

${- 1.78tan(y^{2})sin(z) - 1.8ycos(z)\cdot (tan(y^{2})^{2} + 1) - 3.56z^{3}sin(x^{2})e^{(z^{4})}}$

${35.8ycos(z)\cdot (tan(y^{2})^{2} + 1) - 17.9tan(y^{2})sin(z) - 17.9xcos(x^{2})e^{(z^{4})} - 35.8z^{3}sin(x^{2})e^{(z^{4})}}$

Correct answer:

${- 1.78tan(y^{2})sin(z) - 1.8ycos(z)\cdot (tan(y^{2})^{2} + 1) - 3.56z^{3}sin(x^{2})e^{(z^{4})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-4)^2+(8)^2}=8.94\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{8.94}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{8.94}=-0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8}{8.94}=0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2xcos(x^{2})e^{(z^{4})})+(-0.45)(4ycos(z)\cdot (tan(y^{2})^{2} + 1))+(0.89)(- 2.0tan(y^{2})sin(z) - 4z^{3}sin(x^{2})e^{(z^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.78tan(y^{2})sin(z) - 1.8ycos(z)\cdot (tan(y^{2})^{2} + 1) - 3.56z^{3}sin(x^{2})e^{(z^{4})}\end{align*}$

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Example Question #3702 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- cos(y^{3})ln(x^{2}) - ln(y^{4})sin(x^{2})e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(20,0,-20).\end{align*}$

Possible Answers:

${20ln(y^{4})sin(x^{2})e^{(z)} -\frac{ (40cos(y^{3}))}{x}- 40xcos(x^{2})ln(y^{4})e^{(z)}}$

${-\frac{(226xcos(x^{2})e^{(z)})}{y}}$

${84.8y^{2}ln(x^{2})sin(y^{3}) -\frac{ (113sin(x^{2})e^{(z)})}{y}-\frac{ (56.6cos(y^{3}))}{x}- 28.3ln(y^{4})sin(x^{2})e^{(z)} - 56.6xcos(x^{2})ln(y^{4})e^{(z)}}$

${0.71ln(y^{4})sin(x^{2})e^{(z)} -\frac{ (1.42cos(y^{3}))}{x}- 1.42xcos(x^{2})ln(y^{4})e^{(z)}}$

Correct answer:

${0.71ln(y^{4})sin(x^{2})e^{(z)} -\frac{ (1.42cos(y^{3}))}{x}- 1.42xcos(x^{2})ln(y^{4})e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(0)^2+(-20)^2}=28.28\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{28.28}=0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{28.28}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{28.28}=-0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.71)(-\frac{ (2cos(y^{3}))}{x}- 2xcos(x^{2})ln(y^{4})e^{(z)})+(0)(3y^{2}ln(x^{2})sin(y^{3}) -\frac{ (4sin(x^{2})e^{(z)})}{y})+(-0.71)(-1ln(y^{4})sin(x^{2})e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.71ln(y^{4})sin(x^{2})e^{(z)} -\frac{ (1.42cos(y^{3}))}{x}- 1.42xcos(x^{2})ln(y^{4})e^{(z)}\end{align*}$

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Example Question #275 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(2^xcos(y^{3}))}{z}\\&\text{In the direction of }\overrightarrow{v}=(0,5,-9).\end{align*}$

Possible Answers:

${\frac{(9\cdot 2^xcos(y^{3}))}{z^{2}}-\frac{ (15\cdot 2^xy^{2}sin(y^{3}))}{z}}$

${\frac{(7.14\cdot 2^xcos(y^{3}))}{z}-\frac{ (10.3\cdot 2^xcos(y^{3}))}{z^{2}}-\frac{ (30.9\cdot 2^xy^{2}sin(y^{3}))}{z}}$

${\frac{(21.4\cdot 2^xy^{2}sin(y^{3}))}{z^{2}}}$

${\frac{(0.87\cdot 2^xcos(y^{3}))}{z^{2}}-\frac{ (1.47\cdot 2^xy^{2}sin(y^{3}))}{z}}$

Correct answer:

${\frac{(0.87\cdot 2^xcos(y^{3}))}{z^{2}}-\frac{ (1.47\cdot 2^xy^{2}sin(y^{3}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(-9)^2}=10.3\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.3}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{10.3}=0.49\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-9}{10.3}=-0.87\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(0.693\cdot 2^xcos(y^{3}))}{z})+(0.49)(-\frac{(3\cdot 2^xy^{2}sin(y^{3}))}{z})+(-0.87)(-\frac{(1\cdot 2^xcos(y^{3}))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.87\cdot 2^xcos(y^{3}))}{z^{2}}-\frac{ (1.47\cdot 2^xy^{2}sin(y^{3}))}{z}\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #1333 : Partial Derivatives

Example Question #1334 : Partial Derivatives

Example Question #3701 : Calculus 3

Example Question #1336 : Partial Derivatives

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Example Question #1331 : Partial Derivatives

Example Question #272 : Directional Derivatives

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Example Question #275 : Directional Derivatives

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