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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #191 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(x^{2})}\cdot3^ze^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(19,0,16).\end{align*}$

Possible Answers:

${27.3\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 54.6\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})} + 49.7\cdot3^z\cdot3^{(x^{2})}ye^{(y^{2})}}$

${0.703\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 1.67\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}}$

${120\cdot3^z\cdot3^{(x^{2})}xye^{(y^{2})}}$

${17.6\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 41.7\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}}$

Correct answer:

${0.703\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 1.67\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(19)^2+(0)^2+(16)^2}=24.84\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{19}{24.84}=0.76\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.84}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{24.84}=0.64\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.76)(2.2\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})})+(0)(2\cdot3^z\cdot3^{(x^{2})}ye^{(y^{2})})+(0.64)(1.1\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=0.703\cdot3^z\cdot3^{(x^{2})}e^{(y^{2})} + 1.67\cdot3^z\cdot3^{(x^{2})}xe^{(y^{2})}\end{align*}$

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Example Question #192 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^{(z^{4})}e^{(y^{3})}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,9,-11).\end{align*}$

Possible Answers:

${-236\cdot4^{(z^{4})}y^{2}z^{3}e^{(y^{3})}e^{(x)}}$

${- 1.19\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)} - 3.29\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)}}$

${- 14.2\cdot4^{(z^{4})}e^{(y^{3})}e^{(x)} - 42.6\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)} - 78.8\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)}}$

${4.27\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)} - 1.89\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)}}$

Correct answer:

${4.27\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)} - 1.89\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(-11)^2}=14.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{14.21}=0.63\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{14.21}=-0.77\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1\cdot4^{(z^{4})}e^{(y^{3})}e^{(x)})+(0.63)(-3\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)})+(-0.77)(-5.55\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=4.27\cdot4^{(z^{4})}z^{3}e^{(y^{3})}e^{(x)} - 1.89\cdot4^{(z^{4})}y^{2}e^{(y^{3})}e^{(x)}\end{align*}$

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Example Question #193 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{2}e^{(y^{4})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,16,-16).\end{align*}$

Possible Answers:

${1.01x^{2}ze^{(y^{4})}e^{(z^{2})} + 2.02x^{2}y^{3}e^{(y^{4})}e^{(z^{2})}}$

${45.3xe^{(y^{4})}e^{(z^{2})} + 45.3x^{2}ze^{(y^{4})}e^{(z^{2})} + 90.5x^{2}y^{3}e^{(y^{4})}e^{(z^{2})}}$

${2.84x^{2}y^{3}e^{(y^{4})}e^{(z^{2})} - 1.42x^{2}ze^{(y^{4})}e^{(z^{2})}}$

${362xy^{3}ze^{(y^{4})}e^{(z^{2})}}$

Correct answer:

${2.84x^{2}y^{3}e^{(y^{4})}e^{(z^{2})} - 1.42x^{2}ze^{(y^{4})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(16)^2+(-16)^2}=22.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.63}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{22.63}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{22.63}=-0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xe^{(y^{4})}e^{(z^{2})})+(0.71)(4x^{2}y^{3}e^{(y^{4})}e^{(z^{2})})+(-0.71)(2x^{2}ze^{(y^{4})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=2.84x^{2}y^{3}e^{(y^{4})}e^{(z^{2})} - 1.42x^{2}ze^{(y^{4})}e^{(z^{2})}\end{align*}$

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Example Question #194 : Directional Derivatives

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

Possible Answers:

${- 0.45y^{2}cos(z^{3})sin(x^{3}) - 2.97y^{3}z^{2}sin(x^{3})sin(z^{3})}$

${60.7y^{2}cos(z^{3})sin(x^{3}) + 60.7x^{2}y^{3}cos(x^{3})cos(z^{3}) - 60.7y^{3}z^{2}sin(x^{3})sin(z^{3})}$

${0.0675y^{2}cos(z^{3})sin(x^{3}) - 2.94y^{3}z^{2}sin(x^{3})sin(z^{3})}$

${-546x^{2}y^{2}z^{2}cos(x^{3})sin(z^{3})}$

Correct answer:

${- 0.45y^{2}cos(z^{3})sin(x^{3}) - 2.97y^{3}z^{2}sin(x^{3})sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-3)^2+(20)^2}=20.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20.22}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{20.22}=-0.15\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{20.22}=0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3x^{2}y^{3}cos(x^{3})cos(z^{3}))+(-0.15)(3y^{2}cos(z^{3})sin(x^{3}))+(0.99)(-3y^{3}z^{2}sin(x^{3})sin(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=- 0.45y^{2}cos(z^{3})sin(x^{3}) - 2.97y^{3}z^{2}sin(x^{3})sin(z^{3})\end{align*}$

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Example Question #195 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^y\cdot 2^zcos(x)\\&\text{In the direction of }\overrightarrow{v}=(-15,0,-14).\end{align*}$

Possible Answers:

${-9.86\cdot 2^y\cdot 2^zsin(x)}$

${28.4\cdot 2^y\cdot 2^zcos(x) - 20.5\cdot 2^y\cdot 2^zsin(x)}$

${0.321\cdot 2^y\cdot 2^zcos(x) - 0.533\cdot 2^y\cdot 2^zsin(x)}$

${0.73\cdot 2^y\cdot 2^zsin(x) - 0.471\cdot 2^y\cdot 2^zcos(x)}$

Correct answer:

${0.73\cdot 2^y\cdot 2^zsin(x) - 0.471\cdot 2^y\cdot 2^zcos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(0)^2+(-14)^2}=20.52\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{20.52}=-0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.52}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{20.52}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.73)(-1\cdot 2^y\cdot 2^zsin(x))+(0)(0.693\cdot 2^y\cdot 2^zcos(x))+(-0.68)(0.693\cdot 2^y\cdot 2^zcos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.73\cdot 2^y\cdot 2^zsin(x) - 0.471\cdot 2^y\cdot 2^zcos(x)\end{align*}$

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Example Question #196 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(z^{2})}e^{(x)}ln(y)\\&\text{In the direction of }\overrightarrow{v}=(3,0,13).\end{align*}$

Possible Answers:

${-\frac{ (13.3\cdot 3^{(z^{2})}e^{(x)})}{y}- 13.3\cdot 3^{(z^{2})}e^{(x)}ln(y) - 29.3\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

${-\frac{(29.3\cdot 3^{(z^{2})}ze^{(x)})}{y}}$

${- 0.22\cdot 3^{(z^{2})}e^{(x)}ln(y) - 2.13\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

${- 3\cdot 3^{(z^{2})}e^{(x)}ln(y) - 28.6\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

Correct answer:

${- 0.22\cdot 3^{(z^{2})}e^{(x)}ln(y) - 2.13\cdot 3^{(z^{2})}ze^{(x)}ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(3)^2+(0)^2+(13)^2}=13.34\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{3}{13.34}=0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{13.34}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{13}{13.34}=0.97\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.22)(-1\cdot 3^{(z^{2})}e^{(x)}ln(y))+(0)(-\frac{(1\cdot 3^{(z^{2})}e^{(x)})}{y})+(0.97)(-2.2\cdot 3^{(z^{2})}ze^{(x)}ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=- 0.22\cdot 3^{(z^{2})}e^{(x)}ln(y) - 2.13\cdot 3^{(z^{2})}ze^{(x)}ln(y)\end{align*}$

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Example Question #197 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(x^{3})}y^{4}z^{2}\\&\text{In the direction of }\overrightarrow{v}=(-19,0,-8).\end{align*}$

Possible Answers:

${0.78\cdot 2^{(x^{3})}y^{4}z + 1.91\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

${-343\cdot 2^{(x^{3})}x^{2}y^{3}z}$

${- 0.304\cdot 2^{(x^{3})}y^{4}z - 1.76\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

${- 41.2\cdot 2^{(x^{3})}y^{4}z - 82.5\cdot 2^{(x^{3})}y^{3}z^{2} - 42.9\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

Correct answer:

${0.78\cdot 2^{(x^{3})}y^{4}z + 1.91\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(0)^2+(-8)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{20.62}=-0.92\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{20.62}=-0.39\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.92)(-2.08\cdot 2^{(x^{3})}x^{2}y^{4}z^{2})+(0)(-4\cdot 2^{(x^{3})}y^{3}z^{2})+(-0.39)(-2\cdot 2^{(x^{3})}y^{4}z)\\&D_{\overrightarrow{u}}(x,y,z)=0.78\cdot 2^{(x^{3})}y^{4}z + 1.91\cdot 2^{(x^{3})}x^{2}y^{4}z^{2}\end{align*}$

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Example Question #198 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(x^{3})ln(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(14,0,-20).\end{align*}$

Possible Answers:

${-\frac{(1.71ln(y^{4}))}{x}}$

${-\frac{ (97.6ln(x^{3}))}{y}-\frac{ (73.2ln(y^{4}))}{x}}$

{0}

${-\frac{(0.975ln(y^{4}))}{x}}$

Correct answer:

${-\frac{(1.71ln(y^{4}))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(0)^2+(-20)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{24.41}=0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{24.41}=-0.82\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.57)(-\frac{(3ln(y^{4}))}{x})+(0)(-\frac{(4ln(x^{3}))}{y})+(-0.82)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{(1.71ln(y^{4}))}{x}\end{align*}$

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Example Question #199 : Directional Derivatives

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

Possible Answers:

${42.4xz^{2}sin(z^{3})e^{(x^{2})}}$

${- 0.504cos(z^{3})e^{(x^{2})} - 1.01xycos(z^{3})e^{(x^{2})}}$

${0.71cos(z^{3})e^{(x^{2})} + 1.42xycos(z^{3})e^{(x^{2})}}$

${21.2yz^{2}sin(z^{3})e^{(x^{2})} - 7.07cos(z^{3})e^{(x^{2})} - 14.1xycos(z^{3})e^{(x^{2})}}$

Correct answer:

${0.71cos(z^{3})e^{(x^{2})} + 1.42xycos(z^{3})e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(-5)^2+(0)^2}=7.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{7.07}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{7.07}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{7.07}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-2xycos(z^{3})e^{(x^{2})})+(-0.71)(-1cos(z^{3})e^{(x^{2})})+(0)(3yz^{2}sin(z^{3})e^{(x^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=0.71cos(z^{3})e^{(x^{2})} + 1.42xycos(z^{3})e^{(x^{2})}\end{align*}$

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Example Question #200 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^{3})}x^{2}y^{4}\\&\text{In the direction of }\overrightarrow{v}=(13,0,12).\end{align*}$

Possible Answers:

${1.46\cdot 4^{(z^{3})}xy^{4} + 2.83\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

${26\cdot 4^{(z^{3})}xy^{4} + 49.9\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

${589\cdot 4^{(z^{3})}xy^{3}z^{2}}$

${35.4\cdot 4^{(z^{3})}xy^{4} + 70.8\cdot 4^{(z^{3})}x^{2}y^{3} + 73.6\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

Correct answer:

${1.46\cdot 4^{(z^{3})}xy^{4} + 2.83\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(12)^2}=17.69\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{17.69}=0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.69}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{17.69}=0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.73)(2\cdot 4^{(z^{3})}xy^{4})+(0)(4\cdot 4^{(z^{3})}x^{2}y^{3})+(0.68)(4.16\cdot 4^{(z^{3})}x^{2}y^{4}z^{2})\\&D_{\overrightarrow{u}}(x,y,z)=1.46\cdot 4^{(z^{3})}xy^{4} + 2.83\cdot 4^{(z^{3})}x^{2}y^{4}z^{2}\end{align*}$

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Calculus 3 : Directional Derivatives

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Example Question #191 : Directional Derivatives

Example Question #192 : Directional Derivatives

Example Question #193 : Directional Derivatives

Example Question #194 : Directional Derivatives

Example Question #195 : Directional Derivatives

Example Question #196 : Directional Derivatives

Example Question #197 : Directional Derivatives

Example Question #198 : Directional Derivatives

Example Question #199 : Directional Derivatives

Example Question #200 : Directional Derivatives

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