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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1248 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(x^{4})}cos(z^{4})sin(y)\\&\text{In the direction of }\overrightarrow{v}=(-6,0,10).\end{align*}$

Possible Answers:

${- 3.44\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 2.24\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${- 40\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 26.4\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${-205\cdot3^{(x^{4})}x^{3}z^{3}sin(z^{4})cos(y)}$

${11.7\cdot3^{(x^{4})}cos(z^{4})cos(y) - 46.6\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 51.2\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Correct answer:

${- 3.44\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 2.24\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(0)^2+(10)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{11.66}=-0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{11.66}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.51)(4.39\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y))+(0)(3^{(x^{4})}cos(z^{4})cos(y))+(0.86)(-4\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=- 3.44\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 2.24\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)\end{align*}$

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Example Question #1249 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{3})}ln(z^{3})sin(x^{2})\\&\text{In the direction of }\overrightarrow{v}=(10,0,7).\end{align*}$

Possible Answers:

${\frac{(1.71\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 1.64\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

${\frac{(241\cdot3^{(y^{3})}xy^{2}cos(x^{2}))}{z}}$

${\frac{(21\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 20\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

${\frac{(36.6\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 40.2\cdot3^{(y^{3})}y^{2}ln(z^{3})sin(x^{2}) + 24.4\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

Correct answer:

${\frac{(1.71\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 1.64\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(0)^2+(7)^2}=12.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{12.21}=0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{12.21}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{12.21}=0.57\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.82)(2\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3}))+(0)(3.3\cdot3^{(y^{3})}y^{2}ln(z^{3})sin(x^{2}))+(0.57)(\frac{(3\cdot3^{(y^{3})}sin(x^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.71\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 1.64\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})\end{align*}$

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Example Question #1250 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(y^{3})}x^{2}ln(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,13,10).\end{align*}$

Possible Answers:

${\frac{(1.22\cdot4^{(y^{3})}x^{2})}{z}+ 3.29\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

${\frac{(32.8\cdot4^{(y^{3})}x^{2})}{z}+ 32.8\cdot4^{(y^{3})}xln(z^{2}) + 68.2\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

${\frac{(273\cdot4^{(y^{3})}xy^{2})}{z}}$

${\frac{(0.744\cdot4^{(y^{3})}x^{2})}{z}+ 2.6\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

Correct answer:

${\frac{(1.22\cdot4^{(y^{3})}x^{2})}{z}+ 3.29\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(13)^2+(10)^2}=16.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{16.4}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{13}{16.4}=0.79\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{16.4}=0.61\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2\cdot4^{(y^{3})}xln(z^{2}))+(0.79)(4.16\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2}))+(0.61)(\frac{(2\cdot4^{(y^{3})}x^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.22\cdot4^{(y^{3})}x^{2})}{z}+ 3.29\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})\end{align*}$

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Example Question #1251 : Partial Derivatives

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

Possible Answers:

${17y^{2}z^{4}sin(x) - 12y^{2}z^{3}cos(x)}$

${0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)}$

${17.3y^{2}z^{4}sin(x) - 69y^{2}z^{3}cos(x) - 34.5yz^{4}cos(x)}$

${138yz^{3}sin(x)}$

Correct answer:

${0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17)^2+(0)^2+(3)^2}=17.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17}{17.26}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.26}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{17.26}=0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(y^{2}z^{4}sin(x))+(0)(-2yz^{4}cos(x))+(0.17)(-4y^{2}z^{3}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)\end{align*}$

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Example Question #1252 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{4})}e^{(x^{2})}e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(12,14,0).\end{align*}$

Possible Answers:

${0.845\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 2.54\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${18.4\cdot3^{(y^{4})}e^{(x^{2})}e^{(z)} + 36.9\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 81\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${162\cdot3^{(y^{4})}xy^{3}e^{(x^{2})}e^{(z)}}$

Correct answer:

${1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(12)^2+(14)^2+(0)^2}=18.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{12}{18.44}=0.65\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{18.44}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.65)(2\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)})+(0.76)(4.39\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)})+(0)(3^{(y^{4})}e^{(x^{2})}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}\end{align*}$

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Example Question #1253 : Partial Derivatives

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

Possible Answers:

${\frac{(120\cdot2^{(y^{3})}y^{2}z^{2}cos(z^{3}))}{x}}$

${\frac{(19.2\cdot2^{(y^{3})}sin(z^{3}))}{x}+ 57.7\cdot2^{(y^{3})}z^{2}cos(z^{3})ln(x) + 40\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

${-\frac{ (19\cdot2^{(y^{3})}sin(z^{3}))}{x}- 6.24\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

${-\frac{ (0.99\cdot2^{(y^{3})}sin(z^{3}))}{x}- 0.333\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

Correct answer:

${-\frac{ (0.99\cdot2^{(y^{3})}sin(z^{3}))}{x}- 0.333\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(-3)^2+(0)^2}=19.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{19.24}=-0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{19.24}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{19.24}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{(2^{(y^{3})}sin(z^{3}))}{x})+(-0.16)(2.08\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x))+(0)(3\cdot2^{(y^{3})}z^{2}cos(z^{3})ln(x))\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ (0.99\cdot2^{(y^{3})}sin(z^{3}))}{x}- 0.333\cdot2^{(y^{3})}y^{2}sin(z^{3})ln(x)\end{align*}$

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Example Question #181 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{2}ln(y^{3})sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-13,0,9).\end{align*}$

Possible Answers:

${31.6xln(y^{3})sin(z) + 15.8x^{2}ln(y^{3})cos(z) +\frac{ (47.4x^{2}sin(z))}{y}}$

${0.57x^{2}ln(y^{3})cos(z) - 1.64xln(y^{3})sin(z)}$

${\frac{(94.9xcos(z))}{y}}$

${1.34xln(y^{3})sin(z) + 0.325x^{2}ln(y^{3})cos(z)}$

Correct answer:

${0.57x^{2}ln(y^{3})cos(z) - 1.64xln(y^{3})sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(0)^2+(9)^2}=15.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{15.81}=-0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.81}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{15.81}=0.57\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.82)(2xln(y^{3})sin(z))+(0)(\frac{(3x^{2}sin(z))}{y})+(0.57)(x^{2}ln(y^{3})cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.57x^{2}ln(y^{3})cos(z) - 1.64xln(y^{3})sin(z)\end{align*}$

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Example Question #188 : Directional Derivatives

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

Possible Answers:

${\frac{(20.4e^{(y^{3})}e^{(z^{2})})}{x}+ 20.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} + 30.6y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

${0.08zln(x^{2})e^{(y^{3})}e^{(z^{2})} + 2.88y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

${\frac{(122y^{2}ze^{(y^{3})}e^{(z^{2})})}{x}}$

${- 0.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} - 2.94y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

Correct answer:

${- 0.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} - 2.94y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-10)^2+(-2)^2}=10.2\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.2}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{10.2}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{10.2}=-0.2\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(2e^{(y^{3})}e^{(z^{2})})}{x})+(-0.98)(3y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})})+(-0.2)(2zln(x^{2})e^{(y^{3})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.4zln(x^{2})e^{(y^{3})}e^{(z^{2})} - 2.94y^{2}ln(x^{2})e^{(y^{3})}e^{(z^{2})}\end{align*}$

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Example Question #182 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(z^{4})e^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(1,0,16).\end{align*}$

Possible Answers:

${-\frac{(64e^{(y^{3})})}{z}}$

${-\frac{ (64.1e^{(y^{3})})}{z}- 48.1y^{2}ln(z^{4})e^{(y^{3})}}$

{0}

${-\frac{(4e^{(y^{3})})}{z}}$

Correct answer:

${-\frac{(4e^{(y^{3})})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(0)^2+(16)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{16.03}=0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{16.03}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{16.03}=1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.06)(0.0)+(0)(-3y^{2}ln(z^{4})e^{(y^{3})})+(1)(-\frac{(4e^{(y^{3})})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{(4e^{(y^{3})})}{z}\end{align*}$

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Example Question #190 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(y^{2})}ln(z^{3})\\&\text{In the direction of }\overrightarrow{v}=(0,-7,7).\end{align*}$

Possible Answers:

${-\frac{ (29.7\cdot3^{(y^{2})})}{z}- 21.8\cdot3^{(y^{2})}yln(z^{3})}$

${15.4\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (21\cdot3^{(y^{2})})}{z}}$

{0}

${1.56\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (2.13\cdot3^{(y^{2})})}{z}}$

Correct answer:

${1.56\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (2.13\cdot3^{(y^{2})})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-7)^2+(7)^2}=9.9\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.9}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{9.9}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{9.9}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(-0.71)(-2.2\cdot3^{(y^{2})}yln(z^{3}))+(0.71)(-\frac{(3\cdot3^{(y^{2})})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.56\cdot3^{(y^{2})}yln(z^{3}) -\frac{ (2.13\cdot3^{(y^{2})})}{z}\end{align*}$

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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1248 : Partial Derivatives

Example Question #1249 : Partial Derivatives

Example Question #1250 : Partial Derivatives

Example Question #1251 : Partial Derivatives

Example Question #1252 : Partial Derivatives

Example Question #1253 : Partial Derivatives

Example Question #181 : Directional Derivatives

Example Question #188 : Directional Derivatives

Example Question #182 : Directional Derivatives

Example Question #190 : Directional Derivatives

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