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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #41 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=zsin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(17,20,0).\end{align*}$

Possible Answers:

$60y^2zcos{(y^3)}$

$26.2sin{(y^3)} + 78.8y^2zcos{(y^3)}$

$2.28y^2zcos{(y^3)}$

Correct answer:

$2.28y^2zcos{(y^3)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17)^2+(20)^2+(0)^2}=26.25\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17}{26.25}=0.65\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{26.25}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{26.25}=0\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.65)(0.0)+(0.76)(3y^2zcos{(y^3)})+(0)(sin{(y^3)})\\&D_{\overrightarrow{u}}(x,y,z)=2.28y^2zcos{(y^3)}\end{align*}$

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Example Question #42 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^ycos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-2,18,6).\end{align*}$

Possible Answers:

$21\cdot 3^ycos{(x)} - 19.1\cdot 3^ysin{(x)}$

$19.8\cdot 3^ycos{(x)} + 2\cdot 3^ysin{(x)}$

$1.03\cdot 3^ycos{(x)} + 0.1\cdot 3^ysin{(x)}$

Correct answer:

$1.03\cdot 3^ycos{(x)} + 0.1\cdot 3^ysin{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(18)^2+(6)^2}=19.08\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{19.08}=-0.1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{19.08}=0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{19.08}=0.31\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.1)(-1\cdot 3^ysin{(x)})+(0.94)(1.1\cdot 3^ycos{(x)})+(0.31)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=1.03\cdot 3^ycos{(x)} + 0.1\cdot 3^ysin{(x)}\end{align*}$

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Example Question #43 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^4cos{(y^2)}sin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-16,-2,-8).\end{align*}$

Possible Answers:

$0.22yz^4sin{(x^3)}sin{(y^2)} - 1.76z^3cos{(y^2)}sin{(x^3)} - 2.67x^2z^4cos{(x^3)}cos{(y^2)}$

$72z^3cos{(y^2)}sin{(x^3)} - 36yz^4sin{(x^3)}sin{(y^2)} + 54x^2z^4cos{(x^3)}cos{(y^2)}$

$-432x^2yz^3cos{(x^3)}sin{(y^2)}$

$4yz^4sin{(x^3)}sin{(y^2)} - 32z^3cos{(y^2)}sin{(x^3)} - 48x^2z^4cos{(x^3)}cos{(y^2)}$

Correct answer:

$0.22yz^4sin{(x^3)}sin{(y^2)} - 1.76z^3cos{(y^2)}sin{(x^3)} - 2.67x^2z^4cos{(x^3)}cos{(y^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(-2)^2+(-8)^2}=18\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{18}=-0.89\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{18}=-0.11\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{18}=-0.44\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.89)(3x^2z^4cos{(x^3)}cos{(y^2)})+(-0.11)(-2yz^4sin{(x^3)}sin{(y^2)})+(-0.44)(4z^3cos{(y^2)}sin{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=0.22yz^4sin{(x^3)}sin{(y^2)} - 1.76z^3cos{(y^2)}sin{(x^3)} - 2.67x^2z^4cos{(x^3)}cos{(y^2)}\end{align*}$

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Example Question #44 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^4zsin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-8,-3,-16).\end{align*}$

Possible Answers:

$18.1y^4sin{(x)} + 18.1y^4zcos{(x)} + 72.6y^3zsin{(x)}$

$72.6y^3cos{(x)}$

$- 16y^4sin{(x)} - 8y^4zcos{(x)} - 12y^3zsin{(x)}$

$- 0.88y^4sin{(x)} - 0.44y^4zcos{(x)} - 0.68y^3zsin{(x)}$

Correct answer:

$- 0.88y^4sin{(x)} - 0.44y^4zcos{(x)} - 0.68y^3zsin{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(-3)^2+(-16)^2}=18.14\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{18.14}=-0.44\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{18.14}=-0.17\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{18.14}=-0.88\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.44)(y^4zcos{(x)})+(-0.17)(4y^3zsin{(x)})+(-0.88)(y^4sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.88y^4sin{(x)} - 0.44y^4zcos{(x)} - 0.68y^3zsin{(x)}\end{align*}$

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Example Question #1112 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^xsin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-10,-13,-15).\end{align*}$

Possible Answers:

$- 15\cdot 2^xcos{(z)} - 6.93\cdot 2^xsin{(z)}$

$22.2\cdot 2^xcos{(z)} + 15.4\cdot 2^xsin{(z)}$

$- 0.67\cdot 2^xcos{(z)} - 0.312\cdot 2^xsin{(z)}$

Correct answer:

$- 0.67\cdot 2^xcos{(z)} - 0.312\cdot 2^xsin{(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-10)^2+(-13)^2+(-15)^2}=22.23\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-10}{22.23}=-0.45\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{22.23}=-0.58\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-15}{22.23}=-0.67\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.45)(0.693\cdot 2^xsin{(z)})+(-0.58)(0.0)+(-0.67)(2^xcos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.67\cdot 2^xcos{(z)} - 0.312\cdot 2^xsin{(z)}\end{align*}$

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Example Question #1113 : Partial Derivatives

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

Possible Answers:

$108xycos{(y^2)}cos{(z)}$

$6zcos{(y)} - 38sin{(y)} + 38xsin{(y^2)}sin{(z)} + 19x^2sin{(y^2)}cos{(z)} - 6x^2ycos{(y^2)}sin{(z)}$

$54.1xsin{(y^2)}sin{(z)} - 54.1zcos{(y)} - 54.1sin{(y)} + 27x^2sin{(y^2)}cos{(z)} + 54.1x^2ycos{(y^2)}sin{(z)}$

$0.22zcos{(y)} - 1.4sin{(y)} + 1.4xsin{(y^2)}sin{(z)} + 0.7x^2sin{(y^2)}cos{(z)} - 0.22x^2ycos{(y^2)}sin{(z)}$

Correct answer:

$0.22zcos{(y)} - 1.4sin{(y)} + 1.4xsin{(y^2)}sin{(z)} + 0.7x^2sin{(y^2)}cos{(z)} - 0.22x^2ycos{(y^2)}sin{(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(19)^2+(-3)^2+(19)^2}=27.04\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{19}{27.04}=0.7\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{27.04}=-0.11\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{27.04}=0.7\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.7)(2xsin{(y^2)}sin{(z)})+(-0.11)(2x^2ycos{(y^2)}sin{(z)} - 2zcos{(y)})+(0.7)(x^2sin{(y^2)}cos{(z)} - 2sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=0.22zcos{(y)} - 1.4sin{(y)} + 1.4xsin{(y^2)}sin{(z)} + 0.7x^2sin{(y^2)}cos{(z)} - 0.22x^2ycos{(y^2)}sin{(z)}\end{align*}$

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Example Question #1114 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4x^2y^4e^{(z)} - cos{(x^3)}cos{(y^2)}sin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(-4,-7,4).\end{align*}$

Possible Answers:

${1.76x^2y^4e^{(z)} - 12.5x^2y^3e^{(z)} - 3.52xy^4e^{(z)} - 1.32x^2cos{(y^2)}sin{(x^3)}sin{(z^3)} - 1.56ycos{(x^3)}sin{(y^2)}sin{(z^3)} - 1.32z^2cos{(x^3)}cos{(y^2)}cos{(z^3)} }$

$288xy^3e^{(z)} - 162x^2yz^2cos{(z^3)}sin{(x^3)}sin{(y^2)}$

${16x^2y^4e^{(z)} - 112x^2y^3e^{(z)} - 32xy^4e^{(z)} - 12x^2cos{(y^2)}sin{(x^3)}sin{(z^3)} - 14ycos{(x^3)}sin{(y^2)}sin{(z^3)} - 12z^2cos{(x^3)}cos{(y^2)}cos{(z^3)} }$

$\begin{align*}&72xy^4e^{(z)} + 144x^2y^3e^{(z)} + 36x^2y^4e^{(z)} + 27x^2cos{(y^2)}sin{(x^3)}sin{(z^3)} + 18ycos{(x^3)}sin{(y^2)}sin{(z^3)} - 27z^2cos{(x^3)}cos{(y^2)}cos{(z^3)}\end{align*}$

Correct answer:

${1.76x^2y^4e^{(z)} - 12.5x^2y^3e^{(z)} - 3.52xy^4e^{(z)} - 1.32x^2cos{(y^2)}sin{(x^3)}sin{(z^3)} - 1.56ycos{(x^3)}sin{(y^2)}sin{(z^3)} - 1.32z^2cos{(x^3)}cos{(y^2)}cos{(z^3)} }$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-7)^2+(4)^2}=9\\&\text{The unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{9}=-0.44\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{9}=-0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{9}=0.44\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.44)(8xy^4e^{(z)} + 3x^2cos{(y^2)}sin{(x^3)}sin{(z^3)})+(-0.78)(16x^2y^3e^{(z)} + 2ycos{(x^3)}sin{(y^2)}sin{(z^3)})+(0.44)(4x^2y^4e^{(z)} - 3z^2cos{(x^3)}cos{(y^2)}cos{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.76x^2y^4e^{(z)} - 12.5x^2y^3e^{(z)} - 3.52xy^4e^{(z)} - 1.32x^2cos{(y^2)}sin{(x^3)}sin{(z^3)} - 1.56ycos{(x^3)}sin{(y^2)}sin{(z^3)} - 1.32z^2cos{(x^3)}cos{(y^2)}cos{(z^3)}\end{align*}$

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Example Question #1115 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)={2^{(y^2)}x^4sin{(z^2)} - 3yz^4}\\&\text{In the direction of }\overrightarrow{v}=(-7,-4,2).\end{align*}$

Possible Answers:

${1.44z^4 - 2.88yz^3 - 3.36\cdot 2^{(y^2)}x^3sin{(z^2)} + 0.48\cdot 2^{(y^2)}x^4zcos{(z^2)} - 0.665\cdot 2^{(y^2)}x^4ysin{(z^2)}}$

${92.2\cdot 2^{(y^2)}x^3yzcos{(z^2)}}$

${12z^4 - 24yz^3 - 28\cdot 2^{(y^2)}x^3sin{(z^2)} + 4\cdot 2^{(y^2)}x^4zcos{(z^2)} - 5.55\cdot 2^{(y^2)}x^4ysin{(z^2)}}$

${33.2\cdot 2^{(y^2)}x^3sin{(z^2)} - 24.9z^4 - 99.7yz^3 + 16.6\cdot 2^{(y^2)}x^4zcos{(z^2)} + 11.5\cdot 2^{(y^2)}x^4ysin{(z^2)}}$

Correct answer:

${1.44z^4 - 2.88yz^3 - 3.36\cdot 2^{(y^2)}x^3sin{(z^2)} + 0.48\cdot 2^{(y^2)}x^4zcos{(z^2)} - 0.665\cdot 2^{(y^2)}x^4ysin{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(-4)^2+(2)^2}=8.31\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{8.31}=-0.84\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{8.31}=-0.48\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{8.31}=0.24\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.84)({4\cdot 2^{(y^2)}x^3sin{(z^2)}})+(-0.48)({1.39\cdot 2^{(y^2)}x^4ysin{(z^2)} - 3z^4})+(0.24)({2\cdot 2^{(y^2)}x^4zcos{(z^2)} - 12yz^3})\\&D_{\overrightarrow{u}}(x,y,z)={1.44z^4 - 2.88yz^3 - 3.36\cdot 2^{(y^2)}x^3sin{(z^2)} + 0.48\cdot 2^{(y^2)}x^4zcos{(z^2)} - 0.665\cdot 2^{(y^2)}x^4ysin{(z^2)}}\end{align*}$

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Example Question #1116 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)={x^2y^2z}\\&\text{In the direction of }\overrightarrow{v}=(-13,-20,-8).\end{align*}$

Possible Answers:

${- 8x^2y^2 - 26xy^2z - 40x^2yz}$

${- 0.32x^2y^2 - 1.04xy^2z - 1.58x^2yz}$

101xy

${25.2x^2y^2 + 50.3xy^2z + 50.3x^2yz}$

Correct answer:

${- 0.32x^2y^2 - 1.04xy^2z - 1.58x^2yz}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-20)^2+(-8)^2}=25.16\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{25.16}=-0.52\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{25.16}=-0.79\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{25.16}=-0.32\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.52)({2xy^2z})+(-0.79)({2x^2yz})+(-0.32)({x^2y^2})\\&D_{\overrightarrow{u}}(x,y,z)={- 0.32x^2y^2 - 1.04xy^2z - 1.58x^2yz}\end{align*}$

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Example Question #1117 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)={y^3sin{(x^2)}sin{(z^3)} - 4cos{(y^3)}cos{(z^2)}}\\&\text{In the direction of }\overrightarrow{v}=(0,-6,-17).\end{align*}$

Possible Answers:

${- 136zcos{(y^3)}sin{(z^2)} - 72y^2cos{(z^2)}sin{(y^3)} - 18y^2sin{(x^2)}sin{(z^3)} - 51y^3z^2cos{(z^3)}sin{(x^2)}}$

${144zcos{(y^3)}sin{(z^2)} + 216y^2cos{(z^2)}sin{(y^3)} + 54.1y^2sin{(x^2)}sin{(z^3)} + 36.1xy^3cos{(x^2)}sin{(z^3)} + 54.1y^3z^2cos{(z^3)}sin{(x^2)}}$

${- 7.52zcos{(y^3)}sin{(z^2)} - 3.96y^2cos{(z^2)}sin{(y^3)} - 0.99y^2sin{(x^2)}sin{(z^3)} - 2.82y^3z^2cos{(z^3)}sin{(x^2)}}$

${325xy^2z^2cos{(x^2)}cos{(z^3)}}$

Correct answer:

${- 7.52zcos{(y^3)}sin{(z^2)} - 3.96y^2cos{(z^2)}sin{(y^3)} - 0.99y^2sin{(x^2)}sin{(z^3)} - 2.82y^3z^2cos{(z^3)}sin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-6)^2+(-17)^2}=18.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.03}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-6}{18.03}=-0.33\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{18.03}=-0.94\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)({2xy^3cos{(x^2)}sin{(z^3)}})+(-0.33)({12y^2cos{(z^2)}sin{(y^3)} + 3y^2sin{(x^2)}sin{(z^3)}})+(-0.94)({8zcos{(y^3)}sin{(z^2)} + 3y^3z^2cos{(z^3)}sin{(x^2)}})\\&D_{\overrightarrow{u}}(x,y,z)={- 7.52zcos{(y^3)}sin{(z^2)} - 3.96y^2cos{(z^2)}sin{(y^3)} - 0.99y^2sin{(x^2)}sin{(z^3)} - 2.82y^3z^2cos{(z^3)}sin{(x^2)}}\end{align*}$

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Calculus 3 : Directional Derivatives

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