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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #101 : Directional Derivatives

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

Possible Answers:

${210x^2yzcos{(x^3)}sin{(z^2)}}$

${0.69z^2ln{(x)} -\frac{ {(0.06z^3)}}{x} + 1.94ycos{(z^2)}sin{(x^3)} - 0.46y^2zsin{(x^3)}sin{(z^2)} - 0.18x^2y^2cos{(x^3)}cos{(z^2)}}$

${35y^2zsin{(x^3)}sin{(z^2)} - 52.5z^2ln{(x)} - 35ycos{(z^2)}sin{(x^3)} -\frac{ {(17.5z^3)}}{x} - 52.5x^2y^2cos{(x^3)}cos{(z^2)}}$

${12z^2ln{(x)} -\frac{ {(1z^3)}}{x} + 34ycos{(z^2)}sin{(x^3)} - 8y^2zsin{(x^3)}sin{(z^2)} - 3x^2y^2cos{(x^3)}cos{(z^2)}}$

Correct answer:

${0.69z^2ln{(x)} -\frac{ {(0.06z^3)}}{x} + 1.94ycos{(z^2)}sin{(x^3)} - 0.46y^2zsin{(x^3)}sin{(z^2)} - 0.18x^2y^2cos{(x^3)}cos{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(-17)^2+(-4)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{17.49}=0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{17.49}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-4}{17.49}=-0.23\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.06)(-\frac{ {(1z^3)}}{x} - 3x^2y^2cos{(x^3)}cos{(z^2)})+(-0.97)(-2ycos{(z^2)}sin{(x^3)})+(-0.23)(2y^2zsin{(x^3)}sin{(z^2)} - 3z^2ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.69z^2ln{(x)} -\frac{ {(0.06z^3)}}{x} + 1.94ycos{(z^2)}sin{(x^3)} - 0.46y^2zsin{(x^3)}sin{(z^2)} - 0.18x^2y^2cos{(x^3)}cos{(z^2)}\end{align*}$

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Example Question #102 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4cos{(y^2)}sin{(x)} + e^{(y^2)}cos{(x)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-13,-10,12).\end{align*}$

Possible Answers:

${-40.6ye^{(y^2)}cos{(z)}sin{(x)}}$

${81.3cos{(y^2)}cos{(x)} - 163ysin{(y^2)}sin{(x)} + 20.3e^{(y^2)}cos{(x)}cos{(z)} - 20.3e^{(y^2)}sin{(x)}sin{(z)} + 40.6ye^{(y^2)}cos{(x)}sin{(z)}}$

${3.92ysin{(y^2)}sin{(x)} - 2.56cos{(y^2)}cos{(x)} + 0.59e^{(y^2)}cos{(x)}cos{(z)} + 0.64e^{(y^2)}sin{(x)}sin{(z)} - 0.98ye^{(y^2)}cos{(x)}sin{(z)}}$

${80ysin{(y^2)}sin{(x)} - 52cos{(y^2)}cos{(x)} + 12e^{(y^2)}cos{(x)}cos{(z)} + 13e^{(y^2)}sin{(x)}sin{(z)} - 20ye^{(y^2)}cos{(x)}sin{(z)}}$

Correct answer:

${3.92ysin{(y^2)}sin{(x)} - 2.56cos{(y^2)}cos{(x)} + 0.59e^{(y^2)}cos{(x)}cos{(z)} + 0.64e^{(y^2)}sin{(x)}sin{(z)} - 0.98ye^{(y^2)}cos{(x)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-10)^2+(12)^2}=20.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{20.32}=-0.64\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{20.32}=-0.49\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{20.32}=0.59\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.64)(4cos{(y^2)}cos{(x)} - e^{(y^2)}sin{(x)}sin{(z)})+(-0.49)(2ye^{(y^2)}cos{(x)}sin{(z)} - 8ysin{(y^2)}sin{(x)})+(0.59)(e^{(y^2)}cos{(x)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=3.92ysin{(y^2)}sin{(x)} - 2.56cos{(y^2)}cos{(x)} + 0.59e^{(y^2)}cos{(x)}cos{(z)} + 0.64e^{(y^2)}sin{(x)}sin{(z)} - 0.98ye^{(y^2)}cos{(x)}sin{(z)}\end{align*}$

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Example Question #103 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(4y^4ln{(z)})}}{x} - cos{(y^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-15,-15,-17).\end{align*}$

Possible Answers:

${17cos{(y^2)}cos{(z)} - 30ysin{(y^2)}sin{(z)} -\frac{ {(68y^4)}}{{(xz)}} -\frac{ {(240y^3ln{(z)})}}{x} +\frac{ {(60y^4ln{(z)})}}{x^2}}$

${\frac{-{(435y^3)}}{{(x^2z)}}}$

${54.4ysin{(y^2)}sin{(z)} - 27.2cos{(y^2)}cos{(z)} +\frac{ {(109y^4)}}{{(xz)}} +\frac{ {(435y^3ln{(z)})}}{x} -\frac{ {(109y^4ln{(z)})}}{x^2}}$

${0.63cos{(y^2)}cos{(z)} - 1.1ysin{(y^2)}sin{(z)} -\frac{ {(2.52y^4)}}{{(xz)}} -\frac{ {(8.8y^3ln{(z)})}}{x} +\frac{ {(2.2y^4ln{(z)})}}{x^2}}$

Correct answer:

${0.63cos{(y^2)}cos{(z)} - 1.1ysin{(y^2)}sin{(z)} -\frac{ {(2.52y^4)}}{{(xz)}} -\frac{ {(8.8y^3ln{(z)})}}{x} +\frac{ {(2.2y^4ln{(z)})}}{x^2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(-15)^2+(-17)^2}=27.18\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{27.18}=-0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{27.18}=-0.55\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{27.18}=-0.63\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.55)(\frac{-{(4y^4ln{(z)})}}{x^2})+(-0.55)(2ysin{(y^2)}sin{(z)} +\frac{ {(16y^3ln{(z)})}}{x})+(-0.63)(\frac{{(4y^4)}}{{(xz)}} - cos{(y^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.63cos{(y^2)}cos{(z)} - 1.1ysin{(y^2)}sin{(z)} -\frac{ {(2.52y^4)}}{{(xz)}} -\frac{ {(8.8y^3ln{(z)})}}{x} +\frac{ {(2.2y^4ln{(z)})}}{x^2}\end{align*}$

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Example Question #1171 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- yz - cos{(z^2)}sin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-7,-8,-12).\end{align*}$

Possible Answers:

${12y + 8z - 24zsin{(x^3)}sin{(z^2)} + 21x^2cos{(x^3)}cos{(z^2)}}$

${0.75y + 0.5z - 1.5zsin{(x^3)}sin{(z^2)} + 1.32x^2cos{(x^3)}cos{(z^2)}}$

${32.1zsin{(x^3)}sin{(z^2)} - 16z - 16y - 48.1x^2cos{(x^3)}cos{(z^2)}}$

Correct answer:

${0.75y + 0.5z - 1.5zsin{(x^3)}sin{(z^2)} + 1.32x^2cos{(x^3)}cos{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(-8)^2+(-12)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{16.03}=-0.44\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{16.03}=-0.5\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{16.03}=-0.75\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.44)(-3x^2cos{(x^3)}cos{(z^2)})+(-0.5)(-1z)+(-0.75)(2zsin{(x^3)}sin{(z^2)} - y)\\&D_{\overrightarrow{u}}(x,y,z)=0.75y + 0.5z - 1.5zsin{(x^3)}sin{(z^2)} + 1.32x^2cos{(x^3)}cos{(z^2)}\end{align*}$

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Example Question #1172 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=xy^2\\&\text{In the direction of }\overrightarrow{v}=(20,-14,-10).\end{align*}$

Possible Answers:

${20y^2 - 28xy}$

${0.76y^2 - 1.06xy}$

${52.8xy + 26.4y^2}$

{0}

Correct answer:

${0.76y^2 - 1.06xy}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(-14)^2+(-10)^2}=26.38\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{26.38}=0.76\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{26.38}=-0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{26.38}=-0.38\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.76)(y^2)+(-0.53)(2xy)+(-0.38)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.76y^2 - 1.06xy\end{align*}$

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Example Question #1173 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-e^{(x^2)}cos{(z)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(8,-8,1).\end{align*}$

Possible Answers:

${22.7xe^{(x^2)}cos{(y)}sin{(z)}}$

${0.7e^{(x^2)}cos{(y)}cos{(z)} + 0.09e^{(x^2)}sin{(y)}sin{(z)} - 1.4xe^{(x^2)}cos{(z)}sin{(y)}}$

${0.0081e^{(x^2)}sin{(y)}sin{(z)} - 0.49e^{(x^2)}cos{(y)}cos{(z)} - 0.98xe^{(x^2)}cos{(z)}sin{(y)}}$

${11.4e^{(x^2)}sin{(y)}sin{(z)} - 11.4e^{(x^2)}cos{(y)}cos{(z)} - 22.7xe^{(x^2)}cos{(z)}sin{(y)}}$

Correct answer:

${0.7e^{(x^2)}cos{(y)}cos{(z)} + 0.09e^{(x^2)}sin{(y)}sin{(z)} - 1.4xe^{(x^2)}cos{(z)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(-8)^2+(1)^2}=11.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{11.36}=0.7\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{11.36}=-0.7\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{1}{11.36}=0.09\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.7)(-2xe^{(x^2)}cos{(z)}sin{(y)})+(-0.7)(-1e^{(x^2)}cos{(y)}cos{(z)})+(0.09)(e^{(x^2)}sin{(y)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.7e^{(x^2)}cos{(y)}cos{(z)} + 0.09e^{(x^2)}sin{(y)}sin{(z)} - 1.4xe^{(x^2)}cos{(z)}sin{(y)}\end{align*}$

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Example Question #1174 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^4sin{(x^2)}sin{(y)} - 4sin{(x)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-9,-7,6).\end{align*}$

Possible Answers:

${0.292z^4sin{(x^2)}cos{(y)} - 1.17cos{(y)}sin{(x)} - 1.96cos{(x)}sin{(y)} + 0.884z^3sin{(x^2)}sin{(y)} + 0.98xz^4cos{(x^2)}sin{(y)}}$

${12.9z^4sin{(x^2)}cos{(y)} - 51.5cos{(y)}sin{(x)} - 51.5cos{(x)}sin{(y)} + 51.5z^3sin{(x^2)}sin{(y)} + 25.8xz^4cos{(x^2)}sin{(y)}}$

${103xz^3cos{(x^2)}cos{(y)}}$

${2.8cos{(x)}sin{(y)} + 2.16cos{(y)}sin{(x)} - 0.54z^4sin{(x^2)}cos{(y)} + 1.88z^3sin{(x^2)}sin{(y)} - 1.4xz^4cos{(x^2)}sin{(y)}}$

Correct answer:

${2.8cos{(x)}sin{(y)} + 2.16cos{(y)}sin{(x)} - 0.54z^4sin{(x^2)}cos{(y)} + 1.88z^3sin{(x^2)}sin{(y)} - 1.4xz^4cos{(x^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-9)^2+(-7)^2+(6)^2}=12.88\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-9}{12.88}=-0.7\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{12.88}=-0.54\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{12.88}=0.47\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.7)(2xz^4cos{(x^2)}sin{(y)} - 4cos{(x)}sin{(y)})+(-0.54)(z^4sin{(x^2)}cos{(y)} - 4cos{(y)}sin{(x)})+(0.47)(4z^3sin{(x^2)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.8cos{(x)}sin{(y)} + 2.16cos{(y)}sin{(x)} - 0.54z^4sin{(x^2)}cos{(y)} + 1.88z^3sin{(x^2)}sin{(y)} - 1.4xz^4cos{(x^2)}sin{(y)}\end{align*}$

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Example Question #1175 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^x\cdot4^ycos{(z^3)} - 3x^2ln{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-14,7,16).\end{align*}$

Possible Answers:

${0.408\cdot2^x\cdot4^ycos{(z^3)} - 2.38xln{(z^2)} -\frac{ {(3.02x^2)}}{z} - 1.51\cdot2^x\cdot4^yz^2sin{(z^3)}}$

${-64.5\cdot2^x\cdot4^yz^2sin{(z^3)}}$

${46.5\cdot2^x\cdot4^ycos{(z^3)} - 134xln{(z^2)} -\frac{ {(134x^2)}}{z} - 67.1\cdot2^x\cdot4^yz^2sin{(z^3)}}$

${3.78xln{(z^2)} -\frac{ {(4.26x^2)}}{z} - 0.00693\cdot2^x\cdot4^ycos{(z^3)} - 2.13\cdot2^x\cdot4^yz^2sin{(z^3)}}$

Correct answer:

${3.78xln{(z^2)} -\frac{ {(4.26x^2)}}{z} - 0.00693\cdot2^x\cdot4^ycos{(z^3)} - 2.13\cdot2^x\cdot4^yz^2sin{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(7)^2+(16)^2}=22.38\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{22.38}=-0.63\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{22.38}=0.31\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{22.38}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.63)(0.693\cdot2^x\cdot4^ycos{(z^3)} - 6xln{(z^2)})+(0.31)(1.39\cdot2^x\cdot4^ycos{(z^3)})+(0.71)(-\frac{ {(6x^2)}}{z} - 3\cdot2^x\cdot4^yz^2sin{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=3.78xln{(z^2)} -\frac{ {(4.26x^2)}}{z} - 0.00693\cdot2^x\cdot4^ycos{(z^3)} - 2.13\cdot2^x\cdot4^yz^2sin{(z^3)}\end{align*}$

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Example Question #1176 : Partial Derivatives

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

Possible Answers:

{0}

${0.44ysin{(x^3)}sin{(y^2)} - 0.87x^2cos{(x^3)}cos{(y^2)}}$

${6ysin{(x^3)}sin{(y^2)} - 12x^2cos{(x^3)}cos{(y^2)}}$

${41.8x^2cos{(x^3)}cos{(y^2)} - 27.9ysin{(x^3)}sin{(y^2)}}$

Correct answer:

${0.44ysin{(x^3)}sin{(y^2)} - 0.87x^2cos{(x^3)}cos{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-3)^2+(-13)^2}=13.93\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{13.93}=-0.29\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{13.93}=-0.22\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-13}{13.93}=-0.93\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.29)(3x^2cos{(x^3)}cos{(y^2)})+(-0.22)(-2ysin{(x^3)}sin{(y^2)})+(-0.93)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.44ysin{(x^3)}sin{(y^2)} - 0.87x^2cos{(x^3)}cos{(y^2)}\end{align*}$

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Example Question #1177 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ye^{(x)}e^{(z)} - 3y^3sin{(x^2)}ln{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-18,11,7).\end{align*}$

Possible Answers:

${0.49e^{(x)}e^{(z)} -\frac{ {(0.93y^3sin{(x^2)})}}{z} - 4.41y^2sin{(x^2)}ln{(z)} - 0.5ye^{(x)}e^{(z)} + 4.86xy^3cos{(x^2)}ln{(z)}}$

${22.2e^{(x)}e^{(z)} -\frac{ {(66.7y^3sin{(x^2)})}}{z} - 200y^2sin{(x^2)}ln{(z)} + 44.5ye^{(x)}e^{(z)} - 133xy^3cos{(x^2)}ln{(z)}}$

${0.24e^{(x)}e^{(z)} -\frac{ {(0.288y^3sin{(x^2)})}}{z} - 2.16y^2sin{(x^2)}ln{(z)} + 0.752ye^{(x)}e^{(z)} - 3.94xy^3cos{(x^2)}ln{(z)}}$

${22.2e^{(x)}e^{(z)} -\frac{ {(400xy^2cos{(x^2)})}}{z}}$

Correct answer:

${0.49e^{(x)}e^{(z)} -\frac{ {(0.93y^3sin{(x^2)})}}{z} - 4.41y^2sin{(x^2)}ln{(z)} - 0.5ye^{(x)}e^{(z)} + 4.86xy^3cos{(x^2)}ln{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(11)^2+(7)^2}=22.23\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{22.23}=-0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{22.23}=0.49\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{22.23}=0.31\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.81)(ye^{(x)}e^{(z)} - 6xy^3cos{(x^2)}ln{(z)})+(0.49)(e^{(x)}e^{(z)} - 9y^2sin{(x^2)}ln{(z)})+(0.31)(ye^{(x)}e^{(z)} -\frac{ {(3y^3sin{(x^2)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.49e^{(x)}e^{(z)} -\frac{ {(0.93y^3sin{(x^2)})}}{z} - 4.41y^2sin{(x^2)}ln{(z)} - 0.5ye^{(x)}e^{(z)} + 4.86xy^3cos{(x^2)}ln{(z)}\end{align*}$

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Calculus 3 : Directional Derivatives

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