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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

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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Example Question #1118 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3ze^{(x^2)}e^{(y^2)} - cos{(y^2)}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,-1,12).\end{align*}$

Possible Answers:

$36e^{(x^2)}e^{(y^2)} - 2ysin{(y^2)}e^{(x)} - 6yze^{(x^2)}e^{(y^2)}$

$36.1e^{(x^2)}e^{(y^2)} - 12cos{(y^2)}e^{(x)} + 24.1ysin{(y^2)}e^{(x)} + 72.2xze^{(x^2)}e^{(y^2)} + 72.2yze^{(x^2)}e^{(y^2)}$

$144xye^{(x^2)}e^{(y^2)}$

$3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}$

Correct answer:

$3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-1)^2+(12)^2}=12.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.04}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-1}{12.04}=-0.08\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{12.04}=1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(6xze^{(x^2)}e^{(y^2)} -cos{(y^2)}e^{(x)})+(-0.08)(2ysin{(y^2)}e^{(x)} + 6yze^{(x^2)}e^{(y^2)})+(1)(3e^{(x^2)}e^{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}\end{align*}$

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Example Question #1119 : Partial Derivatives

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

Possible Answers:

$-\frac{ {(19cos{(x)})}}{{(yz)}} - 105\cdot2^{(x^2)}xy^2z^3sin{(y^3)}$

$\frac{{(6ln{(z)}sin{(x)})}}{y} -\frac{ {(18ln{(y)}sin{(x)})}}{z} + 24\cdot2^{(x^2)}z^3cos{(y^3)} + 6\cdot2^{(x^2)}y^2z^4sin{(y^3)}$

${25.3\cdot2^{(x^2)}z^3cos{(y^3)} -\frac{ {(19ln{(z)}sin{(x)})}}{y} -\frac{ {(19ln{(y)}sin{(x)})}}{z} - 19cos{(x)}ln{(y)}ln{(z)} + 8.76\cdot2^{(x^2)}xz^4cos{(y^3)} - 19\cdot2^{(x^2)}y^2z^4sin{(y^3)}}$

$\frac{{(0.96ln{(z)}sin{(x)})}}{y} -\frac{ {(2.85ln{(y)}sin{(x)})}}{z} + 3.8\cdot2^{(x^2)}z^3cos{(y^3)} + 0.96\cdot2^{(x^2)}y^2z^4sin{(y^3)}$

Correct answer:

$\frac{{(0.96ln{(z)}sin{(x)})}}{y} -\frac{ {(2.85ln{(y)}sin{(x)})}}{z} + 3.8\cdot2^{(x^2)}z^3cos{(y^3)} + 0.96\cdot2^{(x^2)}y^2z^4sin{(y^3)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-2)^2+(6)^2}=6.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{6.32}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{6.32}=-0.32\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{6.32}=0.95\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(1.39\cdot2^{(x^2)}xz^4cos{(y^3)} - 3cos{(x)}ln{(y)}ln{(z)})+(-0.32)(-\frac{ {(3ln{(z)}sin{(x)})}}{y} - 3\cdot2^{(x^2)}y^2z^4sin{(y^3)})+(0.95)(4\cdot2^{(x^2)}z^3cos{(y^3)} -\frac{ {(3ln{(y)}sin{(x)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.96ln{(z)}sin{(x)})}}{y} -\frac{ {(2.85ln{(y)}sin{(x)})}}{z} + 3.8\cdot2^{(x^2)}z^3cos{(y^3)} + 0.96\cdot2^{(x^2)}y^2z^4sin{(y^3)}\end{align*}$

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Example Question #1120 : Partial Derivatives

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

Possible Answers:

${120y^3e^{(x^2)}cos{(z)} - 60xy^4e^{(x^2)}cos{(z)} + 20xzcos{(y^2)}sin{(x^2)} - 20yzcos{(x^2)}sin{(y^2)}}$

${8.52y^3e^{(x^2)}cos{(z)} - 4.26xy^4e^{(x^2)}cos{(z)} + 1.42xzcos{(y^2)}sin{(x^2)} - 1.42yzcos{(x^2)}sin{(y^2)}}$

${42.4y^4e^{(x^2)}sin{(z)} - 170y^3e^{(x^2)}cos{(z)} - 14.1cos{(x^2)}cos{(y^2)} - 84.8xy^4e^{(x^2)}cos{(z)} + 28.3xzcos{(y^2)}sin{(x^2)} + 28.3yzcos{(x^2)}sin{(y^2)}}$

${339xy^3e^{(x^2)}sin{(z)} - 56.6xysin{(x^2)}sin{(y^2)}}$

Correct answer:

${8.52y^3e^{(x^2)}cos{(z)} - 4.26xy^4e^{(x^2)}cos{(z)} + 1.42xzcos{(y^2)}sin{(x^2)} - 1.42yzcos{(x^2)}sin{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(-10)^2+(0)^2}=14.14\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{14.14}=0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{14.14}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.14}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.71)(2xzcos{(y^2)}sin{(x^2)} - 6xy^4e^{(x^2)}cos{(z)})+(-0.71)(2yzcos{(x^2)}sin{(y^2)} - 12y^3e^{(x^2)}cos{(z)})+(0)(3y^4e^{(x^2)}sin{(z)} -cos{(x^2)}cos{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=8.52y^3e^{(x^2)}cos{(z)} - 4.26xy^4e^{(x^2)}cos{(z)} + 1.42xzcos{(y^2)}sin{(x^2)} - 1.42yzcos{(x^2)}sin{(y^2)}\end{align*}$

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Example Question #51 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5x^3z^2ln{(y^2)} - yzsin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-6,0,-10).\end{align*}$

Possible Answers:

${\frac{{(700x^2z)}}{y} - 35x^2cos{(x^3)}}$

${10ysin{(x^3)} - 90x^2z^2ln{(y^2)} - 100x^3zln{(y^2)} + 18x^2yzcos{(x^3)}}$

${175x^2z^2ln{(y^2)} - 11.7zsin{(x^3)} - 11.7ysin{(x^3)} +\frac{ {(117x^3z^2)}}{y} + 117x^3zln{(y^2)} - 35x^2yzcos{(x^3)}}$

${0.86ysin{(x^3)} - 7.65x^2z^2ln{(y^2)} - 8.6x^3zln{(y^2)} + 1.53x^2yzcos{(x^3)}}$

Correct answer:

${0.86ysin{(x^3)} - 7.65x^2z^2ln{(y^2)} - 8.6x^3zln{(y^2)} + 1.53x^2yzcos{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(0)^2+(-10)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{11.66}=-0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{11.66}=-0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.51)(15x^2z^2ln{(y^2)} - 3x^2yzcos{(x^3)})+(0)(\frac{{(10x^3z^2)}}{y} -zsin{(x^3)})+(-0.86)(10x^3zln{(y^2)} -ysin{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86ysin{(x^3)} - 7.65x^2z^2ln{(y^2)} - 8.6x^3zln{(y^2)} + 1.53x^2yzcos{(x^3)}\end{align*}$

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Example Question #52 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4cos{(z)}ln{(x)}sin{(y)} - sin{(y^3)}e^{(z)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-18,1,0).\end{align*}$

Possible Answers:

${\frac{{(72cos{(z)}sin{(y)})}}{x} - 18sin{(y^3)}e^{(z)}sin{(x)} - 4cos{(y)}cos{(z)}ln{(x)} - 3y^2cos{(y^3)}e^{(z)}cos{(x)}}$

${\frac{{(72.1cos{(y)}sin{(z)})}}{x} + 54.1y^2cos{(y^3)}e^{(z)}sin{(x)}}$

${72.1ln{(x)}sin{(y)}sin{(z)} -\frac{ {(72.1cos{(z)}sin{(y)})}}{x} - 18sin{(y^3)}e^{(z)}cos{(x)} + 18sin{(y^3)}e^{(z)}sin{(x)} - 72.1cos{(y)}cos{(z)}ln{(x)} - 54.1y^2cos{(y^3)}e^{(z)}cos{(x)}}$

${\frac{{(4cos{(z)}sin{(y)})}}{x} -sin{(y^3)}e^{(z)}sin{(x)} - 0.24cos{(y)}cos{(z)}ln{(x)} - 0.18y^2cos{(y^3)}e^{(z)}cos{(x)}}$

Correct answer:

${\frac{{(4cos{(z)}sin{(y)})}}{x} -sin{(y^3)}e^{(z)}sin{(x)} - 0.24cos{(y)}cos{(z)}ln{(x)} - 0.18y^2cos{(y^3)}e^{(z)}cos{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(1)^2+(0)^2}=18.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{18.03}=-1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{18.03}=0.06\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-1)(sin{(y^3)}e^{(z)}sin{(x)} -\frac{ {(4cos{(z)}sin{(y)})}}{x})+(0.06)(- 4cos{(y)}cos{(z)}ln{(x)} - 3y^2cos{(y^3)}e^{(z)}cos{(x)})+(0)(4ln{(x)}sin{(y)}sin{(z)} -sin{(y^3)}e^{(z)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(4cos{(z)}sin{(y)})}}{x} -sin{(y^3)}e^{(z)}sin{(x)} - 0.24cos{(y)}cos{(z)}ln{(x)} - 0.18y^2cos{(y^3)}e^{(z)}cos{(x)}\end{align*}$

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