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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1401 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4xcos(z^{2}) + cos(z^{2})e^{(x^{4})}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(-11.32,-2.46,7.54).\end{align*}$

Possible Answers:

${-111x^{3}zsin(z^{2})e^{(x^{4})}cos(y)}$

${2.69cos(z^{2}) - 2.42xzsin(z^{2}) + 0.0324cos(z^{2})e^{(x^{4})}cos(y) - 0.605zsin(z^{2})e^{(x^{4})}sin(y) + 2.69x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

${55.3cos(z^{2}) - 111xzsin(z^{2}) + 13.8cos(z^{2})e^{(x^{4})}cos(y) - 27.6zsin(z^{2})e^{(x^{4})}sin(y) + 55.3x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

${- 3.28cos(z^{2}) - 4.4xzsin(z^{2}) - 0.18cos(z^{2})e^{(x^{4})}cos(y) - 1.1zsin(z^{2})e^{(x^{4})}sin(y) - 3.28x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

Correct answer:

${- 3.28cos(z^{2}) - 4.4xzsin(z^{2}) - 0.18cos(z^{2})e^{(x^{4})}cos(y) - 1.1zsin(z^{2})e^{(x^{4})}sin(y) - 3.28x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11.32)^2+(-2.46)^2+(7.54)^2}=13.82\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11.32}{13.82}=-0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2.46}{13.82}=-0.18\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7.54}{13.82}=0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.82)(4cos(z^{2}) + 4x^{3}cos(z^{2})e^{(x^{4})}sin(y))+(-0.18)(cos(z^{2})e^{(x^{4})}cos(y))+(0.55)(- 8xzsin(z^{2}) - 2zsin(z^{2})e^{(x^{4})}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=- 3.28cos(z^{2}) - 4.4xzsin(z^{2}) - 0.18cos(z^{2})e^{(x^{4})}cos(y) - 1.1zsin(z^{2})e^{(x^{4})}sin(y) - 3.28x^{3}cos(z^{2})e^{(x^{4})}sin(y)\end{align*}$

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Example Question #341 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^xe^{(z^{2})}ln(y)\\&\text{In the direction of }\overrightarrow{v}=(5.49,17.87,-11.29).\end{align*}$

Possible Answers:

${\frac{(30.3\cdot 2^xze^{(z^{2})})}{y}}$

${\frac{(0.82\cdot 2^xe^{(z^{2})})}{y}+ 0.173\cdot 2^xe^{(z^{2})}ln(y) - 1.04\cdot 2^xze^{(z^{2})}ln(y)}$

${\frac{(21.8\cdot 2^xe^{(z^{2})})}{y}+ 15.1\cdot 2^xe^{(z^{2})}ln(y) + 43.7\cdot 2^xze^{(z^{2})}ln(y)}$

${\frac{(0.672\cdot 2^xe^{(z^{2})})}{y}+ 0.0433\cdot 2^xe^{(z^{2})}ln(y) + 0.541\cdot 2^xze^{(z^{2})}ln(y)}$

Correct answer:

${\frac{(0.82\cdot 2^xe^{(z^{2})})}{y}+ 0.173\cdot 2^xe^{(z^{2})}ln(y) - 1.04\cdot 2^xze^{(z^{2})}ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5.49)^2+(17.87)^2+(-11.29)^2}=21.84\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5.49}{21.84}=0.25\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17.87}{21.84}=0.82\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11.29}{21.84}=-0.52\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.25)(0.693\cdot 2^xe^{(z^{2})}ln(y))+(0.82)(\frac{(2^xe^{(z^{2})})}{y})+(-0.52)(2\cdot 2^xze^{(z^{2})}ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.82\cdot 2^xe^{(z^{2})})}{y}+ 0.173\cdot 2^xe^{(z^{2})}ln(y) - 1.04\cdot 2^xze^{(z^{2})}ln(y)\end{align*}$

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Example Question #342 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3yln(z^{2}) + ln(y^{4})e^{(x^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-15.17,11.6,3.25).\end{align*}$

Possible Answers:

${1.08ln(z^{2}) +\frac{ (0.173y)}{z}+\frac{ (1.44e^{(x^{4})}sin(z))}{y}+ 0.0289ln(y^{4})e^{(x^{4})}cos(z) + 2.43x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

${1.8ln(z^{2}) +\frac{ (1.02y)}{z}+\frac{ (2.4e^{(x^{4})}sin(z))}{y}+ 0.17ln(y^{4})e^{(x^{4})}cos(z) - 3.12x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

${\frac{(310x^{3}e^{(x^{4})}cos(z))}{y}}$

${58.1ln(z^{2}) +\frac{ (116y)}{z}+\frac{ (77.5e^{(x^{4})}sin(z))}{y}+ 19.4ln(y^{4})e^{(x^{4})}cos(z) + 77.5x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

Correct answer:

${1.8ln(z^{2}) +\frac{ (1.02y)}{z}+\frac{ (2.4e^{(x^{4})}sin(z))}{y}+ 0.17ln(y^{4})e^{(x^{4})}cos(z) - 3.12x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15.17)^2+(11.6)^2+(3.25)^2}=19.37\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15.17}{19.37}=-0.78\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11.6}{19.37}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3.25}{19.37}=0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.78)(4x^{3}ln(y^{4})e^{(x^{4})}sin(z))+(0.6)(3ln(z^{2}) +\frac{ (4e^{(x^{4})}sin(z))}{y})+(0.17)(\frac{(6y)}{z}+ ln(y^{4})e^{(x^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=1.8ln(z^{2}) +\frac{ (1.02y)}{z}+\frac{ (2.4e^{(x^{4})}sin(z))}{y}+ 0.17ln(y^{4})e^{(x^{4})}cos(z) - 3.12x^{3}ln(y^{4})e^{(x^{4})}sin(z)\end{align*}$

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Example Question #341 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(3^{(x^{3})}cos(y^{3}))}{z}\\&\text{In the direction of }\overrightarrow{v}=(-12.37,-17.98,8.28).\end{align*}$

Possible Answers:

${-\frac{(231\cdot 3^{(x^{3})}x^{2}y^{2}sin(y^{3}))}{z^{2}}}$

${\frac{(0.35\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (1.75\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (2.31\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

${\frac{(8.28\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (40.8\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (53.9\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

${\frac{(23.3\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}-\frac{ (76.9\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}+\frac{ (70\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

Correct answer:

${\frac{(0.35\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (1.75\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (2.31\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12.37)^2+(-17.98)^2+(8.28)^2}=23.34\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12.37}{23.34}=-0.53\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17.98}{23.34}=-0.77\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8.28}{23.34}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.53)(-\frac{(3.3\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z})+(-0.77)(\frac{(3\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z})+(0.35)(\frac{(3^{(x^{3})}cos(y^{3}))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.35\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (1.75\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (2.31\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}\end{align*}$

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Example Question #342 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^{4}cos(x^{4})cos(y)\\&\text{In the direction of }\overrightarrow{v}=(19.26,-17.85,-17.23).\end{align*}$

Possible Answers:

${17.9z^{4}cos(x^{4})sin(y) - 68.9z^{3}cos(x^{4})cos(y) - 77x^{3}z^{4}sin(x^{4})cos(y)}$

${126z^{3}cos(x^{4})cos(y) - 31.4z^{4}cos(x^{4})sin(y) - 126x^{3}z^{4}sin(x^{4})cos(y)}$

${503x^{3}z^{3}sin(x^{4})sin(y)}$

${0.57z^{4}cos(x^{4})sin(y) - 2.2z^{3}cos(x^{4})cos(y) - 2.44x^{3}z^{4}sin(x^{4})cos(y)}$

Correct answer:

${0.57z^{4}cos(x^{4})sin(y) - 2.2z^{3}cos(x^{4})cos(y) - 2.44x^{3}z^{4}sin(x^{4})cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(19.26)^2+(-17.85)^2+(-17.23)^2}=31.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{19.26}{31.41}=0.61\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17.85}{31.41}=-0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17.23}{31.41}=-0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.61)(-4x^{3}z^{4}sin(x^{4})cos(y))+(-0.57)(-1z^{4}cos(x^{4})sin(y))+(-0.55)(4z^{3}cos(x^{4})cos(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.57z^{4}cos(x^{4})sin(y) - 2.2z^{3}cos(x^{4})cos(y) - 2.44x^{3}z^{4}sin(x^{4})cos(y)\end{align*}$

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Example Question #343 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(4^{(y^{2})}ln(x^{2}))}{z}\\&\text{In the direction of }\overrightarrow{v}=(-12.26,-16.17,-7.03).\end{align*}$

Possible Answers:

${\frac{(21.5\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}-\frac{ (43\cdot 4^{(y^{2})})}{(xz)}-\frac{ (59.6\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

${\frac{(24.5\cdot 4^{(y^{2})})}{(xz)}-\frac{ (7.03\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (44.8\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

${\frac{(119\cdot 4^{(y^{2})}y)}{(xz^{2})}}$

${\frac{(1.14\cdot 4^{(y^{2})})}{(xz)}-\frac{ (0.33\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (2.08\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

Correct answer:

${\frac{(1.14\cdot 4^{(y^{2})})}{(xz)}-\frac{ (0.33\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (2.08\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12.26)^2+(-16.17)^2+(-7.03)^2}=21.48\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12.26}{21.48}=-0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16.17}{21.48}=-0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7.03}{21.48}=-0.33\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.57)(-\frac{(2\cdot 4^{(y^{2})})}{(xz)})+(-0.75)(-\frac{(2.77\cdot 4^{(y^{2})}yln(x^{2}))}{z})+(-0.33)(\frac{(4^{(y^{2})}ln(x^{2}))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.14\cdot 4^{(y^{2})})}{(xz)}-\frac{ (0.33\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (2.08\cdot 4^{(y^{2})}yln(x^{2}))}{z}\end{align*}$

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Example Question #344 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(y^{2})}ln(x^{4})sin(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(8.74,-13.71,-14.88).\end{align*}$

Possible Answers:

${1.36\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (1.6\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 2.72\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

${30.1\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (35\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 59.5\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

${-\frac{ (88.2\cdot 3^{(y^{2})}sin(z^{4}))}{x}- 48.4\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) - 88.2\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

${-\frac{(775\cdot 3^{(y^{2})}yz^{3}cos(z^{4}))}{x}}$

Correct answer:

${1.36\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (1.6\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 2.72\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8.74)^2+(-13.71)^2+(-14.88)^2}=22.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8.74}{22.04}=0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13.71}{22.04}=-0.62\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14.88}{22.04}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.4)(-\frac{(4\cdot 3^{(y^{2})}sin(z^{4}))}{x})+(-0.62)(-2.2\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}))+(-0.68)(-4\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.36\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (1.6\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 2.72\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})\end{align*}$

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Example Question #345 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 3zsin(y^{3}) - 4^{(y^{4})}xsin(z^{3})\\&\text{In the direction of }\overrightarrow{v}=(-5.63,-7.28,19.88).\end{align*}$

Possible Answers:

${-364\cdot 4^{(y^{4})}y^{3}z^{2}cos(z^{3})}$

${0.26\cdot 4^{(y^{4})}sin(z^{3}) - 2.73sin(y^{3}) + 2.97y^{2}zcos(y^{3}) - 2.73\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) + 1.83\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

${- 65.7sin(y^{3}) - 21.9\cdot 4^{(y^{4})}sin(z^{3}) - 197y^{2}zcos(y^{3}) - 65.7\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) - 121\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

${- 2.48sin(y^{3}) - 0.0676\cdot 4^{(y^{4})}sin(z^{3}) - 0.98y^{2}zcos(y^{3}) - 2.48\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) - 0.604\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

Correct answer:

${0.26\cdot 4^{(y^{4})}sin(z^{3}) - 2.73sin(y^{3}) + 2.97y^{2}zcos(y^{3}) - 2.73\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) + 1.83\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5.63)^2+(-7.28)^2+(19.88)^2}=21.91\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5.63}{21.91}=-0.26\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7.28}{21.91}=-0.33\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19.88}{21.91}=0.91\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.26)(-1\cdot 4^{(y^{4})}sin(z^{3}))+(-0.33)(- 9y^{2}zcos(y^{3}) - 5.55\cdot 4^{(y^{4})}xy^{3}sin(z^{3}))+(0.91)(- 3sin(y^{3}) - 3\cdot 4^{(y^{4})}xz^{2}cos(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=0.26\cdot 4^{(y^{4})}sin(z^{3}) - 2.73sin(y^{3}) + 2.97y^{2}zcos(y^{3}) - 2.73\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) + 1.83\cdot 4^{(y^{4})}xy^{3}sin(z^{3})\end{align*}$

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Example Question #346 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^{3}sin(y^{4})cos(x)\\&\text{In the direction of }\overrightarrow{v}=(-14.68,12.52,7.66).\end{align*}$

Possible Answers:

${-249y^{3}z^{2}cos(y^{4})sin(x)}$

${1.11z^{2}sin(y^{4})cos(x) + 0.71z^{3}sin(y^{4})sin(x) + 2.4y^{3}z^{3}cos(y^{4})cos(x)}$

${0.411z^{2}sin(y^{4})cos(x) - 0.504z^{3}sin(y^{4})sin(x) + 1.44y^{3}z^{3}cos(y^{4})cos(x)}$

${62.3z^{2}sin(y^{4})cos(x) - 20.8z^{3}sin(y^{4})sin(x) + 83y^{3}z^{3}cos(y^{4})cos(x)}$

Correct answer:

${1.11z^{2}sin(y^{4})cos(x) + 0.71z^{3}sin(y^{4})sin(x) + 2.4y^{3}z^{3}cos(y^{4})cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14.68)^2+(12.52)^2+(7.66)^2}=20.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14.68}{20.76}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12.52}{20.76}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7.66}{20.76}=0.37\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-1z^{3}sin(y^{4})sin(x))+(0.6)(4y^{3}z^{3}cos(y^{4})cos(x))+(0.37)(3z^{2}sin(y^{4})cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=1.11z^{2}sin(y^{4})cos(x) + 0.71z^{3}sin(y^{4})sin(x) + 2.4y^{3}z^{3}cos(y^{4})cos(x)\end{align*}$

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Example Question #347 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(z^{2}sin(x^{4}))}{y^{2}}\\&\text{In the direction of }\overrightarrow{v}=(4.93,2.88,2.11).\end{align*}$

Possible Answers:

${\frac{(0.7zsin(x^{4}))}{y^{2}}-\frac{ (0.94z^{2}sin(x^{4}))}{y^{3}}+\frac{ (3.24x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

${-\frac{(97.4x^{3}zcos(x^{4}))}{y^{3}}}$

${\frac{(4.22zsin(x^{4}))}{y^{2}}-\frac{ (5.76z^{2}sin(x^{4}))}{y^{3}}+\frac{ (19.7x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

${\frac{(12.2zsin(x^{4}))}{y^{2}}-\frac{ (12.2z^{2}sin(x^{4}))}{y^{3}}+\frac{ (24.4x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

Correct answer:

${\frac{(0.7zsin(x^{4}))}{y^{2}}-\frac{ (0.94z^{2}sin(x^{4}))}{y^{3}}+\frac{ (3.24x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(4.93)^2+(2.88)^2+(2.11)^2}=6.09\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{4.93}{6.09}=0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{2.88}{6.09}=0.47\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2.11}{6.09}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.81)(\frac{(4x^{3}z^{2}cos(x^{4}))}{y^{2}})+(0.47)(-\frac{(2z^{2}sin(x^{4}))}{y^{3}})+(0.35)(\frac{(2zsin(x^{4}))}{y^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.7zsin(x^{4}))}{y^{2}}-\frac{ (0.94z^{2}sin(x^{4}))}{y^{3}}+\frac{ (3.24x^{3}z^{2}cos(x^{4}))}{y^{2}}\end{align*}$

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Calculus 3 : Directional Derivatives

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