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Calculus 3 : Partial Derivatives

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Example Questions

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:

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

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

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

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

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:

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

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

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

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

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 #341 : 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:

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

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

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

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

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 #342 : 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:

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

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

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

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

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 #343 : 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 #344 : 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:

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

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

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

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

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 #345 : 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:

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

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

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

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

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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Example Question #351 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{4}sin(x^{4})e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-14.93,-4.93,-3.31).\end{align*}$

Possible Answers:

${514x^{3}y^{3}zcos(x^{4})e^{(z^{2})}}$

${64.3y^{3}sin(x^{4})e^{(z^{2})} + 32.1y^{4}zsin(x^{4})e^{(z^{2})} + 64.3x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

${- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

${- 19.7y^{3}sin(x^{4})e^{(z^{2})} - 6.62y^{4}zsin(x^{4})e^{(z^{2})} - 59.7x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

Correct answer:

${- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14.93)^2+(-4.93)^2+(-3.31)^2}=16.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14.93}{16.07}=-0.93\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4.93}{16.07}=-0.31\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3.31}{16.07}=-0.21\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.93)(4x^{3}y^{4}cos(x^{4})e^{(z^{2})})+(-0.31)(4y^{3}sin(x^{4})e^{(z^{2})})+(-0.21)(2y^{4}zsin(x^{4})e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}\end{align*}$

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Example Question #352 : Directional Derivatives

Find the directional derivative D_uf(x,y) for the function:

f(x,y)=ln(xy)+ye^x

in the direction of the vector $\vec{u}=2\hat{i}+\hat{j}$

Possible Answers:

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[\frac{xe^x(x+1)+2x+y}{x} \right]$

$D_uf(x,y) =2\left(\frac{1}{y}+e^x \right)\hat{i}+\left(\frac{1}{x}+ye^x\right)\hat{j}$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]$

$D_uf(x,y) =\frac{\sqrt{7}}{3}\left[ 7e^x(x+1)-\frac{2}{y} \right]$

$D_uf(x,y) =2\left(\frac{1}{x}+ye^x \right)\hat{i}+\left(\frac{1}{y}+e^x\right)\hat{j}$

Correct answer:

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]$

Explanation:

Find the directional derivative for the function:

f(x,y)=ln(xy)+ye^x

in the direction of the vector $\vec{u}=2\hat{i}+\hat{j}$

Find the gradient of

The gradient by definition is the vector:

$\vec{\bigtriangledown} f(x,y)=\frac{\partial}{\partial x}f(x,y)\hat{i}+\frac{\partial}{\partial y}f(x,y)\hat{j}$ (1)

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Differentiate with respect to while holding constant:

$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left[\ln(xy)+ye^x\right ]=\frac{y}{xy}+ye^x$

$\frac{\partial}{\partial x}f(x,y)=\frac{1}{x}+ye^x$

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Differentiate with respect to while holding constant:

$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left[\ln(xy)+ye^x\right ]=\frac{x}{xy}+e^x$

$\frac{\partial}{\partial y}f(x, y)=\frac{1}{y}+e^x$

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Enter the components of the gradient, writing in the form of Equation (1)

$\vec{\bigtriangledown}f(x,y)=\left(\frac{1}{x}+ye^x\right)\hat{i}+ \left(\frac{1}{y}+e^x\right)\hat{j}$

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Find the unit vector in the direction of $\vec{u}$

$\vec{u}=2\hat{i}+\hat{j}$

Divide the vector by its' magnitude $\left \| \vec{u}\right \|$ to obtain the unit vector $\hat{u}$ ,

$\hat{u}=\frac{\vec{u}}{\left \| \vec{u}\right \|} =\frac{\vec{u}}{\sqrt{2^2+1^2}}=\frac{2\hat{i}+\hat{j}}{\sqrt{5}}$

$\hat{u}=\frac{\sqrt{5}}{5}(2\hat{i}+\hat{j})$

You can check the magnitude of the unit vector and see that it is equal to 1 as required. It also has the same direction as the original vector $\vec{u}$ .

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The directional derivative of f(x,y) in the direction of the vector $\hat{u}$ is the dot product of the gradient $\vec{\bigtriangledown} f(x,y)$ and $\hat{u}$ . Dot products are also referred to as "projections," since the give the component of some vector in the direction of another.

$D_uf(x,y)= \vec{\bigtriangledown} f(x,y)\cdot \hat{u}$

$D_uf(x,y)= \left[ \left(\frac{1}{x}+ye^x\right)\hat{i}+ \left(\frac{1}{y}+e^x\right)\hat{j}\right]\cdot\left[\frac{2\sqrt{5}}{5}\hat{i}+\frac{\sqrt{5}}{5}\hat{j}\right]$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[\left(\frac{2}{x}+2ye^x\right)+ \left(\frac{1}{y}+e^x\right) \right ]$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ e^x(2y+1)+\frac{1}{y}+\frac{2}{x} \right]$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]$

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Example Question #1 : Differentials

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

$dz=2xe^{x^2+4y^2}\sin(y)dx$

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

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Calculus 3 : Partial Derivatives

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