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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1261 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ye^{(x^{4})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-4,0,20).\end{align*}$

Possible Answers:

${163x^{3}ze^{(x^{4})}e^{(z^{2})}}$

${1.96yze^{(x^{4})}e^{(z^{2})} - 0.8x^{3}ye^{(x^{4})}e^{(z^{2})}}$

${0.16x^{3}ye^{(x^{4})}e^{(z^{2})} + 1.92yze^{(x^{4})}e^{(z^{2})}}$

${20.4e^{(x^{4})}e^{(z^{2})} + 81.6x^{3}ye^{(x^{4})}e^{(z^{2})} + 40.8yze^{(x^{4})}e^{(z^{2})}}$

Correct answer:

${1.96yze^{(x^{4})}e^{(z^{2})} - 0.8x^{3}ye^{(x^{4})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(20)^2}=20.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{20.4}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.4}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{20.4}=0.98\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(4x^{3}ye^{(x^{4})}e^{(z^{2})})+(0)(e^{(x^{4})}e^{(z^{2})})+(0.98)(2yze^{(x^{4})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.96yze^{(x^{4})}e^{(z^{2})} - 0.8x^{3}ye^{(x^{4})}e^{(z^{2})}\end{align*}$

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Example Question #1261 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(e^{(x^{2})}cos(y))}{z}\\&\text{In the direction of }\overrightarrow{v}=(9,-16,0).\end{align*}$

Possible Answers:

${\frac{(18.4e^{(x^{2})}cos(y))}{z^{2}}+\frac{ (18.4e^{(x^{2})}sin(y))}{z}-\frac{ (36.7xe^{(x^{2})}cos(y))}{z}}$

${-\frac{ (16e^{(x^{2})}sin(y))}{z}-\frac{ (18xe^{(x^{2})}cos(y))}{z}}$

${-\frac{ (0.87e^{(x^{2})}sin(y))}{z}-\frac{ (0.98xe^{(x^{2})}cos(y))}{z}}$

${-\frac{(36.7xe^{(x^{2})}sin(y))}{z^{2}}}$

Correct answer:

${-\frac{ (0.87e^{(x^{2})}sin(y))}{z}-\frac{ (0.98xe^{(x^{2})}cos(y))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(9)^2+(-16)^2+(0)^2}=18.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{9}{18.36}=0.49\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{18.36}=-0.87\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.36}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.49)(-\frac{(2xe^{(x^{2})}cos(y))}{z})+(-0.87)(\frac{(e^{(x^{2})}sin(y))}{z})+(0)(\frac{(e^{(x^{2})}cos(y))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ (0.87e^{(x^{2})}sin(y))}{z}-\frac{ (0.98xe^{(x^{2})}cos(y))}{z}\end{align*}$

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Example Question #1263 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{2}cos(z^{4})sin(x^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,-18,-12).\end{align*}$

Possible Answers:

${692x^{3}yz^{3}cos(x^{4})sin(z^{4})}$

${36ycos(z^{4})sin(x^{4}) - 48y^{2}z^{3}sin(x^{4})sin(z^{4})}$

${86.5y^{2}z^{3}sin(x^{4})sin(z^{4}) - 86.5x^{3}y^{2}cos(x^{4})cos(z^{4}) - 43.3ycos(z^{4})sin(x^{4})}$

${1.66ycos(z^{4})sin(x^{4}) - 2.2y^{2}z^{3}sin(x^{4})sin(z^{4})}$

Correct answer:

${1.66ycos(z^{4})sin(x^{4}) - 2.2y^{2}z^{3}sin(x^{4})sin(z^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(-12)^2}=21.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.63}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{21.63}=-0.83\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{21.63}=-0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-4x^{3}y^{2}cos(x^{4})cos(z^{4}))+(-0.83)(-2ycos(z^{4})sin(x^{4}))+(-0.55)(4y^{2}z^{3}sin(x^{4})sin(z^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.66ycos(z^{4})sin(x^{4}) - 2.2y^{2}z^{3}sin(x^{4})sin(z^{4})\end{align*}$

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Example Question #1271 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^zln(x^{3})e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,11,20).\end{align*}$

Possible Answers:

${1.3\cdot 4^zln(x^{3})e^{(y)}}$

${1.7\cdot 4^zln(x^{3})e^{(y)}}$

${\frac{(68.5\cdot 4^ze^{(y)})}{x}+ 54.5\cdot 4^zln(x^{3})e^{(y)}}$

${\frac{(94.9\cdot 4^ze^{(y)})}{x}}$

Correct answer:

${1.7\cdot 4^zln(x^{3})e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(11)^2+(20)^2}=22.83\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.83}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{22.83}=0.48\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{22.83}=0.88\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(3\cdot 4^ze^{(y)})}{x})+(0.48)(4^zln(x^{3})e^{(y)})+(0.88)(1.39\cdot 4^zln(x^{3})e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=1.7\cdot 4^zln(x^{3})e^{(y)}\end{align*}$

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Example Question #1272 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-e^{(x^{2})}e^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(-10,16,0).\end{align*}$

Possible Answers:

${1.06xe^{(x^{2})}e^{(y^{3})} - 2.55y^{2}e^{(x^{2})}e^{(y^{3})}}$

${- 37.7xe^{(x^{2})}e^{(y^{3})} - 56.6y^{2}e^{(x^{2})}e^{(y^{3})}}$

${20xe^{(x^{2})}e^{(y^{3})} - 48y^{2}e^{(x^{2})}e^{(y^{3})}}$

Correct answer:

${1.06xe^{(x^{2})}e^{(y^{3})} - 2.55y^{2}e^{(x^{2})}e^{(y^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-10)^2+(16)^2+(0)^2}=18.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-10}{18.87}=-0.53\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{18.87}=0.85\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.87}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.53)(-2xe^{(x^{2})}e^{(y^{3})})+(0.85)(-3y^{2}e^{(x^{2})}e^{(y^{3})})+(0)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=1.06xe^{(x^{2})}e^{(y^{3})} - 2.55y^{2}e^{(x^{2})}e^{(y^{3})}\end{align*}$

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Example Question #1271 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-x^{3}ze^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(0,10,8).\end{align*}$

Possible Answers:

${- 0.62x^{3}e^{(y^{3})} - 2.34x^{3}y^{2}ze^{(y^{3})}}$

${-115x^{2}y^{2}e^{(y^{3})}}$

${- 12.8x^{3}e^{(y^{3})} - 38.4x^{2}ze^{(y^{3})} - 38.4x^{3}y^{2}ze^{(y^{3})}}$

${- 8x^{3}e^{(y^{3})} - 30x^{3}y^{2}ze^{(y^{3})}}$

Correct answer:

${- 0.62x^{3}e^{(y^{3})} - 2.34x^{3}y^{2}ze^{(y^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(8)^2}=12.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.81}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{12.81}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8}{12.81}=0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-3x^{2}ze^{(y^{3})})+(0.78)(-3x^{3}y^{2}ze^{(y^{3})})+(0.62)(-1x^{3}e^{(y^{3})})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.62x^{3}e^{(y^{3})} - 2.34x^{3}y^{2}ze^{(y^{3})}\end{align*}$

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Example Question #1274 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-e^{(x^{2})}e^{(z)}ln(y)\\&\text{In the direction of }\overrightarrow{v}=(-8,0,18).\end{align*}$

Possible Answers:

${-\frac{ (19.7e^{(x^{2})}e^{(z)})}{y}- 19.7e^{(x^{2})}e^{(z)}ln(y) - 39.4xe^{(x^{2})}e^{(z)}ln(y)}$

${-\frac{(39.4xe^{(x^{2})}e^{(z)})}{y}}$

${16xe^{(x^{2})}e^{(z)}ln(y) - 18e^{(x^{2})}e^{(z)}ln(y)}$

${0.82xe^{(x^{2})}e^{(z)}ln(y) - 0.91e^{(x^{2})}e^{(z)}ln(y)}$

Correct answer:

${0.82xe^{(x^{2})}e^{(z)}ln(y) - 0.91e^{(x^{2})}e^{(z)}ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(0)^2+(18)^2}=19.7\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{19.7}=-0.41\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{19.7}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{19.7}=0.91\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.41)(-2xe^{(x^{2})}e^{(z)}ln(y))+(0)(-\frac{(e^{(x^{2})}e^{(z)})}{y})+(0.91)(-1e^{(x^{2})}e^{(z)}ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.82xe^{(x^{2})}e^{(z)}ln(y) - 0.91e^{(x^{2})}e^{(z)}ln(y)\end{align*}$

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Example Question #1275 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(3^{(z^{3})}cos(y^{4}))}{x^{2}}\\&\text{In the direction of }\overrightarrow{v}=(14,-3,0).\end{align*}$

Possible Answers:

${\frac{(28.6\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (47.2\cdot 3^{(z^{3})}z^{2}cos(y^{4}))}{x^{2}}+\frac{ (57.3\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

${\frac{(28\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (12\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

${-\frac{(378\cdot 3^{(z^{3})}y^{3}z^{2}sin(y^{4}))}{x^{3}}}$

${\frac{(1.96\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (0.84\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

Correct answer:

${\frac{(1.96\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (0.84\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(-3)^2+(0)^2}=14.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{14.32}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{14.32}=-0.21\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(\frac{(2\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}})+(-0.21)(\frac{(4\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}})+(0)(-\frac{(3.3\cdot 3^{(z^{3})}z^{2}cos(y^{4}))}{x^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.96\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (0.84\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}\end{align*}$

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Example Question #1276 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{e^{(y^{2})}}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-5,16,0).\end{align*}$

Possible Answers:

${\frac{(33.5e^{(y^{2})})}{z^{3}}-\frac{ (33.5ye^{(y^{2})})}{z^{2}}}$

${-\frac{(1.8ye^{(y^{2})})}{z^{2}}}$

${-\frac{(1.9ye^{(y^{2})})}{z^{2}}}$

Correct answer:

${-\frac{(1.9ye^{(y^{2})})}{z^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(16)^2+(0)^2}=16.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{16.76}=-0.3\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{16.76}=0.95\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.76}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.3)(0.0)+(0.95)(-\frac{(2ye^{(y^{2})})}{z^{2}})+(0)(\frac{(2e^{(y^{2})})}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{(1.9ye^{(y^{2})})}{z^{2}}\end{align*}$

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Example Question #1277 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2x^{3}z^{4}cos(y^{3}) +\frac{ (y^{3}ln(x^{2}))}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(16,18,0).\end{align*}$

Possible Answers:

${144x^{2}z^{4}cos(y^{3}) + 193x^{3}z^{3}cos(y^{3}) +\frac{ (72.2y^{2}ln(x^{2}))}{z^{2}}-\frac{ (48.2y^{3}ln(x^{2}))}{z^{3}}+\frac{ (48.2y^{3})}{(xz^{2})}- 144x^{3}y^{2}z^{4}sin(y^{3})}$

${3.96x^{2}z^{4}cos(y^{3}) +\frac{ (2.25y^{2}ln(x^{2}))}{z^{2}}+\frac{ (1.32y^{3})}{(xz^{2})}- 4.5x^{3}y^{2}z^{4}sin(y^{3})}$

${96x^{2}z^{4}cos(y^{3}) +\frac{ (54y^{2}ln(x^{2}))}{z^{2}}+\frac{ (32y^{3})}{(xz^{2})}- 108x^{3}y^{2}z^{4}sin(y^{3})}$

${-\frac{ (289y^{2})}{(xz^{3})}- 1733x^{2}y^{2}z^{3}sin(y^{3})}$

Correct answer:

${3.96x^{2}z^{4}cos(y^{3}) +\frac{ (2.25y^{2}ln(x^{2}))}{z^{2}}+\frac{ (1.32y^{3})}{(xz^{2})}- 4.5x^{3}y^{2}z^{4}sin(y^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(16)^2+(18)^2+(0)^2}=24.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{16}{24.08}=0.66\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{24.08}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{24.08}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.66)(6x^{2}z^{4}cos(y^{3}) +\frac{ (2y^{3})}{(xz^{2})})+(0.75)(\frac{(3y^{2}ln(x^{2}))}{z^{2}}- 6x^{3}y^{2}z^{4}sin(y^{3}))+(0)(8x^{3}z^{3}cos(y^{3}) -\frac{ (2y^{3}ln(x^{2}))}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=3.96x^{2}z^{4}cos(y^{3}) +\frac{ (2.25y^{2}ln(x^{2}))}{z^{2}}+\frac{ (1.32y^{3})}{(xz^{2})}- 4.5x^{3}y^{2}z^{4}sin(y^{3})\end{align*}$

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Calculus 3 : Directional Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #1261 : Partial Derivatives

Example Question #1261 : Partial Derivatives

Example Question #1263 : Partial Derivatives

Example Question #1271 : Partial Derivatives

Example Question #1272 : Partial Derivatives

Example Question #1271 : Partial Derivatives

Example Question #1274 : Partial Derivatives

Example Question #1275 : Partial Derivatives

Example Question #1276 : Partial Derivatives

Example Question #1277 : Partial Derivatives

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