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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #331 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5tan(y^{2})tan(x) - e^{(x^{4})}e^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(-8,-11,-19).\end{align*}$

Possible Answers:

${117.0tan(y^{2})\cdot (tan(x)^{2} + 1) + 234ytan(x)\cdot (tan(y^{2})^{2} + 1) - 93.5x^{3}e^{(x^{4})}e^{(y^{3})} - 70.1y^{2}e^{(x^{4})}e^{(y^{3})}}$

${1.36x^{3}e^{(x^{4})}e^{(y^{3})} - 4.7ytan(x)\cdot (tan(y^{2})^{2} + 1) - 1.7tan(y^{2})\cdot (tan(x)^{2} + 1) + 1.41y^{2}e^{(x^{4})}e^{(y^{3})}}$

${0.578tan(y^{2})\cdot (tan(x)^{2} + 1) + 2.21ytan(x)\cdot (tan(y^{2})^{2} + 1) - 0.462x^{3}e^{(x^{4})}e^{(y^{3})} - 0.663y^{2}e^{(x^{4})}e^{(y^{3})}}$

Correct answer:

${1.36x^{3}e^{(x^{4})}e^{(y^{3})} - 4.7ytan(x)\cdot (tan(y^{2})^{2} + 1) - 1.7tan(y^{2})\cdot (tan(x)^{2} + 1) + 1.41y^{2}e^{(x^{4})}e^{(y^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(-11)^2+(-19)^2}=23.37\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{23.37}=-0.34\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-11}{23.37}=-0.47\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{23.37}=-0.81\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.34)(5.0tan(y^{2})\cdot (tan(x)^{2} + 1) - 4x^{3}e^{(x^{4})}e^{(y^{3})})+(-0.47)(10ytan(x)\cdot (tan(y^{2})^{2} + 1) - 3y^{2}e^{(x^{4})}e^{(y^{3})})+(-0.81)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=1.36x^{3}e^{(x^{4})}e^{(y^{3})} - 4.7ytan(x)\cdot (tan(y^{2})^{2} + 1) - 1.7tan(y^{2})\cdot (tan(x)^{2} + 1) + 1.41y^{2}e^{(x^{4})}e^{(y^{3})}\end{align*}$

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Example Question #332 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(4^xsin(z^{2}))}{y^{2}}\\&\text{In the direction of }\overrightarrow{v}=(6,-19,-1).\end{align*}$

Possible Answers:

${\frac{(39.9\cdot 4^xsin(z^{2}))}{y^{3}}-\frac{ (27.7\cdot 4^xsin(z^{2}))}{y^{2}}-\frac{ (39.9\cdot 4^xzcos(z^{2}))}{y^{2}}}$

${\frac{(1.8\cdot 4^xsin(z^{2}))}{y^{3}}-\frac{ (0.125\cdot 4^xsin(z^{2}))}{y^{2}}-\frac{ (0.005\cdot 4^xzcos(z^{2}))}{y^{2}}}$

${\frac{(0.1\cdot 4^xzcos(z^{2}))}{y^{2}}-\frac{ (1.9\cdot 4^xsin(z^{2}))}{y^{3}}-\frac{ (0.416\cdot 4^xsin(z^{2}))}{y^{2}}}$

${\frac{(111\cdot 4^xzcos(z^{2}))}{y^{3}}}$

Correct answer:

${\frac{(0.1\cdot 4^xzcos(z^{2}))}{y^{2}}-\frac{ (1.9\cdot 4^xsin(z^{2}))}{y^{3}}-\frac{ (0.416\cdot 4^xsin(z^{2}))}{y^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-19)^2+(-1)^2}=19.95\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{19.95}=0.3\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-19}{19.95}=-0.95\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-1}{19.95}=-0.05\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.3)(-\frac{(1.39\cdot 4^xsin(z^{2}))}{y^{2}})+(-0.95)(\frac{(2\cdot 4^xsin(z^{2}))}{y^{3}})+(-0.05)(-\frac{(2\cdot 4^xzcos(z^{2}))}{y^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.1\cdot 4^xzcos(z^{2}))}{y^{2}}-\frac{ (1.9\cdot 4^xsin(z^{2}))}{y^{3}}-\frac{ (0.416\cdot 4^xsin(z^{2}))}{y^{2}}\end{align*}$

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Example Question #333 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(x^{2})}\cdot 3^{(z^{2})}sin(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(5,18,17).\end{align*}$

Possible Answers:

${-308\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xy^{3}zcos(y^{4})}$

${- 35\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xsin(y^{4}) - 55.5\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}zsin(y^{4}) - 101\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}y^{3}cos(y^{4})}$

${- 0.277\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xsin(y^{4}) - 1.47\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}zsin(y^{4}) - 2.84\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}y^{3}cos(y^{4})}$

${- 0.0555\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xsin(y^{4}) - 0.986\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}zsin(y^{4}) - 2.02\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}y^{3}cos(y^{4})}$

Correct answer:

${- 0.277\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xsin(y^{4}) - 1.47\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}zsin(y^{4}) - 2.84\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}y^{3}cos(y^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(18)^2+(17)^2}=25.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{25.26}=0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{25.26}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{17}{25.26}=0.67\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.2)(-1.39\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xsin(y^{4}))+(0.71)(-4\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}y^{3}cos(y^{4}))+(0.67)(-2.2\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}zsin(y^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=- 0.277\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}xsin(y^{4}) - 1.47\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}zsin(y^{4}) - 2.84\cdot 2^{(x^{2})}\cdot 3^{(z^{2})}y^{3}cos(y^{4})\end{align*}$

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Example Question #334 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{4}e^{(z^{2})}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(18,13,-3).\end{align*}$

Possible Answers:

${2.32y^{3}e^{(z^{2})}e^{(x)} + 0.8y^{4}e^{(z^{2})}e^{(x)} - 0.26y^{4}ze^{(z^{2})}e^{(x)}}$

${89.6y^{3}e^{(z^{2})}e^{(x)} + 22.4y^{4}e^{(z^{2})}e^{(x)} + 44.8y^{4}ze^{(z^{2})}e^{(x)}}$

${179y^{3}ze^{(z^{2})}e^{(x)}}$

${52y^{3}e^{(z^{2})}e^{(x)} + 18y^{4}e^{(z^{2})}e^{(x)} - 6y^{4}ze^{(z^{2})}e^{(x)}}$

Correct answer:

${2.32y^{3}e^{(z^{2})}e^{(x)} + 0.8y^{4}e^{(z^{2})}e^{(x)} - 0.26y^{4}ze^{(z^{2})}e^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(13)^2+(-3)^2}=22.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{22.41}=0.8\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{13}{22.41}=0.58\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{22.41}=-0.13\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.8)(y^{4}e^{(z^{2})}e^{(x)})+(0.58)(4y^{3}e^{(z^{2})}e^{(x)})+(-0.13)(2y^{4}ze^{(z^{2})}e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=2.32y^{3}e^{(z^{2})}e^{(x)} + 0.8y^{4}e^{(z^{2})}e^{(x)} - 0.26y^{4}ze^{(z^{2})}e^{(x)}\end{align*}$

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Example Question #335 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 3e^{(y^{2})} - 2^{(z^{3})}sin(x^{4})e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-2,-6,9).\end{align*}$

Possible Answers:

${-91.5\cdot 2^{(z^{3})}x^{3}z^{2}cos(x^{4})e^{(y)}}$

${3.3ye^{(y^{2})} + 0.55\cdot 2^{(z^{3})}sin(x^{4})e^{(y)} + 0.72\cdot 2^{(z^{3})}x^{3}cos(x^{4})e^{(y)} - 1.71\cdot 2^{(z^{3})}z^{2}sin(x^{4})e^{(y)}}$

${- 66ye^{(y^{2})} - 11\cdot 2^{(z^{3})}sin(x^{4})e^{(y)} - 44\cdot 2^{(z^{3})}x^{3}cos(x^{4})e^{(y)} - 22.9\cdot 2^{(z^{3})}z^{2}sin(x^{4})e^{(y)}}$

${36ye^{(y^{2})} + 6\cdot 2^{(z^{3})}sin(x^{4})e^{(y)} + 8\cdot 2^{(z^{3})}x^{3}cos(x^{4})e^{(y)} - 18.7\cdot 2^{(z^{3})}z^{2}sin(x^{4})e^{(y)}}$

Correct answer:

${3.3ye^{(y^{2})} + 0.55\cdot 2^{(z^{3})}sin(x^{4})e^{(y)} + 0.72\cdot 2^{(z^{3})}x^{3}cos(x^{4})e^{(y)} - 1.71\cdot 2^{(z^{3})}z^{2}sin(x^{4})e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(-6)^2+(9)^2}=11\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{11}=-0.18\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-6}{11}=-0.55\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{11}=0.82\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.18)(-4\cdot 2^{(z^{3})}x^{3}cos(x^{4})e^{(y)})+(-0.55)(- 6ye^{(y^{2})} - 1\cdot 2^{(z^{3})}sin(x^{4})e^{(y)})+(0.82)(-2.08\cdot 2^{(z^{3})}z^{2}sin(x^{4})e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=3.3ye^{(y^{2})} + 0.55\cdot 2^{(z^{3})}sin(x^{4})e^{(y)} + 0.72\cdot 2^{(z^{3})}x^{3}cos(x^{4})e^{(y)} - 1.71\cdot 2^{(z^{3})}z^{2}sin(x^{4})e^{(y)}\end{align*}$

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Example Question #336 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2x^{4}e^{(z^{2})} - z^{3}e^{(x^{3})}e^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,-11,7).\end{align*}$

Possible Answers:

${2.16x^{4}ze^{(z^{2})} - 1.62z^{2}e^{(x^{3})}e^{(y^{2})} + 1.68yz^{3}e^{(x^{3})}e^{(y^{2})}}$

${104x^{3}e^{(z^{2})} - 39.1z^{2}e^{(x^{3})}e^{(y^{2})} + 52.2x^{4}ze^{(z^{2})} - 26.1yz^{3}e^{(x^{3})}e^{(y^{2})} - 39.1x^{2}z^{3}e^{(x^{3})}e^{(y^{2})}}$

${1.17x^{4}ze^{(z^{2})} - 0.875z^{2}e^{(x^{3})}e^{(y^{2})} - 1.41yz^{3}e^{(x^{3})}e^{(y^{2})}}$

${-235x^{2}yz^{2}e^{(x^{3})}e^{(y^{2})}}$

Correct answer:

${2.16x^{4}ze^{(z^{2})} - 1.62z^{2}e^{(x^{3})}e^{(y^{2})} + 1.68yz^{3}e^{(x^{3})}e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-11)^2+(7)^2}=13.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.04}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-11}{13.04}=-0.84\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{13.04}=0.54\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(8x^{3}e^{(z^{2})} - 3x^{2}z^{3}e^{(x^{3})}e^{(y^{2})})+(-0.84)(-2yz^{3}e^{(x^{3})}e^{(y^{2})})+(0.54)(4x^{4}ze^{(z^{2})} - 3z^{2}e^{(x^{3})}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=2.16x^{4}ze^{(z^{2})} - 1.62z^{2}e^{(x^{3})}e^{(y^{2})} + 1.68yz^{3}e^{(x^{3})}e^{(y^{2})}\end{align*}$

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Example Question #337 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(y^{4}ln(z^{4}))}{x^{2}}\\&\text{In the direction of }\overrightarrow{v}=(18,-16,15).\end{align*}$

Possible Answers:

${\frac{(113y^{3}ln(z^{4}))}{x^{2}}-\frac{ (56.7y^{4}ln(z^{4}))}{x^{3}}+\frac{ (113y^{4})}{(x^{2}z)}}$

${\frac{(60y^{4})}{(x^{2}z)}-\frac{ (36y^{4}ln(z^{4}))}{x^{3}}-\frac{ (64y^{3}ln(z^{4}))}{x^{2}}}$

${-\frac{(908y^{3})}{(x^{3}z)}}$

${\frac{(2.12y^{4})}{(x^{2}z)}-\frac{ (1.26y^{4}ln(z^{4}))}{x^{3}}-\frac{ (2.24y^{3}ln(z^{4}))}{x^{2}}}$

Correct answer:

${\frac{(2.12y^{4})}{(x^{2}z)}-\frac{ (1.26y^{4}ln(z^{4}))}{x^{3}}-\frac{ (2.24y^{3}ln(z^{4}))}{x^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(-16)^2+(15)^2}=28.37\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{28.37}=0.63\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{28.37}=-0.56\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{28.37}=0.53\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.63)(-\frac{(2y^{4}ln(z^{4}))}{x^{3}})+(-0.56)(\frac{(4y^{3}ln(z^{4}))}{x^{2}})+(0.53)(\frac{(4y^{4})}{(x^{2}z)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(2.12y^{4})}{(x^{2}z)}-\frac{ (1.26y^{4}ln(z^{4}))}{x^{3}}-\frac{ (2.24y^{3}ln(z^{4}))}{x^{2}}\end{align*}$

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Example Question #338 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2y^{3}sin(z) - z^{2}e^{(x^{3})}e^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-7,5,-8).\end{align*}$

Possible Answers:

${- 23.5y^{3}cos(z) - 70.5y^{2}sin(z) - 23.5ze^{(x^{3})}e^{(y^{2})} - 23.5yz^{2}e^{(x^{3})}e^{(y^{2})} - 35.2x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

${16y^{3}cos(z) - 30y^{2}sin(z) + 16ze^{(x^{3})}e^{(y^{2})} - 10yz^{2}e^{(x^{3})}e^{(y^{2})} + 21x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

${-141x^{2}yze^{(x^{3})}e^{(y^{2})}}$

${1.36y^{3}cos(z) - 2.58y^{2}sin(z) + 1.36ze^{(x^{3})}e^{(y^{2})} - 0.86yz^{2}e^{(x^{3})}e^{(y^{2})} + 1.8x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

Correct answer:

${1.36y^{3}cos(z) - 2.58y^{2}sin(z) + 1.36ze^{(x^{3})}e^{(y^{2})} - 0.86yz^{2}e^{(x^{3})}e^{(y^{2})} + 1.8x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(5)^2+(-8)^2}=11.75\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{11.75}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{11.75}=0.43\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{11.75}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(-3x^{2}z^{2}e^{(x^{3})}e^{(y^{2})})+(0.43)(- 6y^{2}sin(z) - 2yz^{2}e^{(x^{3})}e^{(y^{2})})+(-0.68)(- 2y^{3}cos(z) - 2ze^{(x^{3})}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.36y^{3}cos(z) - 2.58y^{2}sin(z) + 1.36ze^{(x^{3})}e^{(y^{2})} - 0.86yz^{2}e^{(x^{3})}e^{(y^{2})} + 1.8x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}\end{align*}$

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Example Question #339 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}\\&\text{In the direction of }\overrightarrow{v}=(15,-4,4).\end{align*}$

Possible Answers:

${0.75y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 1.88xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 0.75z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

${12y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 30xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 12z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

${-289xy^{2}z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

${- 32.1xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 48.1y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 48.1z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

Correct answer:

${0.75y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 1.88xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 0.75z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(15)^2+(-4)^2+(4)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{15}{16.03}=0.94\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{16.03}=-0.25\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{16.03}=0.25\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.94)(-2xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})})+(-0.25)(-3y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})})+(0.25)(-3z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})})\\&D_{\overrightarrow{u}}(x,y,z)=0.75y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 1.88xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 0.75z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}\end{align*}$

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Example Question #340 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(x^{4})sin(z^{4})ln(y)\\&\text{In the direction of }\overrightarrow{v}=(17.01,15.74,-1.27).\end{align*}$

Possible Answers:

${5.08z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (15.7ln(x^{4})sin(z^{4}))}{y}-\frac{ (68sin(z^{4})ln(y))}{x}}$

${-\frac{ (92.8sin(z^{4})ln(y))}{x}-\frac{ (23.2ln(x^{4})sin(z^{4}))}{y}- 92.8z^{3}cos(z^{4})ln(x^{4})ln(y)}$

${-\frac{(371z^{3}cos(z^{4}))}{(xy)}}$

${0.2z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (0.68ln(x^{4})sin(z^{4}))}{y}-\frac{ (2.92sin(z^{4})ln(y))}{x}}$

Correct answer:

${0.2z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (0.68ln(x^{4})sin(z^{4}))}{y}-\frac{ (2.92sin(z^{4})ln(y))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17.01)^2+(15.74)^2+(-1.27)^2}=23.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17.01}{23.21}=0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15.74}{23.21}=0.68\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-1.27}{23.21}=-0.05\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.73)(-\frac{(4sin(z^{4})ln(y))}{x})+(0.68)(-\frac{(ln(x^{4})sin(z^{4}))}{y})+(-0.05)(-4z^{3}cos(z^{4})ln(x^{4})ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.2z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (0.68ln(x^{4})sin(z^{4}))}{y}-\frac{ (2.92sin(z^{4})ln(y))}{x}\end{align*}$

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Calculus 3 : Directional Derivatives

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