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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1281 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(tan(y^{2})ln(z))}{x}- y^{4}sin(x^{3})e^{(z^{4})}\\&\text{In the direction of }\overrightarrow{v}=(0,7,17).\end{align*}$

Possible Answers:

${-\frac{ (36.8y\cdot (tan(y^{2})^{2} + 1))}{(x^{2}z)}- 882x^{2}y^{3}z^{3}cos(x^{3})e^{(z^{4})}}$

${\frac{(18.4\cdot tan(y^{2}))}{(xz)}-\frac{ (18.4\cdot tan(y^{2})ln(z))}{x^{2}}- 73.5y^{3}sin(x^{3})e^{(z^{4})} - 55.1x^{2}y^{4}cos(x^{3})e^{(z^{4})} - 73.5y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (36.8yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

${\frac{(0.92\cdot tan(y^{2}))}{(xz)}- 1.52y^{3}sin(x^{3})e^{(z^{4})} - 3.68y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.76yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

${\frac{(0.846\cdot tan(y^{2}))}{(xz)}- 0.578y^{3}sin(x^{3})e^{(z^{4})} - 3.39y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.289yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

Correct answer:

${\frac{(0.92\cdot tan(y^{2}))}{(xz)}- 1.52y^{3}sin(x^{3})e^{(z^{4})} - 3.68y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.76yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{18.38}=0.38\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{17}{18.38}=0.92\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{ (1\cdot tan(y^{2})ln(z))}{x^{2}}- 3x^{2}y^{4}cos(x^{3})e^{(z^{4})})+(0.38)(\frac{(2yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}- 4y^{3}sin(x^{3})e^{(z^{4})})+(0.92)(\frac{tan(y^{2})}{(xz)}- 4y^{4}z^{3}sin(x^{3})e^{(z^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.92\cdot tan(y^{2}))}{(xz)}- 1.52y^{3}sin(x^{3})e^{(z^{4})} - 3.68y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.76yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}\end{align*}$

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Example Question #1282 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- e^{(x^{2})}e^{(z)} - 4x^{2}cos(z^{3})cos(y)\\&\text{In the direction of }\overrightarrow{v}=(-1,0,-16).\end{align*}$

Possible Answers:

${12x^{2}z^{2}sin(z^{3})cos(y) - 0.0288xcos(z^{3})cos(y) - 0.0072xe^{(x^{2})}e^{(z)} - e^{(x^{2})}e^{(z)}}$

${-385xz^{2}sin(z^{3})sin(y)}$

${e^{(x^{2})}e^{(z)} + 0.48xcos(z^{3})cos(y) + 0.12xe^{(x^{2})}e^{(z)} - 12x^{2}z^{2}sin(z^{3})cos(y)}$

${64.1x^{2}cos(z^{3})sin(y) - 16e^{(x^{2})}e^{(z)} - 128xcos(z^{3})cos(y) - 32.1xe^{(x^{2})}e^{(z)} + 192x^{2}z^{2}sin(z^{3})cos(y)}$

Correct answer:

${e^{(x^{2})}e^{(z)} + 0.48xcos(z^{3})cos(y) + 0.12xe^{(x^{2})}e^{(z)} - 12x^{2}z^{2}sin(z^{3})cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(0)^2+(-16)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{16.03}=-0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{16.03}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{16.03}=-1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.06)(- 8xcos(z^{3})cos(y) - 2xe^{(x^{2})}e^{(z)})+(0)(4x^{2}cos(z^{3})sin(y))+(-1)(12x^{2}z^{2}sin(z^{3})cos(y) - e^{(x^{2})}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=e^{(x^{2})}e^{(z)} + 0.48xcos(z^{3})cos(y) + 0.12xe^{(x^{2})}e^{(z)} - 12x^{2}z^{2}sin(z^{3})cos(y)\end{align*}$

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Example Question #1283 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{3}cos(y^{4})ln(z) - 4ze^{(y^{2})}\cdot tan(x)\\&\text{In the direction of }\overrightarrow{v}=(-14,0,-20).\end{align*}$

Possible Answers:

${- 195ye^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (293x^{2}y^{3}sin(y^{4}))}{z}}$

${3.28e^{(y^{2})}\cdot tan(x) + 2.28ze^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (0.82x^{3}cos(y^{4}))}{z}- 1.71x^{2}cos(y^{4})ln(z)}$

${\frac{(0.672x^{3}cos(y^{4}))}{z}- 1.3ze^{(y^{2})}\cdot (tan(x)^{2} + 1) - 2.69e^{(y^{2})}\cdot tan(x) + 0.975x^{2}cos(y^{4})ln(z)}$

${\frac{(24.4x^{3}cos(y^{4}))}{z}- 97.6ze^{(y^{2})}\cdot (tan(x)^{2} + 1) - 97.6e^{(y^{2})}\cdot tan(x) + 73.2x^{2}cos(y^{4})ln(z) - 97.6x^{3}y^{3}sin(y^{4})ln(z) - 195yze^{(y^{2})}\cdot tan(x)}$

Correct answer:

${3.28e^{(y^{2})}\cdot tan(x) + 2.28ze^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (0.82x^{3}cos(y^{4}))}{z}- 1.71x^{2}cos(y^{4})ln(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(0)^2+(-20)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{24.41}=-0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{24.41}=-0.82\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.57)(3x^{2}cos(y^{4})ln(z) - 4ze^{(y^{2})}\cdot (tan(x)^{2} + 1))+(0)(- 4x^{3}y^{3}sin(y^{4})ln(z) - 8yze^{(y^{2})}\cdot tan(x))+(-0.82)(\frac{(x^{3}cos(y^{4}))}{z}- 4e^{(y^{2})}\cdot tan(x))\\&D_{\overrightarrow{u}}(x,y,z)=3.28e^{(y^{2})}\cdot tan(x) + 2.28ze^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (0.82x^{3}cos(y^{4}))}{z}- 1.71x^{2}cos(y^{4})ln(z)\end{align*}$

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Example Question #1284 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{4}cos(x^{3})e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-1,-17,0).\end{align*}$

Possible Answers:

${68y^{3}cos(x^{3})e^{(z)} - 3x^{2}y^{4}sin(x^{3})e^{(z)}}$

${51.1x^{2}y^{4}sin(x^{3})e^{(z)} - 17y^{4}cos(x^{3})e^{(z)} - 68.1y^{3}cos(x^{3})e^{(z)}}$

${204x^{2}y^{3}sin(x^{3})e^{(z)}}$

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(-17)^2+(0)^2}=17.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{17.03}=-0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{17.03}=-1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.06)(3x^{2}y^{4}sin(x^{3})e^{(z)})+(-1)(-4y^{3}cos(x^{3})e^{(z)})+(0)(-1y^{4}cos(x^{3})e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=4y^{3}cos(x^{3})e^{(z)} - 0.18x^{2}y^{4}sin(x^{3})e^{(z)}\end{align*}$

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Example Question #1285 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{4}sin(z^{3}) + y^{4}z^{3}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(-12,-20,0).\end{align*}$

Possible Answers:

${-280y^{3}z^{2}sin(x)}$

${93.3y^{3}sin(z^{3}) + 70y^{4}z^{2}cos(z^{3}) + 93.3y^{3}z^{3}cos(x) + 70y^{4}z^{2}cos(x) - 23.3y^{4}z^{3}sin(x)}$

${2.96y^{3}sin(z^{3}) + 2.96y^{3}z^{3}cos(x) - 0.26y^{4}z^{3}sin(x)}$

${0.51y^{4}z^{3}sin(x) - 3.44y^{3}z^{3}cos(x) - 3.44y^{3}sin(z^{3})}$

Correct answer:

${0.51y^{4}z^{3}sin(x) - 3.44y^{3}z^{3}cos(x) - 3.44y^{3}sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12)^2+(-20)^2+(0)^2}=23.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12}{23.32}=-0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{23.32}=-0.86\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.51)(-1y^{4}z^{3}sin(x))+(-0.86)(4y^{3}sin(z^{3}) + 4y^{3}z^{3}cos(x))+(0)(3y^{4}z^{2}cos(z^{3}) + 3y^{4}z^{2}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.51y^{4}z^{3}sin(x) - 3.44y^{3}z^{3}cos(x) - 3.44y^{3}sin(z^{3})\end{align*}$

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Example Question #1286 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5ln(x^{2})ln(y^{2}) + 2^{(z^{2})}xy^{2}\\&\text{In the direction of }\overrightarrow{v}=(18,20,0).\end{align*}$

Possible Answers:

${74.6\cdot 2^{(z^{2})}yz}$

${\frac{(269ln(y^{2}))}{x}+\frac{ (269ln(x^{2}))}{y}+ 26.9\cdot 2^{(z^{2})}y^{2} + 53.8\cdot 2^{(z^{2})}xy + 37.3\cdot 2^{(z^{2})}xy^{2}z}$

${\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy}$

${\frac{(4.49ln(y^{2}))}{x}+\frac{ (5.48ln(x^{2}))}{y}+ 0.449\cdot 2^{(z^{2})}y^{2} + 1.1\cdot 2^{(z^{2})}xy}$

Correct answer:

${\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(20)^2+(0)^2}=26.91\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{26.91}=0.67\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{26.91}=0.74\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{26.91}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.67)(\frac{(10ln(y^{2}))}{x}+ 2^{(z^{2})}y^{2})+(0.74)(\frac{(10ln(x^{2}))}{y}+ 2\cdot 2^{(z^{2})}xy)+(0)(1.39\cdot 2^{(z^{2})}xy^{2}z)\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy\end{align*}$

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Example Question #1287 : Partial Derivatives

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

Possible Answers:

${\frac{(12.6ln(y^{3})sin(z^{2}))}{x^{3}}-\frac{ (12.6cos(y^{3})sin(x^{3}))}{z^{2}}-\frac{ (19sin(z^{2}))}{(x^{2}y)}-\frac{ (12.6zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (37.9x^{2}cos(x^{3})cos(y^{3}))}{z}-\frac{ (37.9y^{2}sin(x^{3})sin(y^{3}))}{z}}$

${\frac{(12cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (4ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (12zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (12x^{2}cos(x^{3})cos(y^{3}))}{z}}$

${\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}}$

${\frac{(75.8zcos(z^{2}))}{(x^{3}y)}+\frac{ (114x^{2}y^{2}cos(x^{3})sin(y^{3}))}{z^{2}}}$

Correct answer:

${\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}}$

Explanation:

$\begin{align*}\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{6.32}=0.32\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{6.32}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{6.32}=-0.95\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.32)(\frac{(2ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (6x^{2}cos(x^{3})cos(y^{3}))}{z})+(0)(-\frac{ (3sin(z^{2}))}{(x^{2}y)}-\frac{ (6y^{2}sin(x^{3})sin(y^{3}))}{z})+(-0.95)(-\frac{ (2cos(y^{3})sin(x^{3}))}{z^{2}}-\frac{ (2zcos(z^{2})ln(y^{3}))}{x^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}\end{align*}$

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Example Question #1288 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=yln(x^{2})ln(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,13,-16).\end{align*}$

Possible Answers:

${13ln(x^{2})ln(z^{4}) -\frac{ (64yln(x^{2}))}{z}}$

${0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}}$

${20.6ln(x^{2})ln(z^{4}) +\frac{ (82.5yln(x^{2}))}{z}+\frac{ (41.2yln(z^{4}))}{x}}$

${\frac{165.0}{(xz)}}$

Correct answer:

${0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(13)^2+(-16)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{13}{20.62}=0.63\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{20.62}=-0.78\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(2yln(z^{4}))}{x})+(0.63)(ln(x^{2})ln(z^{4}))+(-0.78)(\frac{(4yln(x^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}\end{align*}$

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Example Question #1289 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3ln(z^{2})tan(x) + x^{2}y^{3}ln(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-13,4).\end{align*}$

Possible Answers:

${\frac{(163xy^{2})}{z}}$

${\frac{(24.0tan(x))}{z}- 39x^{2}y^{2}ln(z^{2}) +\frac{ (8x^{2}y^{3})}{z}}$

${\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}}$

${\frac{(81.6tan(x))}{z}+ 40.8ln(z^{2})\cdot (tan(x)^{2} + 1) + 40.8x^{2}y^{2}ln(z^{2}) +\frac{ (27.2x^{2}y^{3})}{z}+ 27.2xy^{3}ln(z^{2})}$

Correct answer:

${\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(4)^2}=13.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.6}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{13.6}=-0.96\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{13.6}=0.29\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3ln(z^{2})\cdot (tan(x)^{2} + 1) + 2xy^{3}ln(z^{2}))+(-0.96)(3x^{2}y^{2}ln(z^{2}))+(0.29)(\frac{(6.0tan(x))}{z}+\frac{ (2x^{2}y^{3})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}\end{align*}$

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Example Question #1290 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(x^{4}cos(z))}{y}\\&\text{In the direction of }\overrightarrow{v}=(0,20,-7).\end{align*}$

Possible Answers:

${\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}}$

${-\frac{(84.8x^{3}sin(z))}{y^{2}}}$

${\frac{(20x^{4}cos(z))}{y^{2}}-\frac{ (7x^{4}sin(z))}{y}}$

${\frac{(21.2x^{4}cos(z))}{y^{2}}-\frac{ (84.8x^{3}cos(z))}{y}+\frac{ (21.2x^{4}sin(z))}{y}}$

Correct answer:

${\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(20)^2+(-7)^2}=21.19\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.19}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{21.19}=0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{21.19}=-0.33\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(4x^{3}cos(z))}{y})+(0.94)(\frac{(x^{4}cos(z))}{y^{2}})+(-0.33)(\frac{(x^{4}sin(z))}{y})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Question #1281 : Partial Derivatives

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