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Calculus 3 : Directional Derivatives

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Example Question #291 : Directional Derivatives

Determine the directional derivative of the following function in the direction of $\vec{v}=\left \langle 1, 3, 1\right \rangle$ :

f(x, y, z)=zx+xy+z^2y^2

Possible Answers:

$\frac{1}{\sqrt{11}}(4x+ z+y+8yz^2)$

4x+ z+y+6yz^2+2zy^2

$\frac{1}{\sqrt{11}}(4x+ z+y+6yz^2+2zy^2)$

$\frac{1}{\sqrt{11}}(x+ z+y+6yz^2+2zy^2)$

2x+ z+y+6yz^2+2zy^2

Correct answer:

$\frac{1}{\sqrt{11}}(4x+ z+y+6yz^2+2zy^2)$

Explanation:

The directional derivative is given by

$D_{\vec{u}}f(x, y, z)=f_x(x, y, z)a+f_y(x, y, z)b+f_c(x, y, z)c$ , where $\left \langle a, b, c\right \rangle$ is the unit directional vector.

Now, we must find the gradient of f, $\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$ :

f_x=z+y , f_y=x+2yz^2 , f_z=x+2zy^2

The partial derivatives were found by keeping the other variables constant, and using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the unit directional vector, which is given by

$\frac{\vec{v}}{\left | \vec{v}\right |}=\frac{\left \langle 1, 3, 1\right \rangle}{\sqrt{1^2+3^2+1^2}}=\left \langle \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}} \right \rangle$

Plugging all of this into the above formula, and after simplifying, we get

$\frac{1}{\sqrt{11}}(4x+ z+y+6yz^2+2zy^2)$

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Example Question #292 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{3}cos(z^{4})e^{(x^{4})}\\&\text{In the direction of }\overrightarrow{v}=(0,-15,-12).\end{align*}$

Possible Answers:

${48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 45y^{2}cos(z^{4})e^{(x^{4})}}$

${2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}}$

${57.6y^{2}cos(z^{4})e^{(x^{4})} + 76.8x^{3}y^{3}cos(z^{4})e^{(x^{4})} - 76.8y^{3}z^{3}sin(z^{4})e^{(x^{4})}}$

${-922x^{3}y^{2}z^{3}sin(z^{4})e^{(x^{4})}}$

Correct answer:

${2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(-12)^2}=19.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{19.21}=-0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{19.21}=-0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}y^{3}cos(z^{4})e^{(x^{4})})+(-0.78)(3y^{2}cos(z^{4})e^{(x^{4})})+(-0.62)(-4y^{3}z^{3}sin(z^{4})e^{(x^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=2.48y^{3}z^{3}sin(z^{4})e^{(x^{4})} - 2.34y^{2}cos(z^{4})e^{(x^{4})}\end{align*}$

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Example Question #292 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{4})}cos(x^{4})cos(z^{3}) - sin(x^{3})e^{(y^{2})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-12,0,6).\end{align*}$

Possible Answers:

${708\cdot 3^{(y^{4})}x^{3}y^{3}z^{2}sin(x^{4})sin(z^{3}) - 161x^{2}yzcos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

${3.56\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 1.35\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 0.9zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 2.67x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

${48\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 18\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 12zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 36x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

${59\cdot 3^{(y^{4})}y^{3}cos(x^{4})cos(z^{3}) - 53.7\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 40.3\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 26.8ysin(x^{3})e^{(y^{2})}e^{(z^{2})} - 26.8zsin(x^{3})e^{(y^{2})}e^{(z^{2})} - 40.3x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

Correct answer:

${3.56\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 1.35\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 0.9zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 2.67x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{13.42}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{13.42}=0.45\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.89)(- 4\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 3x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})})+(0)(4.39\cdot 3^{(y^{4})}y^{3}cos(x^{4})cos(z^{3}) - 2ysin(x^{3})e^{(y^{2})}e^{(z^{2})})+(0.45)(- 3\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 2zsin(x^{3})e^{(y^{2})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=3.56\cdot 3^{(y^{4})}x^{3}cos(z^{3})sin(x^{4}) - 1.35\cdot 3^{(y^{4})}z^{2}cos(x^{4})sin(z^{3}) - 0.9zsin(x^{3})e^{(y^{2})}e^{(z^{2})} + 2.67x^{2}cos(x^{3})e^{(y^{2})}e^{(z^{2})}\end{align*}$

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Example Question #291 : Directional 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:

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

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

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

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

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 #294 : Directional 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:

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

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

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

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

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:

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

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Example Question #291 : Directional 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:

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

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

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

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

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 #295 : Directional 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 #291 : Directional 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:

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

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

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

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

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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Example Question #297 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^{(y^{3})}e^{(x^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,10,-20).\end{align*}$

Possible Answers:

${-0.842\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

${- 44.7\cdot 4^{(y^{3})}xe^{(x^{2})} - 93\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

${-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

Correct answer:

${-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(-20)^2}=22.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.36}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{22.36}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{22.36}=-0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2\cdot 4^{(y^{3})}xe^{(x^{2})})+(0.45)(-4.16\cdot 4^{(y^{3})}y^{2}e^{(x^{2})})+(-0.89)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}\end{align*}$

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Example Question #298 : Directional Derivatives

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

Possible Answers:

${18.9e^{(y^{3})}e^{(x)}cos(z) + 18.9e^{(y^{3})}e^{(x)}sin(z) + 56.6y^{2}e^{(y^{3})}e^{(x)}sin(z)}$

${1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)}$

${0.723e^{(y^{3})}e^{(x)}cos(z) + 0.843y^{2}e^{(y^{3})}e^{(x)}sin(z)}$

${56.6y^{2}e^{(y^{3})}e^{(x)}cos(z)}$

Correct answer:

${1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(-16)^2}=18.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.87}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{18.87}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{18.87}=-0.85\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(e^{(y^{3})}e^{(x)}sin(z))+(0.53)(3y^{2}e^{(y^{3})}e^{(x)}sin(z))+(-0.85)(e^{(y^{3})}e^{(x)}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)\end{align*}$

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Calculus 3 : Directional Derivatives

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