Directional Derivatives - AP Calculus BC

Card 0 of 1760

Question

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=x^2y+xz^{10}+xy^2z^5$ in the direction of $\vec{v}=(1,2,3)$ .

Answer

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14} eq1$

$\vec{u}=\frac{1}{\sqrt{14}}\vec{v}=\Big(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\ \ D_{\vec{u}}f(x,y,z)= \ \ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+0+2xyz^5)+\frac{3}{\sqrt{14}}(0+10xz^9+5xy^2z^4)$

$\ \ D_{\vec{u}}f(x,y,z)= \ \ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+2xyz^5)+\frac{3}{\sqrt{14}}(10xz^9+5xy^2z^4)$

$\ \ D_{\vec{u}}f(x,y,z)= \ \ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

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Question

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=z^2y\ln(x)+xy^2z^3$ in the direction of $\vec{v}=(0,1,1)$ .

Answer

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{0^2+1^2+1^2}=\sqrt{0+1+1}=\sqrt{2} eq1$

$\vec{u}=\frac{1}{\sqrt{2}}\vec{v}=\Big(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\ \ D_{\vec{u}}f(x,y,z)= \ \ 0(\frac{yz^2}{x}+y^2z^3)+\frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\ \ D_{\vec{u}}f(x,y,z)= \ \ \frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\ \ D_{\vec{u}}f(x,y,z)= \ \ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-5,1,4)\text{ where }f(x,y,z)=y^3cos{(x^2)}sin{(z)}\&\text{In the direction of }\overrightarrow{v}=(-2,20,-14).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-5,1,4)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(20)^2+(-14)^2}=24.49\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{24.49}=-0.08\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{24.49}=0.82\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{24.49}=-0.57\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-5,1,4)=(-0.08)(-2xy^3sin{(x^2)}sin{(z)})+(0.82)(3y^2cos{(x^2)}sin{(z)})+(-0.57)(y^3cos{(x^2)}cos{(z)})=-1.55\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(4,2,-1)\text{ where }f(x,y,z)=x^3sin{(y^2)}sin{(z^3)}\&\text{In the direction of }\overrightarrow{v}=(-16,4,-11).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(4,2,-1)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(4)^2+(-11)^2}=19.82\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{19.82}=-0.81\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{19.82}=0.2\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{19.82}=-0.55\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(4,2,-1)=(-0.81)(3x^2sin{(y^2)}sin{(z^3)})+(0.2)(2x^3ycos{(y^2)}sin{(z^3)})+(-0.55)(3x^3z^2cos{(z^3)}sin{(y^2)})=47.3\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-1,1,-3)\text{ where }f(x,y,z)=x^4z^4cos{(y^2)}\&\text{In the direction of }\overrightarrow{v}=(13,20,9).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-1,1,-3)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(20)^2+(9)^2}=25.5\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{25.5}=0.51\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{25.5}=0.78\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{25.5}=0.35\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-1,1,-3)=(0.51)(4x^3z^4cos{(y^2)})+(0.78)(-2x^4yz^4sin{(y^2)})+(0.35)(4x^4z^3cos{(y^2)})=-216.8\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-2,-1,-3)\text{ where }f(x,y,z)=cos{(x^2)}cos{(z^3)}sin{(y^3)}\&\text{In the direction of }\overrightarrow{v}=(8,6,-19).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-1,-3)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(6)^2+(-19)^2}=21.47\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{21.47}=0.37\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{6}{21.47}=0.28\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{21.47}=-0.88\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-2,-1,-3)=(0.37)(-2xcos{(z^3)}sin{(x^2)}sin{(y^3)})+(0.28)(3y^2cos{(x^2)}cos{(y^3)}cos{(z^3)})+(-0.88)(-3z^2cos{(x^2)}sin{(y^3)}sin{(z^3)})=-12.76\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-3,1,2)\text{ where }f(x,y,z)=y^2z^4sin{(x^2)}\&\text{In the direction of }\overrightarrow{v}=(-14,-9,-2).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-3,1,2)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(-9)^2+(-2)^2}=16.76\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{16.76}=-0.84\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{16.76}=-0.54\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{16.76}=-0.12\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-3,1,2)=(-0.84)(2xy^2z^4cos{(x^2)})+(-0.54)(2yz^4sin{(x^2)})+(-0.12)(4y^2z^3sin{(x^2)})=-81.7\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-3,-3,-2)\text{ where }f(x,y,z)=2zcos{(x^2)}cos{(y^2)}\&\text{In the direction of }\overrightarrow{v}=(9,10,-18).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-3,-3,-2)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(9)^2+(10)^2+(-18)^2}=22.47\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{9}{22.47}=0.4\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{22.47}=0.44\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{22.47}=-0.8\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-3,-3,-2)=(0.4)(-4xzcos{(y^2)}sin{(x^2)})+(0.44)(-4yzcos{(x^2)}sin{(y^2)})+(-0.8)(2cos{(x^2)}cos{(y^2)})=6.29\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-2,-1,3)\text{ where }f(x,y,z)=2^{(1/y^2)}e^{(1/x^2)}cos{(z)} + 2^ye^{(x)}cos{(z)}\&\text{In the direction of }\overrightarrow{v}=(-8,-8,-7).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-1,3)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(-8)^2+(-7)^2}=13.3\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{13.3}=-0.6\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{13.3}=-0.6\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{13.3}=-0.53\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-2,-1,3)=(-0.6)(2^ye^{(x)}cos{(z)} - {((2)2^{(1/y^2)}e^{(1/x^2)}cos{(z)})}/x^3)+(-0.6)(2^ye^{(x)}ln{(2)}cos{(z)} - {((2)2^{(1/y^2)}e^{(1/x^2)}ln{(2)}cos{(z)})}/y^3)+(-0.53)(- 2^ye^{(x)}sin{(z)} - 2^{(1/y^2)}e^{(1/x^2)}sin{(z)})=2.77\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(1,3,2)\text{ where }f(x,y,z)=2^zsin{(x^2)}sin{(y^3)}\&\text{In the direction of }\overrightarrow{v}=(-18,1,-7).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(1,3,2)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(1)^2+(-7)^2}=19.34\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{19.34}=-0.93\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{19.34}=0.05\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{19.34}=-0.36\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(1,3,2)=(-0.93)((2)2^zxcos{(x^2)}sin{(y^3)})+(0.05)((3)2^zy^2cos{(y^3)}sin{(x^2)})+(-0.36)(2^zsin{(x^2)}sin{(y^3)}ln{(2)})=-6.03\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-2,-3,-1)\text{ where }f(x,y,z)=y^2sin{(x^3)}e^{(z^2)}\&\text{In the direction of }\overrightarrow{v}=(-18,-8,-19).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-3,-1)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(-8)^2+(-19)^2}=27.37\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{27.37}=-0.66\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{27.37}=-0.29\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{27.37}=-0.69\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-2,-3,-1)=(-0.66)(3x^2y^2cos{(x^3)}e^{(z^2)})+(-0.29)(2y*sin{(x^3)}e^{(z^2)})+(-0.69)(2y^2zsin{(x^3)}e^{(z^2)})=-10.23\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-3,2,-2)\text{ where }f(x,y,z)=2^zy^2cos{(x^3)}\&\text{In the direction of }\overrightarrow{v}=(6,10,3).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-3,2,-2)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(10)^2+(3)^2}=12.04\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{12.04}=0.5\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{12.04}=0.83\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{12.04}=0.25\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-3,2,-2)=(0.5)(-32^zx^2y^2sin{(x^3)})+(0.83)(22^zycos{(x^3)})+(0.25)(2^zy^2cos{(x^3)}ln{(2)})=12.57\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-1,-1,2)\text{ where }f(x,y,z)=z^2e^{(x^2)}e^{(y)}\&\text{In the direction of }\overrightarrow{v}=(4,12,-5).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-1,-1,2)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(4)^2+(12)^2+(-5)^2}=13.6\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{4}{13.6}=0.29\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{13.6}=0.88\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{13.6}=-0.37\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-1,-1,2)=(0.29)(2x*z^2e^{(x^2)}e^{(y)})+(0.88)(z^2e^{(x^2)}e^{(y)})+(-0.37)(2ze^{(x^2)}e^{(y)})=-0.29\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(-2,-3,2)\text{ where }f(x,y,z)=x^3sin{(y^2)}cos{(z)}\&\text{In the direction of }\overrightarrow{v}=(6,-1,-14).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-3,2)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-1)^2+(-14)^2}=15.26\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{15.26}=0.39\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-1}{15.26}=-0.07\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{15.26}=-0.92\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(-2,-3,2)=(0.39)(3x^2sin{(y^2)}cos{(z)})+(-0.07)(2x^3y*cos{(y^2)}cos{(z)})+(-0.92)(-x^3sin{(y^2)}sin{(z)})=-4.75\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^2cos{(y^2)}sin{(x^2)}\&\text{In the direction of }\overrightarrow{v}=(-14,11,15).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(11)^2+(15)^2}=23.28\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{23.28}=-0.6\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{23.28}=0.47\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{23.28}=0.64\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(2x*z^2cos{(x^2)}cos{(y^2)})+(0.47)(-2yz^2sin{(x^2)}sin{(y^2)})+(0.64)(2zcos{(y^2)}sin{(x^2)})\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ycos{(x^3)}\&\text{In the direction of }\overrightarrow{v}=(7,9,6).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(9)^2+(6)^2}=12.88\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{12.88}=0.54\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{12.88}=0.7\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{12.88}=0.47\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(-3x^2ysin{(x^3)})+(0.7)(cos{(x^3)})+(0.47)(0)\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5cos{(y^2)}ln{(z)}\&\text{In the direction of }\overrightarrow{v}=(-12,-13,-19).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12)^2+(-13)^2+(-19)^2}=25.96\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12}{25.96}=-0.46\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{25.96}=-0.5\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{25.96}=-0.73\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.46)(0)+(-0.5)(-10ysin{(y^2)}ln{(z)})+(-0.73)({(5cos{(y^2)})}/z)\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=9x^4e^{(z^2)}sin{(y)}\&\text{In the direction of }\overrightarrow{v}=(7,-10,-8).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(-10)^2+(-8)^2}=14.59\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{14.59}=0.48\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{14.59}=-0.69\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{14.59}=-0.55\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0.48)(36x^3e^{(z^2)}sin{(y)})+(-0.69)(9x^4e^{(z^2)}cos{(y)})+(-0.55)(18x^4ze^{(z^2)}sin{(y)})\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3y^2zsin{(x)}\&\text{In the direction of }\overrightarrow{v}=(14,17,11).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(17)^2+(11)^2}=24.62\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{24.62}=0.57\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{24.62}=0.69\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{24.62}=0.45\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0.57)(3y^2zcos{(x)})+(0.69)(6yzsin{(x)})+(0.45)(3y^2sin{(x)})\end{align}$

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Question

$\begin{align}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=7y^2z^3cos{(x^3)}\&\text{In the direction of }\overrightarrow{v}=(14,11,18).\end{align}$

Answer

$\begin{align}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(11)^2+(18)^2}=25.32\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{25.32}=0.55\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{25.32}=0.43\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{25.32}=0.71\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(-21x^2y^2z^3sin{(x^3)})+(0.43)(14yz^3cos{(x^3)})+(0.71)(21y^2z^2cos{(x^3)})\&D_{\overrightarrow{u}}(x,y,z)={14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}\end{align}$

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