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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1391 : Partial Derivatives

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

Possible Answers:

${134x^{3}y^{3}sin(z^{3}) - 15.5\cdot 4^yz - 11.2\cdot 4^y + 101x^{4}y^{2}sin(z^{3}) + 101x^{4}y^{3}z^{2}cos(z^{3})}$

${11.5x^{3}y^{3}sin(z^{3}) - 0.0449\cdot 4^yz + 0.292x^{4}y^{2}sin(z^{3})}$

${0.25\cdot 4^yz - 11.8x^{3}y^{3}sin(z^{3}) - 1.62x^{4}y^{2}sin(z^{3})}$

${1211x^{3}y^{2}z^{2}cos(z^{3})}$

Correct answer:

${0.25\cdot 4^yz - 11.8x^{3}y^{3}sin(z^{3}) - 1.62x^{4}y^{2}sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11)^2+(-2)^2+(0)^2}=11.18\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11}{11.18}=-0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{11.18}=-0.18\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{11.18}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.98)(12x^{3}y^{3}sin(z^{3}))+(-0.18)(9x^{4}y^{2}sin(z^{3}) - 1.39\cdot 4^yz)+(0)(9x^{4}y^{3}z^{2}cos(z^{3}) - 1\cdot 4^y)\\&D_{\overrightarrow{u}}(x,y,z)=0.25\cdot 4^yz - 11.8x^{3}y^{3}sin(z^{3}) - 1.62x^{4}y^{2}sin(z^{3})\end{align*}$

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Example Question #1392 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{ cos(y^{4})}{x}- 3xz^{4}e^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-4,19,0).\end{align*}$

Possible Answers:

${\frac{(19.4cos(y^{4}))}{x^{2}}- 58.3z^{4}e^{(y^{2})} +\frac{ (77.7y^{3}sin(y^{4}))}{x}- 233xz^{3}e^{(y^{2})} - 117xyz^{4}e^{(y^{2})}}$

${\frac{(0.0441cos(y^{4}))}{x^{2}}- 0.132z^{4}e^{(y^{2})} +\frac{ (3.84y^{3}sin(y^{4}))}{x}- 5.76xyz^{4}e^{(y^{2})}}$

${0.63z^{4}e^{(y^{2})} -\frac{ (0.21cos(y^{4}))}{x^{2}}+\frac{ (3.92y^{3}sin(y^{4}))}{x}- 5.88xyz^{4}e^{(y^{2})}}$

${-466yz^{3}e^{(y^{2})}}$

Correct answer:

${0.63z^{4}e^{(y^{2})} -\frac{ (0.21cos(y^{4}))}{x^{2}}+\frac{ (3.92y^{3}sin(y^{4}))}{x}- 5.88xyz^{4}e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(19)^2+(0)^2}=19.42\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{19.42}=-0.21\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{19.42}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{19.42}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.21)(\frac{cos(y^{4})}{x^{2}}- 3z^{4}e^{(y^{2})})+(0.98)(\frac{(4y^{3}sin(y^{4}))}{x}- 6xyz^{4}e^{(y^{2})})+(0)(-12xz^{3}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=0.63z^{4}e^{(y^{2})} -\frac{ (0.21cos(y^{4}))}{x^{2}}+\frac{ (3.92y^{3}sin(y^{4}))}{x}- 5.88xyz^{4}e^{(y^{2})}\end{align*}$

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Example Question #1393 : Partial Derivatives

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

Possible Answers:

${-\frac{(452\cdot 4^{(y^{4})}y^{3}zcos(z^{2}))}{x}}$

${- 11.5y^{3}e^{(z)} -\frac{ (0.08\cdot 4^{(y^{4})}sin(z^{2}))}{x}- 5.33\cdot 4^{(y^{4})}y^{3}ln(x^{2})sin(z^{2})}$

${- 245y^{3}e^{(z)} - 61.2y^{4}e^{(z)} -\frac{ (40.8\cdot 4^{(y^{4})}sin(z^{2}))}{x}- 113\cdot 4^{(y^{4})}y^{3}ln(x^{2})sin(z^{2}) - 40.8\cdot 4^{(y^{4})}zcos(z^{2})ln(x^{2})}$

${\frac{(0.4\cdot 4^{(y^{4})}sin(z^{2}))}{x}- 11.8y^{3}e^{(z)} - 5.43\cdot 4^{(y^{4})}y^{3}ln(x^{2})sin(z^{2})}$

Correct answer:

${\frac{(0.4\cdot 4^{(y^{4})}sin(z^{2}))}{x}- 11.8y^{3}e^{(z)} - 5.43\cdot 4^{(y^{4})}y^{3}ln(x^{2})sin(z^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(20)^2+(0)^2}=20.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{20.4}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{20.4}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.4}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(-\frac{(2\cdot 4^{(y^{4})}sin(z^{2}))}{x})+(0.98)(- 12y^{3}e^{(z)} - 5.55\cdot 4^{(y^{4})}y^{3}ln(x^{2})sin(z^{2}))+(0)(- 3y^{4}e^{(z)} - 2\cdot 4^{(y^{4})}zcos(z^{2})ln(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.4\cdot 4^{(y^{4})}sin(z^{2}))}{x}- 11.8y^{3}e^{(z)} - 5.43\cdot 4^{(y^{4})}y^{3}ln(x^{2})sin(z^{2})\end{align*}$

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Example Question #1394 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos(y^{3})sin(z^{3}) + 3^{(y^{3})}\cdot 4^ze^{(x^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-1,0,-9).\end{align*}$

Possible Answers:

${1.36\cdot 3^{(y^{3})}\cdot 4^ze^{(x^{2})} + 2.94z^{2}cos(y^{3})cos(z^{3}) + 0.0242\cdot 3^{(y^{3})}\cdot 4^zxe^{(x^{2})}}$

${12.6\cdot 3^{(y^{3})}\cdot 4^ze^{(x^{2})} + 27.2z^{2}cos(y^{3})cos(z^{3}) - 27.2y^{2}sin(y^{3})sin(z^{3}) + 29.9\cdot 3^{(y^{3})}\cdot 4^zy^{2}e^{(x^{2})} + 18.1\cdot 3^{(y^{3})}\cdot 4^zxe^{(x^{2})}}$

${- 1.37\cdot 3^{(y^{3})}\cdot 4^ze^{(x^{2})} - 2.97z^{2}cos(y^{3})cos(z^{3}) - 0.22\cdot 3^{(y^{3})}\cdot 4^zxe^{(x^{2})}}$

${82.8\cdot 3^{(y^{3})}\cdot 4^zxy^{2}e^{(x^{2})}}$

Correct answer:

${- 1.37\cdot 3^{(y^{3})}\cdot 4^ze^{(x^{2})} - 2.97z^{2}cos(y^{3})cos(z^{3}) - 0.22\cdot 3^{(y^{3})}\cdot 4^zxe^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(0)^2+(-9)^2}=9.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{9.06}=-0.11\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{9.06}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-9}{9.06}=-0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.11)(2\cdot 3^{(y^{3})}\cdot 4^zxe^{(x^{2})})+(0)(3.3\cdot 3^{(y^{3})}\cdot 4^zy^{2}e^{(x^{2})} - 3y^{2}sin(y^{3})sin(z^{3}))+(-0.99)(1.39\cdot 3^{(y^{3})}\cdot 4^ze^{(x^{2})} + 3z^{2}cos(y^{3})cos(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=- 1.37\cdot 3^{(y^{3})}\cdot 4^ze^{(x^{2})} - 2.97z^{2}cos(y^{3})cos(z^{3}) - 0.22\cdot 3^{(y^{3})}\cdot 4^zxe^{(x^{2})}\end{align*}$

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Example Question #1395 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{4}cos(x^{3})cos(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(18,16,-6).\end{align*}$

Possible Answers:

${-1199x^{2}y^{3}z^{3}sin(x^{3})sin(z^{4})}$

${74.5x^{2}y^{4}cos(z^{4})sin(x^{3}) - 99.3y^{3}cos(x^{3})cos(z^{4}) + 99.3y^{4}z^{3}cos(x^{3})sin(z^{4})}$

${2.19x^{2}y^{4}cos(z^{4})sin(x^{3}) - 2.56y^{3}cos(x^{3})cos(z^{4}) - 0.96y^{4}z^{3}cos(x^{3})sin(z^{4})}$

${1.6x^{2}y^{4}cos(z^{4})sin(x^{3}) - 1.64y^{3}cos(x^{3})cos(z^{4}) + 0.23y^{4}z^{3}cos(x^{3})sin(z^{4})}$

Correct answer:

${2.19x^{2}y^{4}cos(z^{4})sin(x^{3}) - 2.56y^{3}cos(x^{3})cos(z^{4}) - 0.96y^{4}z^{3}cos(x^{3})sin(z^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(16)^2+(-6)^2}=24.82\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{24.82}=0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{24.82}=0.64\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{24.82}=-0.24\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.73)(3x^{2}y^{4}cos(z^{4})sin(x^{3}))+(0.64)(-4y^{3}cos(x^{3})cos(z^{4}))+(-0.24)(4y^{4}z^{3}cos(x^{3})sin(z^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=2.19x^{2}y^{4}cos(z^{4})sin(x^{3}) - 2.56y^{3}cos(x^{3})cos(z^{4}) - 0.96y^{4}z^{3}cos(x^{3})sin(z^{4})\end{align*}$

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Example Question #1396 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 3y^{3}sin(z^{3}) - xe^{(z^{4})}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(-18,5,-1).\end{align*}$

Possible Answers:

${18e^{(z^{4})}sin(y) - 45y^{2}sin(z^{3}) + 9y^{3}z^{2}cos(z^{3}) - 5xe^{(z^{4})}cos(y) + 4xz^{3}e^{(z^{4})}sin(y)}$

${- 168y^{2}sin(z^{3}) - 18.7e^{(z^{4})}sin(y) - 168y^{3}z^{2}cos(z^{3}) - 18.7xe^{(z^{4})}cos(y) - 74.8xz^{3}e^{(z^{4})}sin(y)}$

${-74.8z^{3}e^{(z^{4})}cos(y)}$

${0.96e^{(z^{4})}sin(y) - 2.43y^{2}sin(z^{3}) + 0.45y^{3}z^{2}cos(z^{3}) - 0.27xe^{(z^{4})}cos(y) + 0.2xz^{3}e^{(z^{4})}sin(y)}$

Correct answer:

${0.96e^{(z^{4})}sin(y) - 2.43y^{2}sin(z^{3}) + 0.45y^{3}z^{2}cos(z^{3}) - 0.27xe^{(z^{4})}cos(y) + 0.2xz^{3}e^{(z^{4})}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(5)^2+(-1)^2}=18.71\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{18.71}=-0.96\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{18.71}=0.27\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-1}{18.71}=-0.05\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.96)(-1e^{(z^{4})}sin(y))+(0.27)(- 9y^{2}sin(z^{3}) - xe^{(z^{4})}cos(y))+(-0.05)(- 9y^{3}z^{2}cos(z^{3}) - 4xz^{3}e^{(z^{4})}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.96e^{(z^{4})}sin(y) - 2.43y^{2}sin(z^{3}) + 0.45y^{3}z^{2}cos(z^{3}) - 0.27xe^{(z^{4})}cos(y) + 0.2xz^{3}e^{(z^{4})}sin(y)\end{align*}$

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Example Question #1397 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{ (4yz)}{x}-\frac{ (z^{4}e^{(x^{3})})}{y}\\&\text{In the direction of }\overrightarrow{v}=(-6,-18,7).\end{align*}$

Possible Answers:

${\frac{(3.56z)}{x}-\frac{ (1.4y)}{x}-\frac{ (1.2yz)}{x^{2}}-\frac{ (1.4z^{3}e^{(x^{3})})}{y}-\frac{ (0.89z^{4}e^{(x^{3})})}{y^{2}}+\frac{ (0.9x^{2}z^{4}e^{(x^{3})})}{y}}$

${\frac{80.9}{x^{2}}+\frac{ (243x^{2}z^{3}e^{(x^{3})})}{y^{2}}}$

${\frac{(0.36yz)}{x^{2}}-\frac{ (3.17z)}{x}-\frac{ (0.49y)}{x}-\frac{ (0.49z^{3}e^{(x^{3})})}{y}+\frac{ (0.792z^{4}e^{(x^{3})})}{y^{2}}-\frac{ (0.27x^{2}z^{4}e^{(x^{3})})}{y}}$

${\frac{(80.9yz)}{x^{2}}-\frac{ (80.9z)}{x}-\frac{ (80.9y)}{x}-\frac{ (80.9z^{3}e^{(x^{3})})}{y}+\frac{ (20.2z^{4}e^{(x^{3})})}{y^{2}}-\frac{ (60.7x^{2}z^{4}e^{(x^{3})})}{y}}$

Correct answer:

${\frac{(3.56z)}{x}-\frac{ (1.4y)}{x}-\frac{ (1.2yz)}{x^{2}}-\frac{ (1.4z^{3}e^{(x^{3})})}{y}-\frac{ (0.89z^{4}e^{(x^{3})})}{y^{2}}+\frac{ (0.9x^{2}z^{4}e^{(x^{3})})}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(-18)^2+(7)^2}=20.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{20.22}=-0.3\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{20.22}=-0.89\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{20.22}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.3)(\frac{(4yz)}{x^{2}}-\frac{ (3x^{2}z^{4}e^{(x^{3})})}{y})+(-0.89)(\frac{(z^{4}e^{(x^{3})})}{y^{2}}-\frac{ (4z)}{x})+(0.35)(-\frac{ (4y)}{x}-\frac{ (4z^{3}e^{(x^{3})})}{y})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.56z)}{x}-\frac{ (1.4y)}{x}-\frac{ (1.2yz)}{x^{2}}-\frac{ (1.4z^{3}e^{(x^{3})})}{y}-\frac{ (0.89z^{4}e^{(x^{3})})}{y^{2}}+\frac{ (0.9x^{2}z^{4}e^{(x^{3})})}{y}\end{align*}$

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Example Question #1398 : Partial 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:

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

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

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

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 #1399 : Partial 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:

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

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

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

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

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 #1400 : Partial 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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Calculus 3 : Partial Derivatives

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