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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #74 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=e^{(z)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-8,0,-13).\end{align*}$

Possible Answers:

${0.52e^{(z)}sin{(x)} - 0.85e^{(z)}cos{(x)}}$

${15.3e^{(z)}cos{(x)} - 15.3e^{(z)}sin{(x)}}$

${8e^{(z)}sin{(x)} - 13e^{(z)}cos{(x)}}$

Correct answer:

${0.52e^{(z)}sin{(x)} - 0.85e^{(z)}cos{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(0)^2+(-13)^2}=15.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{15.26}=-0.52\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.26}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-13}{15.26}=-0.85\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.52)(-1e^{(z)}sin{(x)})+(0)(0.0)+(-0.85)(e^{(z)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.52e^{(z)}sin{(x)} - 0.85e^{(z)}cos{(x)}\end{align*}$

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Example Question #72 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(x^3)}cos{(y^3)}cos{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(13,0,19).\end{align*}$

Possible Answers:

${1.66zcos{(x^3)}cos{(y^3)}sin{(z^2)} + 1.68x^2cos{(y^3)}cos{(z^2)}sin{(x^3)}}$

${38zcos{(x^3)}cos{(y^3)}sin{(z^2)} + 39x^2cos{(y^3)}cos{(z^2)}sin{(x^3)}}$

${46zcos{(x^3)}cos{(y^3)}sin{(z^2)} + 69.1x^2cos{(y^3)}cos{(z^2)}sin{(x^3)} + 69.1y^2cos{(x^3)}cos{(z^2)}sin{(y^3)}}$

${414x^2y^2zsin{(x^3)}sin{(y^3)}sin{(z^2)}}$

Correct answer:

${1.66zcos{(x^3)}cos{(y^3)}sin{(z^2)} + 1.68x^2cos{(y^3)}cos{(z^2)}sin{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(19)^2}=23.02\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{23.02}=0.56\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{23.02}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{23.02}=0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.56)(3x^2cos{(y^3)}cos{(z^2)}sin{(x^3)})+(0)(3y^2cos{(x^3)}cos{(z^2)}sin{(y^3)})+(0.83)(2zcos{(x^3)}cos{(y^3)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.66zcos{(x^3)}cos{(y^3)}sin{(z^2)} + 1.68x^2cos{(y^3)}cos{(z^2)}sin{(x^3)}\end{align*}$

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Example Question #71 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2sin{(x^2)}sin{(y^3)}sin{(z^2)} - 2^xsin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(1,18,0).\end{align*}$

Possible Answers:

${433xy^2zcos{(x^2)}cos{(y^3)}cos{(z^2)}}$

${6y^2cos{(y^3)}sin{(x^2)}sin{(z^2)} - 0.0416\cdot2^xsin{(y)} - 1\cdot2^xcos{(y)} + 0.24xcos{(x^2)}sin{(y^3)}sin{(z^2)}}$

${108y^2cos{(y^3)}sin{(x^2)}sin{(z^2)} - 0.693\cdot2^xsin{(y)} - 18\cdot2^xcos{(y)} + 4xcos{(x^2)}sin{(y^3)}sin{(z^2)}}$

${108y^2cos{(y^3)}sin{(x^2)}sin{(z^2)} - 12.5\cdot2^xsin{(y)} - 18\cdot2^xcos{(y)} + 72.1xcos{(x^2)}sin{(y^3)}sin{(z^2)} + 72.1zcos{(z^2)}sin{(x^2)}sin{(y^3)}}$

Correct answer:

${6y^2cos{(y^3)}sin{(x^2)}sin{(z^2)} - 0.0416\cdot2^xsin{(y)} - 1\cdot2^xcos{(y)} + 0.24xcos{(x^2)}sin{(y^3)}sin{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(18)^2+(0)^2}=18.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{18.03}=0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{18.03}=1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.06)(4xcos{(x^2)}sin{(y^3)}sin{(z^2)} - 0.693\cdot2^xsin{(y)})+(1)(6y^2cos{(y^3)}sin{(x^2)}sin{(z^2)} - 1\cdot2^xcos{(y)})+(0)(4zcos{(z^2)}sin{(x^2)}sin{(y^3)})\\&D_{\overrightarrow{u}}(x,y,z)=6y^2cos{(y^3)}sin{(x^2)}sin{(z^2)} - 0.0416\cdot2^xsin{(y)} - 1\cdot2^xcos{(y)} + 0.24xcos{(x^2)}sin{(y^3)}sin{(z^2)}\end{align*}$

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Example Question #77 : Directional Derivatives

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

Possible Answers:

${4z^3ln{(y^2)} + 1.74xcos{(y^3)}sin{(x^2)}}$

{0}

${32z^3ln{(y^2)} + 14xcos{(y^3)}sin{(x^2)}}$

${\frac{{(32.2z^4)}}{y} + 64.5z^3ln{(y^2)} + 16.1xcos{(y^3)}sin{(x^2)} + 24.2y^2cos{(x^2)}sin{(y^3)}}$

Correct answer:

${4z^3ln{(y^2)} + 1.74xcos{(y^3)}sin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(0)^2+(4)^2}=8.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{8.06}=0.87\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{8.06}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{8.06}=0.5\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.87)(2xcos{(y^3)}sin{(x^2)})+(0)(\frac{{(4z^4)}}{y} + 3y^2cos{(x^2)}sin{(y^3)})+(0.5)(8z^3ln{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=4z^3ln{(y^2)} + 1.74xcos{(y^3)}sin{(x^2)}\end{align*}$

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Example Question #78 : Directional Derivatives

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

Possible Answers:

${- 8.06\cdot4^zsin{(y^2)}cos{(x)} - 11.2\cdot4^zsin{(y^2)}sin{(x)} - 16.1\cdot4^zycos{(y^2)}sin{(x)}}$

${14\cdot4^zycos{(y^2)}sin{(x)} - 5.55\cdot4^zsin{(y^2)}sin{(x)}}$

${-22.3\cdot4^zycos{(y^2)}cos{(x)}}$

${1.74\cdot4^zycos{(y^2)}sin{(x)} - 0.693\cdot4^zsin{(y^2)}sin{(x)}}$

Correct answer:

${1.74\cdot4^zycos{(y^2)}sin{(x)} - 0.693\cdot4^zsin{(y^2)}sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-7)^2+(4)^2}=8.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{8.06}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{8.06}=-0.87\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{8.06}=0.5\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1\cdot4^zsin{(y^2)}cos{(x)})+(-0.87)(-2\cdot4^zycos{(y^2)}sin{(x)})+(0.5)(-1.39\cdot4^zsin{(y^2)}sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=1.74\cdot4^zycos{(y^2)}sin{(x)} - 0.693\cdot4^zsin{(y^2)}sin{(x)}\end{align*}$

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Example Question #79 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^yx^3z^2\\&\text{In the direction of }\overrightarrow{v}=(6,7,0).\end{align*}$

Possible Answers:

${- 1.95\cdot2^yx^2z^2 - 0.527\cdot2^yx^3z^2}$

${-38.3\cdot2^yx^2z}$

${- 18\cdot2^yx^2z^2 - 4.85\cdot2^yx^3z^2}$

${- 27.7\cdot2^yx^2z^2 - 6.39\cdot2^yx^3z^2 - 18.4\cdot2^yx^3z}$

Correct answer:

${- 1.95\cdot2^yx^2z^2 - 0.527\cdot2^yx^3z^2}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(7)^2+(0)^2}=9.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{9.22}=0.65\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{9.22}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.22}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.65)(-3\cdot2^yx^2z^2)+(0.76)(-0.693\cdot2^yx^3z^2)+(0)(-2\cdot2^yx^3z)\\&D_{\overrightarrow{u}}(x,y,z)=- 1.95\cdot2^yx^2z^2 - 0.527\cdot2^yx^3z^2\end{align*}$

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Example Question #80 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-sin{(z^2)}cos{(y)}\\&\text{In the direction of }\overrightarrow{v}=(14,0,-10).\end{align*}$

Possible Answers:

${1.16zcos{(z^2)}cos{(y)}}$

{0}

${20zcos{(z^2)}cos{(y)}}$

${17.2sin{(z^2)}sin{(y)} - 34.4zcos{(z^2)}cos{(y)}}$

Correct answer:

${1.16zcos{(z^2)}cos{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(0)^2+(-10)^2}=17.2\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{17.2}=0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.2}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{17.2}=-0.58\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.81)(0.0)+(0)(sin{(z^2)}sin{(y)})+(-0.58)(-2zcos{(z^2)}cos{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=1.16zcos{(z^2)}cos{(y)}\end{align*}$

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Example Question #81 : Directional Derivatives

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

Possible Answers:

{0}

${6x^2y^4 + 36x^3y^3}$

${27.7x^2y^4 + 36.9x^3y^3}$

${0.66x^2y^4 + 3.92x^3y^3}$

Correct answer:

${0.66x^2y^4 + 3.92x^3y^3}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(9)^2+(0)^2}=9.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{9.22}=0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{9.22}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.22}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.22)(3x^2y^4)+(0.98)(4x^3y^3)+(0)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.66x^2y^4 + 3.92x^3y^3\end{align*}$

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Example Question #82 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4e^{(z)}ln{(y)} - sin{(z^2)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,-15,8).\end{align*}$

Possible Answers:

${\frac{{(68e^{(z)})}}{y} + 68e^{(z)}ln{(y)} - 17sin{(z^2)}cos{(y)} - 34zcos{(z^2)}sin{(y)}}$

${32e^{(z)}ln{(y)} -\frac{ {(60e^{(z)})}}{y} + 15sin{(z^2)}cos{(y)} - 16zcos{(z^2)}sin{(y)}}$

${1.88e^{(z)}ln{(y)} -\frac{ {(3.52e^{(z)})}}{y} + 0.88sin{(z^2)}cos{(y)} - 0.94zcos{(z^2)}sin{(y)}}$

Correct answer:

${1.88e^{(z)}ln{(y)} -\frac{ {(3.52e^{(z)})}}{y} + 0.88sin{(z^2)}cos{(y)} - 0.94zcos{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(8)^2}=17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{17}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8}{17}=0.47\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(-0.88)(\frac{{(4e^{(z)})}}{y} - sin{(z^2)}cos{(y)})+(0.47)(4e^{(z)}ln{(y)} - 2zcos{(z^2)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=1.88e^{(z)}ln{(y)} -\frac{ {(3.52e^{(z)})}}{y} + 0.88sin{(z^2)}cos{(y)} - 0.94zcos{(z^2)}sin{(y)}\end{align*}$

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Example Question #83 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(z^2)}sin{(x^2)}sin{(y)} - 3x^4ycos{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-19,0,-11).\end{align*}$

Possible Answers:

${263x^3sin{(z)} - 87.8xzcos{(x^2)}sin{(z^2)}cos{(y)}}$

${228x^3ycos{(z)} - 33x^4ysin{(z)} - 38xcos{(x^2)}cos{(z^2)}sin{(y)} + 22zsin{(x^2)}sin{(z^2)}sin{(y)}}$

${10.4x^3ycos{(z)} - 1.5x^4ysin{(z)} - 1.74xcos{(x^2)}cos{(z^2)}sin{(y)} + zsin{(x^2)}sin{(z^2)}sin{(y)}}$

${65.8x^4ysin{(z)} - 263x^3ycos{(z)} - 65.8x^4cos{(z)} + 22cos{(z^2)}sin{(x^2)}cos{(y)} + 43.9xcos{(x^2)}cos{(z^2)}sin{(y)} - 43.9zsin{(x^2)}sin{(z^2)}sin{(y)}}$

Correct answer:

${10.4x^3ycos{(z)} - 1.5x^4ysin{(z)} - 1.74xcos{(x^2)}cos{(z^2)}sin{(y)} + zsin{(x^2)}sin{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(0)^2+(-11)^2}=21.95\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{21.95}=-0.87\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{21.95}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{21.95}=-0.5\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.87)(2xcos{(x^2)}cos{(z^2)}sin{(y)} - 12x^3ycos{(z)})+(0)(cos{(z^2)}sin{(x^2)}cos{(y)} - 3x^4cos{(z)})+(-0.5)(3x^4ysin{(z)} - 2zsin{(x^2)}sin{(z^2)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=10.4x^3ycos{(z)} - 1.5x^4ysin{(z)} - 1.74xcos{(x^2)}cos{(z^2)}sin{(y)} + zsin{(x^2)}sin{(z^2)}sin{(y)}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #74 : Directional Derivatives

Example Question #72 : Directional Derivatives

Example Question #71 : Directional Derivatives

Example Question #77 : Directional Derivatives

Example Question #78 : Directional Derivatives

Example Question #79 : Directional Derivatives

Example Question #80 : Directional Derivatives

Example Question #81 : Directional Derivatives

Example Question #82 : Directional Derivatives

Example Question #83 : Directional Derivatives

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