Directional Derivatives

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AP Calculus BC › Directional Derivatives

Questions 1 - 10

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

${\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}}$

${\frac{(3\cdot 2^{(z^{4})}x)}{y^{2}}- 9zsin(y) -\frac{ (20\cdot 2^{(z^{4})})}{y}}$

${-\frac{(56.1\cdot 2^{(z^{4})}z^{3})}{y^{2}}}$

${\frac{(20.2\cdot 2^{(z^{4})})}{y}- 60.7cos(y) + 60.7zsin(y) -\frac{ (20.2\cdot 2^{(z^{4})}x)}{y^{2}}+\frac{ (56.1\cdot 2^{(z^{4})}xz^{3})}{y}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-3)^2+(0)^2}=20.22\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{20.22}=-0.99\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{20.22}=-0.15\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.22}=0\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{2^{(z^{4})}}{y})+(-0.15)(3zsin(y) -\frac{ (1\cdot 2^{(z^{4})}x)}{y^{2}})+(0)(\frac{(2.77\cdot 2^{(z^{4})}xz^{3})}{y}- 3cos(y))\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(e^{(x)}ln(z))}{y}- 5e^{(x^{2})}e^{(y)}cos(z)\&\text{In the direction of }\overrightarrow{v}=(-4,-15,0).\end{align*}$

${4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)}$

${75e^{(x^{2})}e^{(y)}cos(z) -\frac{ (4e^{(x)}ln(z))}{y}+\frac{ (15e^{(x)}ln(z))}{y^{2}}+ 40xe^{(x^{2})}e^{(y)}cos(z)}$

${155xe^{(x^{2})}e^{(y)}sin(z) -\frac{ (15.5e^{(x)})}{(y^{2}z)}}$

${77.6e^{(x^{2})}e^{(y)}sin(z) - 77.6e^{(x^{2})}e^{(y)}cos(z) +\frac{ (15.5e^{(x)})}{(yz)}+\frac{ (15.5e^{(x)}ln(z))}{y}-\frac{ (15.5e^{(x)}ln(z))}{y^{2}}- 155xe^{(x^{2})}e^{(y)}cos(z)}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-15)^2+(0)^2}=15.52\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{15.52}=-0.26\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{15.52}=-0.97\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15.52}=0\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.26)(\frac{(e^{(x)}ln(z))}{y}- 10xe^{(x^{2})}e^{(y)}cos(z))+(-0.97)(- 5e^{(x^{2})}e^{(y)}cos(z) -\frac{ (e^{(x)}ln(z))}{y^{2}})+(0)(5e^{(x^{2})}e^{(y)}sin(z) +\frac{ e^{(x)}}{(yz)})\&D_{\overrightarrow{u}}(x,y,z)=4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2cos(z^{3})tan(y) - 3^xsin(z^{2})e^{(y^{2})}\&\text{In the direction of }\overrightarrow{v}=(-4,-18,0).\end{align*}$

${1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${36cos(z^{3})\cdot (tan(y)^{2} + 1) + 4.39\cdot 3^xsin(z^{2})e^{(y^{2})} + 36\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${-81\cdot 3^xyzcos(z^{2})e^{(y^{2})}}$

${111z^{2}sin(z^{3})tan(y) - 20.3\cdot 3^xsin(z^{2})e^{(y^{2})} - 36.9cos(z^{3})\cdot (tan(y)^{2} + 1) - 36.9\cdot 3^xzcos(z^{2})e^{(y^{2})} - 36.9\cdot 3^xysin(z^{2})e^{(y^{2})}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-18)^2+(0)^2}=18.44\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{18.44}=-0.22\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.44}=-0.98\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.22)(-1.1\cdot 3^xsin(z^{2})e^{(y^{2})})+(-0.98)(- 2cos(z^{3})\cdot (tan(y)^{2} + 1) - 2\cdot 3^xysin(z^{2})e^{(y^{2})})+(0)(6z^{2}sin(z^{3})tan(y) - 2\cdot 3^xzcos(z^{2})e^{(y^{2})})\&D_{\overrightarrow{u}}(x,y,z)=1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{-{(cos{(x^3)}cos{(y^2)})}}{z}\&\text{In the direction of }\overrightarrow{v}=(0,1,15).\end{align*}$

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}}$

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.0098ycos{(x^3)}sin{(y^2)})}}{z}}$

${\frac{{(90.2x^2ysin{(x^3)}sin{(y^2)})}}{z^2}}$

${\frac{{(15cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(30.1ycos{(x^3)}sin{(y^2)})}}{z} +\frac{ {(45.1x^2cos{(y^2)}sin{(x^3)})}}{z}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(1)^2+(15)^2}=15.03\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.03}=0\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{15.03}=0.07\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{15.03}=1\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(3x^2cos{(y^2)}sin{(x^3)})}}{z})+(0.07)(\frac{{(2ycos{(x^3)}sin{(y^2)})}}{z})+(1)(\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2})\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(x^2)}cos{(z)})}}{y^2}\&\text{In the direction of }\overrightarrow{v}=(0,6,-11).\end{align*}$

${\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}}$

${-\frac{ {(0.461cos{(x^2)}cos{(z)})}}{y^3} -\frac{ {(0.774cos{(x^2)}sin{(z)})}}{y^2}}$

${\frac{-{(50.1xsin{(x^2)}sin{(z)})}}{y^3}}$

${-\frac{ {(25.1cos{(x^2)}cos{(z)})}}{y^3} -\frac{ {(12.5cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(25.1xsin{(x^2)}cos{(z)})}}{y^2}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(6)^2+(-11)^2}=12.53\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.53}=0\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{6}{12.53}=0.48\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{12.53}=-0.88\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(2xsin{(x^2)}cos{(z)})}}{y^2})+(0.48)(\frac{-{(2cos{(x^2)}cos{(z)})}}{y^3})+(-0.88)(\frac{-{(1cos{(x^2)}sin{(z)})}}{y^2})\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(x^2)}ln{(z^2)})}}{y^2}\&\text{In the direction of }\overrightarrow{v}=(-15,0,9).\end{align*}$

${\frac{{(1.02cos{(x^2)})}}{{(y^2z)}} +\frac{ {(1.72xln{(z^2)}sin{(x^2)})}}{y^2}}$

${\frac{{(0.52cos{(x^2)})}}{{(y^2z)}} -\frac{ {(1.48xln{(z^2)}sin{(x^2)})}}{y^2}}$

${\frac{{(140xsin{(x^2)})}}{{(y^3z)}}}$

${\frac{{(35cos{(x^2)})}}{{(y^2z)}} -\frac{ {(35cos{(x^2)}ln{(z^2)})}}{y^3} -\frac{ {(35xln{(z^2)}sin{(x^2)})}}{y^2}}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(0)^2+(9)^2}=17.49\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{17.49}=-0.86\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.86)(\frac{-{(2xln{(z^2)}sin{(x^2)})}}{y^2})+(0)(\frac{-{(2cos{(x^2)}ln{(z^2)})}}{y^3})+(0.51)(\frac{{(2cos{(x^2)})}}{{(y^2z)}})\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(1.02cos{(x^2)})}}{{(y^2z)}} +\frac{ {(1.72xln{(z^2)}sin{(x^2)})}}{y^2}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3ze^{(x^2)}e^{(y^2)} - cos{(y^2)}e^{(x)}\&\text{In the direction of }\overrightarrow{v}=(0,-1,12).\end{align*}$

$3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}$

$36e^{(x^2)}e^{(y^2)} - 2ysin{(y^2)}e^{(x)} - 6yze^{(x^2)}e^{(y^2)}$

$144xye^{(x^2)}e^{(y^2)}$

$36.1e^{(x^2)}e^{(y^2)} - 12cos{(y^2)}e^{(x)} + 24.1ysin{(y^2)}e^{(x)} + 72.2xze^{(x^2)}e^{(y^2)} + 72.2yze^{(x^2)}e^{(y^2)}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-1)^2+(12)^2}=12.04\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.04}=0\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-1}{12.04}=-0.08\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{12.04}=1\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(0)(6xze^{(x^2)}e^{(y^2)} -cos{(y^2)}e^{(x)})+(-0.08)(2ysin{(y^2)}e^{(x)} + 6yze^{(x^2)}e^{(y^2)})+(1)(3e^{(x^2)}e^{(y^2)})\&D_{\overrightarrow{u}}(x,y,z)=3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}\end{align*}$

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

$0.92sin{(y^2)}sin{(x)} + 0.76ycos{(y^2)}cos{(x)}$

$17sin{(y^2)}sin{(x)} + 14ycos{(y^2)}cos{(x)}$

$36.8ycos{(y^2)}cos{(x)} - 18.4sin{(y^2)}sin{(x)}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-17)^2+(7)^2+(0)^2}=18.38\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-17}{18.38}=-0.92\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{18.38}=0.38\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.38}=0\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.92)(-1sin{(y^2)}sin{(x)})+(0.38)(2ycos{(y^2)}cos{(x)})+(0)(0.0)\&D_{\overrightarrow{u}}(x,y,z)=0.92sin{(y^2)}sin{(x)} + 0.76ycos{(y^2)}cos{(x)}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=6y^2cos{(z^2)}e^{(x^2)}\&\text{In the direction of }\overrightarrow{v}=(-20,-17,19).\end{align*}$

${- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}$

${- 204ycos{(z^2)}e^{(x^2)} - 240xy^2cos{(z^2)}e^{(x^2)} - 228y^2zsin{(z^2)}e^{(x^2)}}$

${-1555.2xyzsin{(z^2)}e^{(x^2)}}$

${388.8ycos{(z^2)}e^{(x^2)} + 388.8xy^2cos{(z^2)}e^{(x^2)} - 388.8y^2zsin{(z^2)}e^{(x^2)}}$

Explanation

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\&\text{a good first step is to make sure that }\overrightarrow{v}\&\text{is in unit vector form. First, find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-17)^2+(19)^2}=32.4\&\text{With this, unit vector components can be found:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{32.4}=-0.62\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{32.4}=-0.52\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{32.4}=0.59\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.62)(12xy^2cos{(z^2)}e^{(x^2)})+(-0.52)(12ycos{(z^2)}e^{(x^2)})+(0.59)(-12y^2zsin{(z^2)}e^{(x^2)})\&D_{\overrightarrow{u}}(x,y,z)={- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}\end{align*}$

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-xsin{(z)}\&\text{In the direction of }\overrightarrow{v}=(-13,19,-10).\end{align*}$

Explanation

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\&\text{First make sure that }\overrightarrow{v}\&\text{is in unit vector form. Find the magnitude:}\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(19)^2+(-10)^2}=25.1\&\text{Unit vector components are:}\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{25.1}=-0.52\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{25.1}=0.76\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{25.1}=-0.4\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\&\text{Therefore, for our values:}\&D_{\overrightarrow{u}}(x,y,z)=(-0.52)(-1sin{(z)})+(0.76)(0.0)+(-0.4)(-1xcos{(z)})\&D_{\overrightarrow{u}}(x,y,z)=0.52sin{(z)} + 0.4xcos{(z)}\end{align*}$

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