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Calculus 3 : Gradient Vector, Tangent Planes, and Normal Lines

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Example Questions

Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to the following function at $\displaystyle (2, 5, 4)$ :

$\displaystyle f(x, y, z)=xy+4z$

Possible Answers:

$\displaystyle 5x+2y+4z=36$

$\displaystyle 2x+5y+4z=36$

$\displaystyle 5x+2y+4z=0$

$\displaystyle 2x+5y+4z=0$

Correct answer:

$\displaystyle 5x+2y+4z=36$

Explanation:

The equation of the tangent plane is given by

$\displaystyle f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0$

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=y $\displaystyle f_x=y$

f_y=x $\displaystyle f_y=x$

f_z=4 $\displaystyle f_z=4$

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluated at the given point, the partial derivatives are

f_x=5 $\displaystyle f_x=5$

f_y=2 $\displaystyle f_y=2$

f_z=4 $\displaystyle f_z=4$

Note that the partial derivative with respect to z was 4 to begin with; the fact that the point has a z coordinate of 4 is a coincidence.

Now, plug all of this into our given formula:

$\displaystyle 5(x-2)+2(y-5)+4(z-4)=0$

which simplified becomes

$\displaystyle 5x+2y+4z=36$

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Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ $\displaystyle \nabla f$ of the following function:

$\displaystyle f(x, y, z)=ze^z+xy$

Possible Answers:

$\left \langle x, y, e^z+ze^z\right \rangle$ $\displaystyle \left \langle x, y, e^z+ze^z\right \rangle$

$\left \langle y, x, 2e^z\right \rangle$ $\displaystyle \left \langle y, x, 2e^z\right \rangle$

$\left \langle y, x, e^z\right \rangle$ $\displaystyle \left \langle y, x, e^z\right \rangle$

$\left \langle y, x, e^z+ze^z\right \rangle$ $\displaystyle \left \langle y, x, e^z+ze^z\right \rangle$

Correct answer:

$\left \langle y, x, e^z+ze^z\right \rangle$ $\displaystyle \left \langle y, x, e^z+ze^z\right \rangle$

Explanation:

The gradient of a function is given by

$\nabla f=\left \langle f_x, f_y, f_z\right \rangle$ $\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=y $\displaystyle f_x=y$

f_y=x $\displaystyle f_y=x$

$\displaystyle f_z=e^z+ze^z$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Example Question #52 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ $\displaystyle \nabla f$ of the following function:

$\displaystyle f(x, y, z)=xyz^4+ze^y$

Possible Answers:

$\displaystyle yz^4+xz^4+4xyz^3+2ze^y$

$\left \langle yz^4, xz^4+ze^y, 4xyz^3+ze^y\right \rangle$ $\displaystyle \left \langle yz^4, xz^4+ze^y, 4xyz^3+ze^y\right \rangle$

$\left \langle yz^4, xz^4+ze^y, 4xyz^3+e^y\right \rangle$ $\displaystyle \left \langle yz^4, xz^4+ze^y, 4xyz^3+e^y\right \rangle$

$\left \langle yz^4, xz^4+ze^y, 4xyz+ze^y\right \rangle$ $\displaystyle \left \langle yz^4, xz^4+ze^y, 4xyz+ze^y\right \rangle$

Correct answer:

$\left \langle yz^4, xz^4+ze^y, 4xyz^3+e^y\right \rangle$ $\displaystyle \left \langle yz^4, xz^4+ze^y, 4xyz^3+e^y\right \rangle$

Explanation:

The gradient of a function is given by

$\nabla f=\left \langle f_x, f_y, f_z\right \rangle$ $\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=yz^4$

$\displaystyle f_y=xz^4+ze^y$

$\displaystyle f_z=4xyz^3+ze^y$

The derivatives were found using the following rules:

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

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Example Question #53 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ $\displaystyle \nabla f$ of the following function:

$f(x,y,z)=\sec(xyz^2)$ $\displaystyle f(x,y,z)=\sec(xyz^2)$

Possible Answers:

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz^2\sec(xyz^2)\tan(xyz^2)\right \rangle$ $\displaystyle \left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz^2\sec(xyz^2)\tan(xyz^2)\right \rangle$

$\left \langle x\sec(xyz^2)\tan(xyz^2), y\sec(xyz^2)\tan(xyz^2), 2z\sec(xyz^2)\tan(xyz^2)\right \rangle$ $\displaystyle \left \langle x\sec(xyz^2)\tan(xyz^2), y\sec(xyz^2)\tan(xyz^2), 2z\sec(xyz^2)\tan(xyz^2)\right \rangle$

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$ $\displaystyle \left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$ $\displaystyle \left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$

Correct answer:

Explanation:

The gradient of a function is given by

$\nabla f=\left \langle f_x, f_y, f_z\right \rangle$ $\displaystyle \nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=yz^2\sec(xyz^2)\tan(xyz^2)$ $\displaystyle f_x=yz^2\sec(xyz^2)\tan(xyz^2)$

$f_y=xz^2\sec(xyz^2)\tan(xyz^2)$ $\displaystyle f_y=xz^2\sec(xyz^2)\tan(xyz^2)$

$f_z=2xyz\sec(xyz^2)\tan(xyz^2)$ $\displaystyle f_z=2xyz\sec(xyz^2)\tan(xyz^2)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #54 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to the given function at $\displaystyle (2, 3, 0)$ :

$\displaystyle f(x,y,z)=xyz+y^2z$

Possible Answers:

z=0 $\displaystyle z=0$

z=6 $\displaystyle z=6$

x+y+z=0 $\displaystyle x+y+z=0$

$z=\frac{1}{6}$ $\displaystyle z=\frac{1}{6}$

Correct answer:

z=0 $\displaystyle z=0$

Explanation:

The equation of the tangent plane is given by

$\displaystyle f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0$

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=yz $\displaystyle f_x=yz$

$\displaystyle f_y=xz+2yz^2$

$\displaystyle f_z=xy+2y^2z$

The partial derivatives evaluated at the given point are

f_x=0 $\displaystyle f_x=0$

f_y=0 $\displaystyle f_y=0$

f_z=6 $\displaystyle f_z=6$

Plugging all of this into the above formula, we get

$\displaystyle 6(z-0)=0$

z=0 $\displaystyle z=0$

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Example Question #55 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to the given function at $\displaystyle (1, 1, 3)$ :

$f(x, y, z)=\frac{2xy}{z}$ $\displaystyle f(x, y, z)=\frac{2xy}{z}$

Possible Answers:

$\displaystyle 3x+3y+z=2$

$\displaystyle 3x+3y-z=2$

$\displaystyle 6x+6y-2z=2$

$\displaystyle 3x+3y-z=-2$

Correct answer:

$\displaystyle 3x+3y-z=2$

Explanation:

The equation of the tangent plane is given by

$\displaystyle f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0$

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=\frac{2y}{z}$ $\displaystyle f_x=\frac{2y}{z}$

$f_y=\frac{2x}{z}$ $\displaystyle f_y=\frac{2x}{z}$

$f_z=\frac{-2xy}{z^2}$ $\displaystyle f_z=\frac{-2xy}{z^2}$

The derivatives evaluated at the given point are

$f_x=\frac{2}{3}$ $\displaystyle f_x=\frac{2}{3}$

$f_y=\frac{2}{3}$ $\displaystyle f_y=\frac{2}{3}$

$f_z=\frac{-2}{9}$ $\displaystyle f_z=\frac{-2}{9}$

Plugging all of this into the above formula, we get

$\frac{2}{3}(x-1)+\frac{2}{3}(y-1)-\frac{2}{9}(z-3)=0$ $\displaystyle \frac{2}{3}(x-1)+\frac{2}{3}(y-1)-\frac{2}{9}(z-3)=0$

which simplifies to

$\displaystyle 3x+3y-z=2$

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Example Question #81 : Applications Of Partial Derivatives

Find $\bigtriangledown f$ $\displaystyle \bigtriangledown f$ of the following function:

$f(x, y, z)=x\sqrt{y+z^2}$ $\displaystyle f(x, y, z)=x\sqrt{y+z^2}$

Possible Answers:

$\left \langle \sqrt{y+z^2}, \frac{x}{\sqrt{y+z^2}}, -\frac{z}{\sqrt{y+z^2}} \right \rangle$ $\displaystyle \left \langle \sqrt{y+z^2}, \frac{x}{\sqrt{y+z^2}}, -\frac{z}{\sqrt{y+z^2}} \right \rangle$

$\left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, -\frac{2z}{\sqrt{y+z^2}} \right \rangle$ $\displaystyle \left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, -\frac{2z}{\sqrt{y+z^2}} \right \rangle$

$\left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, \frac{z}{\sqrt{y+z^2}} \right \rangle$ $\displaystyle \left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, \frac{z}{\sqrt{y+z^2}} \right \rangle$

$\left \langle \sqrt{y+z^2}, -\frac{x}{2\sqrt{y+z^2}}, \frac{2z}{\sqrt{y+z^2}} \right \rangle$ $\displaystyle \left \langle \sqrt{y+z^2}, -\frac{x}{2\sqrt{y+z^2}}, \frac{2z}{\sqrt{y+z^2}} \right \rangle$

Correct answer:

$\left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, \frac{z}{\sqrt{y+z^2}} \right \rangle$ $\displaystyle \left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, \frac{z}{\sqrt{y+z^2}} \right \rangle$

Explanation:

The gradient of the function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$ $\displaystyle \bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x= \sqrt{y+z^2}$ $\displaystyle f_x= \sqrt{y+z^2}$

$f_y=\frac{x}{2\sqrt{y+z^2}}$ $\displaystyle f_y=\frac{x}{2\sqrt{y+z^2}}$

$f_z=\frac{z}{\sqrt{y+z^2}}$ $\displaystyle f_z=\frac{z}{\sqrt{y+z^2}}$

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Example Question #57 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the plane tangent to the following function at $\displaystyle (10, 1, 2)$ :

$\displaystyle f(x, y, z)=x^2+y^2+2z^3$

Possible Answers:

$\displaystyle 10x+y+12z=119$

$\displaystyle 10x+y+6z=113$

$\displaystyle 10x+y+6z=226$

$\displaystyle 20x+2y+12z=113$

Correct answer:

$\displaystyle 10x+y+6z=113$

Explanation:

The equation of the tangent plane is given by

$\displaystyle f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0$

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=2x $\displaystyle f_x=2x$

f_y=2y $\displaystyle f_y=2y$

$\displaystyle f_z=3z^2$

The partial derivatives evaluated at the given point are

f_x=20 $\displaystyle f_x=20$

f_y=2 $\displaystyle f_y=2$

f_z=12 $\displaystyle f_z=12$

Plugging this into the formula above, we get

$\displaystyle 20(x-10)+2(y-1)+12(z-2)=0$

which simplifies to

$\displaystyle 10x+y+6z=113$

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Example Question #58 : Gradient Vector, Tangent Planes, And Normal Lines

Write the equation of the plane tangent to the given function at the point $\displaystyle (2, 2, 1)$ :

$f(x, y, z)=\frac{xz^2}{y^2}$ $\displaystyle f(x, y, z)=\frac{xz^2}{y^2}$

Possible Answers:

$\displaystyle x+2y+4z=2$

$\displaystyle 4x-2y+4z=2$

$\displaystyle x-2y+4z=2$

$\displaystyle x-2y+4z=4$

Correct answer:

$\displaystyle x-2y+4z=2$

Explanation:

The equation of the tangent plane is given by

$\displaystyle f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0$

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=\frac{z^2}{y^2}$ $\displaystyle f_x=\frac{z^2}{y^2}$ , $f_y=-\frac{2xz^2}{y^2}$ $\displaystyle f_y=-\frac{2xz^2}{y^2}$ , $f_z=\frac{2xz}{y^2}$ $\displaystyle f_z=\frac{2xz}{y^2}$

The partial derivatives evaluated at the given point are

$f_x=\frac{1}{4}$ $\displaystyle f_x=\frac{1}{4}$ , $f_y=-\frac{1}{2}$ $\displaystyle f_y=-\frac{1}{2}$ , f_z=1 $\displaystyle f_z=1$

Plugging this into the above formula, we get

$\frac{1}{4}(x-2)-\frac{1}{2}(y-2)+(z-1)=0$ $\displaystyle \frac{1}{4}(x-2)-\frac{1}{2}(y-2)+(z-1)=0$

which simplified becomes

$\displaystyle x-2y+4z=2$

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Example Question #59 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ $\displaystyle \bigtriangledown f$ of the following function:

$f(x, y, z)=e^{\cos(xy)+z^2}$ $\displaystyle f(x, y, z)=e^{\cos(xy)+z^2}$

Possible Answers:

$\left \langle -y\sin(xy)e^{\cos(xy)+z^2}, -x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$ $\displaystyle \left \langle -y\sin(xy)e^{\cos(xy)+z^2}, -x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

$\left \langle y\cos(xy)e^{\cos(xy)+z^2}, x\cos(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$ $\displaystyle \left \langle y\cos(xy)e^{\cos(xy)+z^2}, x\cos(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

$\left \langle -x\sin(xy)e^{\cos(xy)+z^2}, -y\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$ $\displaystyle \left \langle -x\sin(xy)e^{\cos(xy)+z^2}, -y\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

$\left \langle y\sin(xy)e^{\cos(xy)+z^2}, x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$ $\displaystyle \left \langle y\sin(xy)e^{\cos(xy)+z^2}, x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

Correct answer:

Explanation:

The gradient of a function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$ $\displaystyle \bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x= -y\sin(xy)e^{\cos(xy)+z^2}$ $\displaystyle f_x= -y\sin(xy)e^{\cos(xy)+z^2}$

$f_y= -x\sin(xy)e^{\cos(xy)+z^2}$ $\displaystyle f_y= -x\sin(xy)e^{\cos(xy)+z^2}$

$f_z= 2ze^{\cos(xy)+z^2}$ $\displaystyle f_z= 2ze^{\cos(xy)+z^2}$

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Calculus 3 : Gradient Vector, Tangent Planes, and Normal Lines

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #52 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #53 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #54 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #55 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #81 : Applications Of Partial Derivatives

Example Question #57 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #58 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #59 : Gradient Vector, Tangent Planes, And Normal Lines

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