AP Calculus BC - Gradient Vector, Tangent Planes, and Normal Lines

Find the gradient, $\bigtriangledown F$ , of the function .

$\left \langle 2xy+yz,x^2+xz,xy\right \rangle$

$\left \langle 2x,2xy+yz,1\right \rangle$

$\left \langle 2xy,yz,z\right \rangle$

$\left \langle x^3y,2x,xyz\right \rangle$

Explanation

The gradient of a function is as follows:

$\bigtriangledown F=\left \langle \frac{\mathrm{dF} }{\mathrm{d} x},\frac{\mathrm{dF} }{\mathrm{d} y},\frac{\mathrm{dF} }{\mathrm{d} z} \right \rangle$ .

We compute the derivative of the function with respect to each of the variables and treat the others like constants.

Using the rule

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ ,

we obtain

$\frac{\mathrm{dF} }{\mathrm{d} x}=2xy+yz$ ,

$\frac{\mathrm{dF} }{\mathrm{d} y}=x^2+xz$ ,

and

$\frac{\mathrm{dF} }{\mathrm{d} z}=xy$ .

Putting these expressions into the vector completes the problem, and you obtain

$\left \langle 2xy+yz,x^2+xz,xy\right \rangle$ .

Find $\bigtriangledown F$ of the function .

$\left \langle 2xy+3y^2z,x^2+6xyz,3xy^2+4z^3\right \rangle$

$\left \langle 2x^2+4y^3z,x^3+6xy^2z,3xy^2+4z^2\right \rangle$

$\left \langle 2x+3y^2z,6yz,6y\right \rangle$

$\left \langle 3x+6y^3,xyz,4x^2z^3\right \rangle$

Explanation

The gradient of a function is as follows:

$\bigtriangledown F=\left \langle \frac{\mathrm{dF} }{\mathrm{d} x},\frac{\mathrm{dF} }{\mathrm{d} y},\frac{\mathrm{dF} }{\mathrm{d} z} \right \rangle$ .

We compute the derivative of the function with respect to each of the variables and treat the others like constants.

Using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , we obtain

$\frac{\mathrm{dF} }{\mathrm{d} x}=2xy+3y^2z$ ,

$\frac{\mathrm{dF} }{\mathrm{d} y}=x^2+6xyz$ ,

and

$\frac{\mathrm{dF} }{\mathrm{d} z}=3xy^2+4z^3$ .

Putting these expressions into the vector completes the problem, and you obtain

$\left \langle 2xy+3y^2z,x^2+6xyz,3xy^2+4z^3\right \rangle$ .

Find the gradient, $\bigtriangledown F$ , of the function .

$\left \langle y^3+yz,3xy^2+4y^3+xz,xy\right \rangle$

$\left \langle y^2+xyz,3xy^3+4y^2,xz\right \rangle$

$\left \langle y^4,3xy^3+4y^3+xy,yz\right \rangle$

$\left \langle y^4+yz,3xy^2+4y^3,xy\right \rangle$

Explanation

The gradient of a function is as follows:

$\bigtriangledown F=\left \langle \frac{\mathrm{dF} }{\mathrm{d} x},\frac{\mathrm{dF} }{\mathrm{d} y},\frac{\mathrm{dF} }{\mathrm{d} z} \right \rangle$ .

We compute the derivative of the function with respect to each of the variables and treat the others like constants.

Using the rule

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ ,

we obtain

$\frac{\mathrm{dF} }{\mathrm{d} x}=y^3+yz$ ,

$\frac{\mathrm{dF} }{\mathrm{d} y}=3xy^2+4y^3+xz$ ,

and

$\frac{\mathrm{dF} }{\mathrm{d} z}=xy$ .

Putting these expressions into the vector completes the problem, and you obtain

$\left \langle y^3+yz,3xy^2+4y^3+xz,xy\right \rangle$

Find the gradient vector for the following function:

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, x^2e^z(4z^3+z^4)\right \rangle$

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, x^2z^2e^z(6z^2)\right \rangle$

$\left \langle 3x^2y^3+2xz^4e^z, 3x^3y^2, e^z(4x^2z^3+z^4)\right \rangle$

$\left \langle 3x^2y^3+2xz^4e^z, 3x^2y^2, x^2e^z(4z^3+z^4)\right \rangle$

Explanation

The gradient vector of the function is given by

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

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

The partial derivatives are

, ,

Find the gradient vector of the following function:

$f(x,y,z)=x^2+x+4+y^2\sin(z^2)$

$\left \langle 2x+1, 2y\sin(z^2), 2y^2z\cos(z^2)\right \rangle$

$\left \langle 2x+1, -2y\sin(z^2), 2y^2z\cos(z^2)\right \rangle$

$\left \langle 2x+1, 2y\sin(z^2), -2y^2z\cos(z^2)\right \rangle$

$\left \langle 2x+1, 2y\sin(z^2), 2z\cos(z^2)\right \rangle$

$\left \langle 2x+1, 2y\sin(z^2), y^2\cos(z^2)\right \rangle$

Explanation

The gradient vector of the function is given by

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

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

The partial derivatives are

, $f_y=2y\sin(z^2)$ , $f_z=2y^2z\cos(z^2)$

Find the equation of the plane tangent to the following function at :

$f(x,y,z)=z^3\ln(y)+\frac{x^2}{2}$

Explanation

The equation of the tangent plane is given by

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

, $f_y=\frac{z^3}{y}$ , $f_z=3z^2\ln(y)$

Evaluated at the given point, the partial derivatives are

, ,

Plugging this into our equation, we get

which simplified becomes

Find the gradient vector of the following function:

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

$\left \langle 1, \cos(z^2), -2yz\sin(z^2)+3z^2\right \rangle$

$\left \langle 1, \cos(z^2), 2yz\sin(z^2)+3z^2\right \rangle$

$\left \langle x, \cos(z^2), -2yz\sin(z^2)+3z^2\right \rangle$

$\left \langle 1, \cos(z^2), -2z\sin(z^2)+3z^2\right \rangle$

Explanation

The gradient vector is given by

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

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

The partial derivatives are

, $f_y=-2yz\sin(z^2)$ ,

Find the equation of the tangent plane to the following function at :

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

$z=\frac{1}{2}$

Explanation

The equation of the tangent plane is given by

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

$f_x=y(\sin(x)+x\cos(x))$

$f_y=x\sin(x)$

The partial derivatives evaluated at the given point are

, ,

Plugging this into the equation above, we get

which simplifies to

Find the equation to the tangent plane to the following function at the point :

$f(x, y, z)=\ln(xy^2z^3)$

Explanation

The equation of the tangent plane is given by

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

$f_x=\frac{1}{x}$ , $f_y=\frac{2}{y}$ , $f_z=\frac{3}{z}$

The partial derivatives evaluated at the given point are

$f_x=-\frac{1}{2}$ , ,

Plugging all of this into the formula above, we get

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

which simplified becomes

Find of the following function:

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

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

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

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

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

Explanation

The gradient of a function is given by

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

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

The partial derivatives are

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

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

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

The derivatives were found using the following rules:

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

Gradient Vector, Tangent Planes, and Normal Lines

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