Applications of Partial Derivatives

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AP Calculus BC › Applications of Partial Derivatives

Questions 1 - 10

Find $\bigtriangledown f$ of the following function:

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

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

$\left \langle z^3\cos(y), xz^3\sin(y), 3xz^2\cos(y)\right \rangle$

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

$\left \langle 0, -\sin(y), z^2\right \rangle$

Explanation

The gradient of the function is

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

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

The partial derivatives are

$f_x=z^3\cos(y)$

$f_y=-xz^3\sin(y)$

$f_z=3xz^2\cos(y)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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 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 $\nabla f$ of the following function:

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

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

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

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

Explanation

The gradient of a function is given by

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

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

The partial derivatives are

The derivatives were found using the following rules:

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

Find $\bigtriangledown f$ of the following function:

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

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

$\left \langle z^3\cos(y), xz^3\sin(y), 3xz^2\cos(y)\right \rangle$

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

$\left \langle 0, -\sin(y), z^2\right \rangle$

Explanation

The gradient of the function is

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

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

The partial derivatives are

$f_x=z^3\cos(y)$

$f_y=-xz^3\sin(y)$

$f_z=3xz^2\cos(y)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Find $\nabla f$ 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

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

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

The partial derivatives are

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

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

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

The derivatives were found using the following rules:

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

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 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 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 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

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