Math Problem Statement

Find the gradient of the function F(x,y,z)=xy3z⎯⎯⎯⎯⎯√ at the point (−6,7,7) . Write the exact answer. Do not round.

Solution

To find the gradient of the function F(x,y,z)=xy3zF(x, y, z) = x y^3 \sqrt{z} at the point (6,7,7)(-6, 7, 7), we need to compute the partial derivatives with respect to xx, yy, and zz and then evaluate them at the given point.

Step 1: Compute the partial derivatives

The function is F(x,y,z)=xy3zF(x, y, z) = x y^3 \sqrt{z}. Let's find each of the partial derivatives:

  1. Partial derivative with respect to xx: Fx=x(xy3z)=y3z\frac{\partial F}{\partial x} = \frac{\partial}{\partial x} \left( x y^3 \sqrt{z} \right) = y^3 \sqrt{z}

  2. Partial derivative with respect to yy: Fy=y(xy3z)=x3y2z\frac{\partial F}{\partial y} = \frac{\partial}{\partial y} \left( x y^3 \sqrt{z} \right) = x \cdot 3y^2 \sqrt{z}

  3. Partial derivative with respect to zz: Fz=z(xy3z)=xy312z\frac{\partial F}{\partial z} = \frac{\partial}{\partial z} \left( x y^3 \sqrt{z} \right) = x y^3 \cdot \frac{1}{2\sqrt{z}}

Step 2: Evaluate at the point (6,7,7)(-6, 7, 7)

Now, we substitute the values x=6x = -6, y=7y = 7, and z=7z = 7 into the partial derivatives:

  1. Evaluate Fx\frac{\partial F}{\partial x} at (6,7,7)(-6, 7, 7): Fx=y3z=737=3437\frac{\partial F}{\partial x} = y^3 \sqrt{z} = 7^3 \cdot \sqrt{7} = 343 \cdot \sqrt{7}

  2. Evaluate Fy\frac{\partial F}{\partial y} at (6,7,7)(-6, 7, 7): [ \frac{\partial F}{\partial y} = x \cdot 3y^2 \sqrt{z} = (-6) \cdot 3 \cdot 7^2 \cdot \sqrt{7} =

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Gradient
Partial Derivatives

Formulas

Gradient of a function: ∇F = (∂F/∂x, ∂F/∂y, ∂F/∂z)
Partial derivatives of F(x, y, z) = x y^3 √z

Theorems

Chain rule for partial derivatives

Suitable Grade Level

Grades 11-12