Problem 11 - Calculus Early Transcendentals

Question

Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$g(x, y)=x^{2}-4 x^{2} y-8 x y^{2} ; P(-1,2)$$

Step-by-Step Solution

Answer

Question: Compute the gradient of the function \(g(x, y)=x^{2}-4 x^{2} y-8 x y^{2}\) and evaluate it at point P(-1,2). Answer: The gradient of the function at point P(-1,2) is \(\begin{bmatrix}18\\ 28\end{bmatrix}\).

Step 1: Find the partial derivative with respect to x

Compute the partial derivative of the function \(g(x,y)\) with respect to x: $$\frac{\partial{g}}{\partial{x}} = \frac{\partial{(x^{2}-4 x^{2} y-8 x y^{2})}}{\partial{x}}$$ Take the derivative term by term: $$\frac{\partial{g}}{\partial{x}} = 2x - 8xy - 8y^2$$

Step 2: Find the partial derivative with respect to y

Compute the partial derivative of the function \(g(x,y)\) with respect to y: $$\frac{\partial{g}}{\partial{y}} = \frac{\partial{(x^{2}-4 x^{2} y-8 x y^{2})}}{\partial{y}}$$ Take the derivative term by term: $$\frac{\partial{g}}{\partial{y}} = -4x^2 - 16xy$$

Step 3: Form the gradient

Combine the results from step 1 and step 2 to form the gradient \(\nabla g(x,y)\): $$\nabla g(x,y) = \begin{bmatrix}2x - 8xy - 8y^2\\ - 4x^2 -16xy\end{bmatrix}$$

Step 4: Evaluate the gradient at point P

Evaluate the gradient of \(g(x,y)\) at point \(P(-1, 2)\). Plug the coordinates of point P into the gradient vector: $$\nabla g(-1, 2) = \begin{bmatrix}2(-1) - 8(-1)(2) - 8(2)^2\\ - 4(-1)^2 - 16(-1)(2)\end{bmatrix}$$ Simplify and compute each component of the gradient: $$\nabla g(-1, 2) = \begin{bmatrix}18\\ 28\end{bmatrix}$$ The gradient of the function \(g(x, y)=x^{2}-4 x^{2} y-8 x y^{2}\) at point P(-1,2) is \(\begin{bmatrix}18\\ 28\end{bmatrix}\).