Problem 9 - Calculus Early Transcendentals

Question

Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$f(x, y)=2+3 x^{2}-5 y^{2} ; P(2,-1)$$

Step-by-Step Solution

Answer

Answer: The gradient of the function at point \(P(2, -1)\) is \((12, 10)\).

Step 1: Understanding the gradient

The gradient of a function is a vector that points in the direction of the greatest increase of the function. It consists of partial derivatives of that function with respect to its variables (in this case, the variables are x and y). The gradient is represented as \(\nabla f(x, y)\) or as the vector \(\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).

Step 2: Calculate the gradient of the function

We have to find the partial derivatives of the given function \(f(x, y) = 2 + 3x^2 - 5y^2\). To find the partial derivative with respect to x (\(\frac{\partial f}{\partial x}\)), differentiate the function with respect to x while treating y as a constant: $$\frac{\partial f}{\partial x} = 6x$$ To find the partial derivative with respect to y (\(\frac{\partial f}{\partial y}\)), differentiate the function with respect to y while treating x as a constant: $$\frac{\partial f}{\partial y} = -10y$$ Thus, the gradient of \(f(x, y)\) is: $$\nabla f(x, y) = \left(6x, -10y\right)$$

Step 3: Evaluate the gradient at the given point P

Now, we need to evaluate the gradient at the given point \(P(2, -1)\). Plug the coordinates of P into the gradient we found in Step 2: $$\nabla f(2, -1) = \left(6(2), -10(-1)\right) = \left(12, 10\right)$$ So, the gradient of the function \(f(x, y) = 2 + 3x^2 - 5y^2\) is \((12, 10)\) at the point \(P(2, -1)\).