Problem 32 - Calculus Early Transcendentals

Question

Use differentials to approximate the change in \(z\) for the given changes in the independent variables. \(z=-x^{2}+3 y^{2}+2\) when \((x, y)\) changes from (-1,2) to (-1.05,1.9)

Step-by-Step Solution

Answer

Answer: The approximate change in z is -1.3.

Step 1: Find the partial derivatives with respect to x and y

To find the partial derivatives, differentiate the function z with respect to x and y as follows: \(\frac{\partial z}{\partial x} = -2x\) \(\frac{\partial z}{\partial y} = 6y\)

Step 2: Plug the initial point into the partial derivatives

Now we use the initial point (-1,2) and plug it into the partial derivatives obtained in step 1 to get: \(\left.\frac{\partial z}{\partial x}\right|_{(-1,\,2)} = -2(-1) = 2\) \(\left.\frac{\partial z}{\partial y}\right|_{(-1,\,2)} = 6(2) = 12\)

Step 3: Calculate the change in x and y

The change in x and y is given as the change from point 1 to point 2, so we find them as follows: \(\Delta x = - 1.05 - (-1) = -0.05\) \(\Delta y = 1.9 - 2 = -0.1\)

Step 4: Use the differentials to approximate the change in z

Now that we have all the required values, we can use the differentials to approximate the change in z. The formula for the change in z is given by: \(\Delta z \approx \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y\) Plugging in the values, we get: \(\Delta z \approx (2)(-0.05) + (12)\,(-0.1) = -0.1 - 1.2 = -1.3\) So, the approximate change in z is -1.3 when \((x,\,y)\) changes from \((-1,\,2)\) to \((-1.05,\,1.9)\).