Problem 33 - Calculus Early Transcendentals

Question

Use differentials to approximate the change in \(z\) for the given changes in the independent variables. \(z=e^{x+y}\) when \((x, y)\) changes from (0,0) to (0.1,-0.05)

Step-by-Step Solution

Answer

Answer: The approximate change in \(z\) is 0.05.

Step 1: Compute the partial derivatives of z

Let's compute the partial derivatives of \(z\) with respect to \(x\) and \(y\) from the given function \(z=e^{x+y}\): - Partial derivative with respect to \(x\): \(\frac{∂z}{∂x} = e^{x+y}\) - Partial derivative with respect to \(y\): \(\frac{∂z}{∂y} = e^{x+y}\)

Step 2: Evaluate the partial derivatives at the initial point

Now, we want to evaluate the partial derivatives when \((x, y) = (0, 0)\). - Evaluate \(\frac{∂z}{∂x}\): \(\frac{∂z}{∂x}(0, 0) = e^{0+0} = 1\) - Evaluate \(\frac{∂z}{∂y}\): \(\frac{∂z}{∂y}(0, 0) = e^{0+0} = 1\)

Step 3: Use the differential of z to approximate the change in z

Finally, let's use the differentials to approximate the change in \(z\) for the given changes in the independent variables \(∆x=0.1\) and \(∆y=-0.05\). We can write the differential of \(z\) as: $$dz = \frac{∂z}{∂x} dx + \frac{∂z}{∂y} dy$$ Now, plug in the values for the partial derivatives, and the changes in \(x\) and \(y\) to compute \(dz\): $$dz = (1)(0.1) + (1)(-0.05) = 0.1 - 0.05 = 0.05$$ So, the approximate change in \(z\) as the independent variables change as described is \(dz = 0.05\).