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Chapter 13: Problem 33

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)

Short Answer

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

Step by step solution

01

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 \(x\): \(\frac{鈭倆}{鈭倄} = e^{x+y}\) - Partial derivative with respect to \(y\): \(\frac{鈭倆}{鈭倅} = e^{x+y}\)
02

Evaluate the partial derivatives at the initial point

Now, we want to evaluate the partial derivatives when \((x, y) = (0, 0)\). - Evaluate \(\frac{鈭倆}{鈭倄}\): \(\frac{鈭倆}{鈭倄}(0, 0) = e^{0+0} = 1\) - Evaluate \(\frac{鈭倆}{鈭倅}\): \(\frac{鈭倆}{鈭倅}(0, 0) = e^{0+0} = 1\)
03

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 \(鈭唜=0.1\) and \(鈭唝=-0.05\). We can write the differential of \(z\) as: $$dz = \frac{鈭倆}{鈭倄} dx + \frac{鈭倆}{鈭倅} dy$$ 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 \(dz = 0.05\).

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