Section 14.4: Tangent Planes

SOLUTION
(a) Definition 10 gives

\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy = (2x + 3y)\,dx + (3x - 2y)\,dy \]

2. Putting \(x = 2\), \(dx = \Delta x = 0.05\), \(y = 3\), and \(dy = \Delta y = -0.04\), we get

\[ dz = [2(2) + 3(3)](0.05) + [3(2) - 2(3)](-0.04) = 0.65 \]

The increment of \(z\) is

\[ \Delta z = f(2.05, 2.96) - f(2, 3) \]

\[ = [(2.05)^2 + 3(2.05)(2.96) - (2.96)^2] - [2^2 + 3(2)(3) - 3^2] \]

\[ = 0.6449 \]

Notice that \(\Delta z \approx dz\) but \(dz\) is easier to compute.

SOLUTION
The volume \(V\) of a cone with base radius \(r\) and height \(h\) is \(V = \frac{\pi r^2 h}{3}\).
So the differential of \(V\) is

\[ dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh = \frac{2\pi rh}{3} dr + \frac{\pi r^2}{3} dh \]

Since each error is at most 0.1 cm, we have \(|dr| \leq 0.1\), \(|dh| \leq 0.1\).
To estimate the largest error in the volume we take the largest error in the measurement of \(r\) and of \(h\).
Therefore we take \(dr = 0.1\) and \(dh = 0.1\) along with \(r = 10\), \(h = 25\). This gives

\[ dV = \frac{500\pi}{3}(0.1) + \frac{100\pi}{3}(0.1) = 20\pi \]

Thus the maximum error in the calculated volume is about \(20\pi \,\text{cm}^3 \approx 63\,\text{cm}^3\).