Activity 11.5.3.
In this activity, we will find the linearization of several different functions that are given in algebraic, tabular, or graphical form.
(a)
Find the linearization \(L(x,y)\) for the function \(g\) defined by
\begin{equation*}
g(x,y) = \frac{x}{x^2+y^2}
\end{equation*}
at the point \((1,2)\text{.}\) Then use the linearization to estimate the value of \(g(0.8, 2.3)\text{.}\)
(b)
Table 11.5.12 provides a collection of values of the wind chill \(w(v,T)\text{,}\) in degrees Fahrenheit, as a function of wind speed, in miles per hour, and temperature, also in degrees Fahrenheit.
| \(v \backslash T\) | \(-20\) | \(-15\) | \(-10\) | \(-5\) | \(0\) | \(5\) | \(10\) |
| \(10\) | \(-41\) | \(-35\) | \(-28\) | \(-22\) | \(-16\) | \(-10\) | \(-4\) |
| \(15\) | \(-45\) | \(-39\) | \(-32\) | \(-26\) | \(-19\) | \(-13\) | \(-7\) |
| \(20\) | \(-48\) | \(-42\) | \(-35\) | \(-29\) | \(-22\) | \(-15\) | \(-9\) |
| \(25\) | \(-51\) | \(-44\) | \(-37\) | \(-31\) | \(-24\) | \(-17\) | \(-11\) |
| \(30\) | \(-53\) | \(-46\) | \(-39\) | \(-33\) | \(-26\) | \(-19\) | \(-12\) |
| \(35\) | \(-55\) | \(-48\) | \(-41\) | \(-34\) | \(-27\) | \(-21\) | \(-14\) |
Use the data to first estimate the appropriate partial derivatives, and then find the linearization \(L(v,T)\) at the point \((20,-10)\text{.}\) Finally, use the linearization to estimate \(w(10,-10)\text{,}\) \(w(20,-12)\text{,}\) and \(w(18,-12)\text{.}\) Compare your results to what you obtained in Activity 11.3.5.
(c)
Figure 11.5.13 gives a contour plot of a continuously differentiable function \(f\text{.}\) After estimating appropriate partial derivatives, determine the linearization \(L(x,y)\) at the point \((2,1)\text{,}\) and use it to estimate \(f(2.2, 1)\text{,}\) \(f(2, 0.8)\text{,}\) and \(f(2.2, 0.8)\text{.}\)
