Find a function that gives the vertical distance $v$ between the line $y=x+56$ and the parabola $y=x^{2}$ for $-7 \leq x \leq 8$.
\[
v(x)=
\]
Find $v^{\prime}(x)$
\[
v^{\prime}(x)=
\]
What is the maximum vertical distance between the line $y=x+56$ and the parabola $y=x^{2}$ for $-7 \leq x \leq 8$ ?

Step 1 ：The vertical distance between the line and the parabola is given by the absolute value of their difference, which is \(|x+56-x^{2}|\). However, since the parabola \(y=x^{2}\) is always below the line \(y=x+56\) for \(-7 \leq x \leq 8\), we can ignore the absolute value and simply subtract \(x^{2}\) from \(x+56\) to get \(v(x)=x+56-x^{2}\).

Step 2 ：To find the maximum vertical distance, we need to find the maximum value of \(v(x)\) in the interval \(-7 \leq x \leq 8\). This can be done by finding the derivative \(v^{\prime}(x)\), setting it equal to zero, and solving for \(x\). The solutions will give the x-coordinates of the local maxima and minima of \(v(x)\).

Step 3 ：We then evaluate \(v(x)\) at these x-coordinates, as well as at the endpoints of the interval, to find the maximum value.

Step 4 ：The derivative of \(v(x)\) is \(v^{\prime}(x)=1-2x\). Setting this equal to zero gives the critical point \(x=\frac{1}{2}\).

Step 5 ：Evaluating \(v(x)\) at the critical point and the endpoints of the interval \(-7 \leq x \leq 8\) gives the values \(v(-7)\), \(v(\frac{1}{2})\), and \(v(8)\).

Step 6 ：The maximum of these values is the maximum vertical distance between the line and the parabola.

Step 7 ：The maximum vertical distance between the line \(y=x+56\) and the parabola \(y=x^{2}\) for \(-7 \leq x \leq 8\) is \(\boxed{\frac{225}{4}}\).