Find a function that gives the vertical distance $v$ between the line $y=x+56$ and the parabola $y=x^2$ for $-7\le x\le 8$.
$$v(x)=$$
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 $-7\le x\le 8$ ?

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 $-7\le x\le 8$, 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 $-7\le x\le 8$. 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 $-7\le x\le 8$ 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 $-7\le x\le 8$ is $\boxed{{\displaystyle \frac{225}{4}}}$.