# A football punt follows the path y = 1/15x (60-x) yards. (A) How far did the punt go...

## Question:

A football punt follows the path {eq}y = \frac{1}{15}x \; (60-x) {/eq} yards.

(A) How far did the punt go horizontally? How far did it go vertically?

(B) Set up an integral to solve the distance traveled by the football.

## Derivatives:

In order to have the ability in identifying where maximum and minimum values occur, derivatives are applied to the original function. Once a function's derivative is found, the resulting expression is set equal to {eq}0 {/eq} to represent the transition from when a function is changing from increasing its value to decreasing its value or vice versa.

Given: {eq}y = \frac{x}{15}(60-x) {/eq}

a. To find how far the punt goes horizontally, the strategy is set {eq}y = 0 {/eq} which indicates the two points when the football is on the ground and subtract the {eq}x {/eq} values that correspond to this solution. The difference will be the distance that the punt travels in the horizontal direction.

{eq}\begin{align*} y = 0 &\Rightarrow \frac{x}{15}(60-x) = 0 \\ &\Rightarrow \frac{x}{15} = 0, 60-x = 0 &\text{ [If either expression is equal to 0, the whole side of the equation holds a value of 0]} \\ &\Rightarrow \frac{x}{15}\cdot 15 = 0\cdot 15, 60-x+x = 0+x \\ &\Rightarrow x = 0, x = 60 \\ \end{align*} {/eq}

From the two {eq}x {/eq} values, the horizontal distance can be calculated.

{eq}\begin{align*} \text{ Horizontal Punt Distance } &= x_2-x_1 \\ &= 60-0 \\ &= 60 \\ \end{align*} {/eq}

Therefore, the punt travels horizontally for {eq}60 {/eq} yards.

To find how far the football punt travels vertically, the strategy is to set the derivative of the path equation equal to zero before solving for the {eq}x {/eq} and {eq}y {/eq} that solves for this condition. This will give the maximum height that the punt reaches. Subtracting the initial height ({eq}0 {/eq} yards in this case since the punt started from the ground) will give the total distance the punt traveled vertically.

{eq}\begin{align*} y' &= \frac{d}{dx}(y) \\ &= \frac{d}{dx}(\frac{x}{15}(60-x)) \\ &= \frac{d}{dx}(\frac{x}{15})(60-x)+(\frac{x}{15})(\frac{d}{dx}(60-x)) &\text{ [Product Rule Applied]} \\ &= (\frac{1}{15}\cdot 1)(60-x)+(\frac{x}{15})(0-1) \\ &= (\frac{1}{15})(60-x)+(\frac{x}{15})(-1) \\ &= 4-\frac{x}{15}-\frac{x}{15} \\ &= 4-\frac{2x}{15} \\ \end{align*} {/eq}

When {eq}y' = 0 {/eq}:

{eq}\begin{align*} y' = 0 &\Rightarrow 4-\frac{2x}{15} = 0 \\ &\Rightarrow 4-\frac{2x}{15}+\frac{2x}{15} = 0+\frac{2x}{15} \\ &\Rightarrow 4\cdot \frac{15}{2} = \frac{2x}{15}\cdot \frac{15}{2} \\ &\Rightarrow x = 30 \\ \end{align*} {/eq}

The associated {eq}y {/eq} value for this is:

{eq}\begin{align*} y &= \frac{30}{15}(60-30) \\ &= 2(30) \\ &= 60 \\ \end{align*} {/eq}

From the two {eq}y {/eq} values, the vertical distance can be calculated.

{eq}\begin{align*} \text{ Vertical Punt Distance } &= y_2-y_1 \\ &= 60-0 \\ &= 60 \\ \end{align*} {/eq}

Therefore, the punt travels vertically for {eq}60 {/eq} yards.

b. To set up an integral to solve the distance traveled by the football, the strategy is to integrate the absolute of the punt's velocity expression from the two {eq}x {/eq} values found in part a since that gives the duration of the punt. The punt's velocity expression was already determined in part a because this is the derivative expression of the given equation. Mathetmatically, this is represented as {eq}\int_a^b |v(x)| dx {/eq}

Applying the given and determined information to this model, the integral setup to find the distance traveled by the football is {eq}\int_0^{60} |4-\frac{2x}{15}| dx {/eq} 