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


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.


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.

Answer and Explanation:

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}

Learn more about this topic:

How to Determine Maximum and Minimum Values of a Graph

from Math 104: Calculus

Chapter 9 / Lesson 3

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