The path of a large stone fired from a torsion catapult can be modeled by y = -0.0054x^2 +...


The path of a large stone fired from a torsion catapult can be modeled by {eq}y = -0.0054x^2 + 1.145x {/eq}, where {eq}x {/eq} is the distance the stone traveled (in yards) and y is the height of the stone (in yards).

1. find the distance the stone traveled.

2. find the maximum height of the stone.


We need to know how to find the maxima of a function using differentiation in order to solve this question. For a function f(x), its maxima can be found by first differentiating f(x) by x and then equating it to 0, the value of x we will get, gives us the maxima of the function f(x)

Answer and Explanation:

Given: {eq}y = -0.0054x^2 + 1.145x {/eq}

1) The distance stone traveled can be found bu putting y=0,

$$\begin{align} y &= -0.0054x^2 + 1.145x \\ 0 &= -0.0054x^2 + 1.145x \\ x(-0.0054x + 1.145)&=0 \\ x&=\frac{1.145}{0.0054} \\ x&\approx 212 \end{align} $$

Hence, the distance traveled by stone is 212 yards.

2) Differentiating y with respect to x and equating it to zero, will give the value of x where the distance is maximum.

$$\begin{align} \frac{dy}{dx}&=\frac{d( -0.0054x^2 + 1.145x)}{dx} \\ \frac{dy}{dx}&=-0.0108x+1.145 \\ \mbox{ Equating it to zero} \\ -0.0108x+1.145 &=0 \\ x&=\frac{1.145}{0.0108} \\ x&\approx 106 \end{align} $$

Putting the value of x=106 in the equation, we will get the maximum height.

$$\begin{align} y&= -0.0054(106)^2 + 1.145(106) \\ y&=-0.0054(11236) + 1.145(106) \\ y&=-60.6744+121.37 \\ y&=60.6956 \\ y&\approx 61 \end{align} $$

Hence, the maximum distance is 61 yards.

Learn more about this topic:

Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2

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