Question

A baseball is hit along the third base line with a speed of 100 ft/s. At what rate is the balls distance from first base changing when it crosses third base?

Answer 1

From the triangle,

`$x^2 = s^2 + 90^2$`

`$x = √(s^2 + 90^2)$`

Now differentiating w.r.t "t", we get

`$(dx)/(dt)=(d)/(dt)\left(s^2 +90^2 \right)^(1/2)$`

`$=(1)/(2)\left(s^2 + 90^2 \right)^{(1)/(2)-1} \cdot 2s \cdot (ds)/(dt)$` [chain rule]

`$=(1)/(2)\left(s^2 + 90^2 \right)^{-(1)/(2)} \cdot 2s \cdot (ds)/(dt)$`

`$=\frac{s}{\left(s^2 + 90^2\right)^{(1)/(2)}} . (ds)/(dt)$`

But given that
`$(ds)/(dt)$` = 100 ft/s

When all the ball crosses the third base, this means that s = 90. So substituting the value of s and
`$(ds)/(dt)$` in
`$(dx)/(dt)$` , we get

`$(dx)/(dt) = (90)/(\left(90^2 +90^2\right)^(1/2)) * 100$`

`$(dx)/(dt) = (90)/(√(8100+8100)) * 100$`

`$(dx)/(dt) = 50\sqrt2$`

= 70.710 ft/s