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Sagot :
Solution :
From the triangle,
[tex]$x^2 = s^2 + 90^2$[/tex]
[tex]$x = \sqrt{s^2 + 90^2}$[/tex]
Now differentiating w.r.t "t", we get
[tex]$\frac{dx}{dt}=\frac{d}{dt}\left(s^2 +90^2 \right)^{1/2}$[/tex]
[tex]$=\frac{1}{2}\left(s^2 + 90^2 \right)^{\frac{1}{2}-1} \cdot 2s \cdot \frac{ds}{dt}$[/tex] [chain rule]
[tex]$=\frac{1}{2}\left(s^2 + 90^2 \right)^{-\frac{1}{2}} \cdot 2s \cdot \frac{ds}{dt}$[/tex]
[tex]$=\frac{s}{\left(s^2 + 90^2\right)^{\frac{1}{2}}} . \frac{ds}{dt}$[/tex]
But given that [tex]$\frac{ds}{dt}$[/tex] = 100 ft/s
When all the ball crosses the third base, this means that s = 90. So substituting the value of s and [tex]$\frac{ds}{dt}$[/tex] in [tex]$\frac{dx}{dt}$[/tex] , we get
[tex]$\frac{dx}{dt} = \frac{90}{\left(90^2 +90^2\right)^{1/2}} \times 100$[/tex]
[tex]$\frac{dx}{dt} = \frac{90}{\sqrt{8100+8100}} \times 100$[/tex]
[tex]$\frac{dx}{dt} = 50\sqrt2$[/tex]
= 70.710 ft/s
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