Tuesday, December 24, 2019

Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 18

a.) By using Pythagorean Theorem, we have...
x2+902=z2; when x=45ft; z=452+902=455ft


Taking the derivative with respect to time,
2xdxdt+0=2zdzdt

xdxdt=zdzdtdzdt=xzdxdt

Plugging in all the values we have,

dzdt=\cancel45\cancel455(24)dzdt=245 or 2455fts


The distance of the battler from the second base is decreasing at a rate of 2455fts
b.)



Again, by using Pythagorean Theorem,
x2+902=z2; when x=45ft; z=452+902=455ft
Taking the derivative with respect to time,

0+2xdxdt=2zdzdtxdxdt=zdzdtdzdt=xzdxdt


Plugging all the values we obtain,


dzdt=45455(24)dzdt=245 or 2455fts

Thus shows that the distance of the batter from the third base is increasing at a rate equal to the decreasing rate of the batter's distance from the second base.

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