Question 517244
A car leaves an intersection traveling east. Its position t sec later is given by
x = t^2 + t
ft. At the same time, another car leaves the same intersection heading north, traveling
y = t^2 + 3t ft
in t sec. Find the rate at which the distance between the two cars will be changing 3 sec later.?
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{{{d(t) = sqrt((t^2+t)^2 + (t^2+3t)^2)}}}
= {{{sqrt(2t^4 + 8t^3 + 10t^2)}}}
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d'(t) = {{{(1/2)*(2t^4 + 8t^3 + 10t^2)^(-1/2)*(8t^3 + 24t^2 + 20t)}}}
d'(3) = (216 + 216 + 60)/(2(162 + 216 + 90))
= 492/936
= 41/78 ft/sec