Please help me with this Calc Related Rates problem.?
After Coming To The Rescue Of A Distressed Calc Student, Math Man Flew Oft Help Another Student. Math Man Can Reach Speeds Of 225ft/s. If Math Man Is Traveling Horizontally At An Altitude Of 3528 Ft And Flying As Fast As He Can, How Fast Is The Distance Between Math Man And The Student Changing When Math Man Is...
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answers:
Captain Matticus, LandPiratesInc say: dx/dt = 225
y = 3528
dy/dt = 0
d^2 = x^2 + y^2
d = 5260

5260^2 = x^2 + 3528^2
5260^2 - 3528^2 = x^2
(5260 - 3528) * (5260 + 3528) = x^2
1732 * 8788 = x^2
4 * 433 * 4 * 2197 = x^2
x = 4 * sqrt(433 * 2197)

d^2 = x^2 + y^2
2d * dd/dt = 2x * dx/dt + 2y * dy/dt
d * dd/dt = x * dx/dt + y * dy/dt
5260 * dd/dt = 4 * sqrt(433 * 2197) * 225 + 3528 * 0
5260 * dd/dt = 4 * 225 * sqrt(433 * 2197)
1315 * dd/dt = 225 * sqrt(433 * 2197)
263 * dd/dt = 45 * sqrt(433 * 2197)
dd/dt = 45 * sqrt(433 * 2197) / 263
dd/dt = 166.88440010514183537503458502038...

The distance is increasing at 166.8844 ft/sec or 167 ft/sec
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