Brian J.

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Calculus

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Question:

If V is a function of x, y, and z given V=(x-y)/(z+2y), and x, y, and z are all functions of s and t given x=3t+s, y=6t-s, and z=-4t+5s, what is the derivative of V when s=1 and t=2?

Brian J.

Answer:

V = (x - y) / (z + 2y) x = 3t + s x' = xt [derivative of x with respect to t] * t' + xs [derivative of x with respect to s] * s' x' = 3 + 1 = 4 y = 6t - s y' = yt [derivative of y with respect to t] * t' + ys [derivative of y with respect to s] * s' y' = 6 + -1 = 5 z = -4t + 5s z' = zt [derivative of z with respect to t] * t' + zs [derivative of z with respect to s] * s' z' = -4 + 5 = 1 V' = Vx [derivative of V with respect to x] * x' + Vy [derivative of V with respect to y] * y' + Vz [derivative of V with respect to z] * z' Vx = 1 / (z + 2y) Vy = (-1(z + 2y) - 2(x - y) ) / ( (z+2y)^2) = (-2x - y - z) / ( (z + 2y)^2) [quotient rule] Vz = (-x + y) / ( (z + 2y)^2) V' = 4 / (z + 2y) + 5(-2x - y -z) / ( (z + 2y)^2) + (y - x) / ( (z + 2y)^2) V' = (4(z + 2y) + 5(-2x - y - z) + (y - x) ) / ( (z + 2y)^2) V' = (4z + 8y - 10x - 5y - 5z + y -x) / ( (z + 2y)^2) V' = (-11x + 4y - z) / ( (z + 2y)^2) V' = (-11(3t + s) + 4(6t - s) - (-4t + 5s) ) / ( (-4t + 5s + 2(6t -s) )^2) V' = (-33t - 11s + 24t - 4s + 4t - 5s) / ( (-4t + 5s + 12t - 2s)^2) V' = (-5t - 20s) / ( (8t +3s)^2) s = 1 t = 2 V' = (-5(2) - 20(1) )/ ( (8(2) + 3(1) )^2) V' = (-10 - 20) / ( (16 + 3)^2) V' = (-30) / (19^2) V' = -30/361 The derivative of V when s=1 and t=2 is -30/361

Calculus

TutorMe

Question:

A parachute jumps out of an airplane at 10,000 feet above the ground. The jumper falls until they are 3,600 feet above the ground, where they release their parachute. Then, they coast down to the ground at a constant 16 feet/s until they reach the ground. If gravitational acceleration is 32 feet/second/second, how long does it take for the parachute jumper to reach the ground?

Brian J.

Answer:

up is positive, down is negative xt=10000=xo1 xo2=3600 vo1=0 vo2=-16 a1=-32 a2=0 a1(t)=a1=-32 v1(t)=-32t+vo1=-32t x1=-16t^2+xo1=-16t^2+10000 3600=-16t1^2+10000 16t1^2=6400 t1^2=400 t1=20 a2(t)=ao2=0 v2(t)=vo2=-16 x2(t)=-16t+xo2=3600-16t 0=3600-16t2 16t2=3600 t2=225 t=t1+t2=20+225=245 It takes the parachute jumper 245 seconds to reach the ground.

Calculus

TutorMe

Question:

Given that positive numbers a, b, and c add up to 57, and that b is three times larger than c, what is the maximum possible product of the three numbers?

Brian J.

Answer:

a+b+c=57 abc=A b=3c a+3c+c=57 a=57-4c A=(57-4c)(3c)c A=228c^2-16c^3 A'=456c-48c^2 Since area is a maximum, A'=0 0=P'=456c-48c^2 0=24c(19-2c) c=0 or 9.5 If c=0, P=0, which is not a maximum c=9.5 b=3*9.5=28.5 a=57-4*9.5=57-38=19 Check a+b+c=19+28.5+9.5=57 P=abc=19*28.5*9.5=5144.25 The maximum product is 5144.25

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