Yahoo Answers: Answers and Comments for Calculus help Related Rates Problems!!?!?!? [Mathematics]
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From Anonymous
enUS
Sun, 14 Dec 2008 13:19:30 +0000
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Yahoo Answers: Answers and Comments for Calculus help Related Rates Problems!!?!?!? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20081214131930AATQtGU
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From simplicitus: 1
If A is the altitude of the plane above t...
https://answers.yahoo.com/question/index?qid=20081214131930AATQtGU
https://answers.yahoo.com/question/index?qid=20081214131930AATQtGU
Wed, 17 Dec 2008 06:36:49 +0000
1
If A is the altitude of the plane above the antenna, and
H(t) is the horizontal distance from the plane to the antenna,
then the distance from the plane to the antenna is:
s(t) = sqrt(H(t)^2 + A^2)
For any functions f and g, if h(x) = f(g(x)) then we can use the chain rule:
let u = g(x), then
dh/dx = (df/du)(du/dx) = f' g'
Since s = (H(t)^2 + A^2)^(1/2), assuming A is constant
s'(t) = (1/2)(H(t)^2 + A^2)^(1/2) d((H(t)^2 + A^2))/dt
= (1/2)(H(t)^2 + A^2)^(1/2)(2)(H(t))H'(t)
= H(t) H'(t) / sqrt(s(t))
We know s(t) and s'(t) at the time of interest T, and we want to find H'(T). That means we need to find H(T) first
since s(t) = (H(t)^2 + A^2)^(1/2) we have
s(t)^2 = H(t)^2 + A^2
s(t)^2  A^2 = H(t)^2 or H(t) = sqrt(s^2  A^2)
so we have the single equation with a single unknown:
s'(T) = H'(T) sqrt(s(T)^2  A^2)) / sqrt(s(T))
so you can solve for H'(T)
#2 has the same flavor: you are given dr/dt and are being asked about dV/dt at a point specified by r = R
V = (pi)(r^2)(L), where L is the length of the cylinder (constant)
thus dV/dt = (dV/dr) (dr/dt) = (2)(pi)(rL) dr/dt
#3 is messier but the principle is the same. You are given dV/dt and are interested in the depth y of the water which is related to the volume by:
V= (pi/3)(y^2)(3Ry)
Differentiating both sides gives:
dV/dt = (pi/3)[6Ry  3y^2] dy/dt so
dy/dt = dV/dt (1 / [6Ry  3y^2]
You are given dV/dt, R, and y at the time of interest so you can determine dy/dt at the time of interest.
the radius r of the water surface is related to the depth y by:
r^2 + (R  y)^2 = R^2
r^2 + R^2  2Ry + y^2 = R^2
r = sqrt(2Ry  y^2)
so you can compute dr/dt in terms of R, y, and dy/dt for part b and then evaluate it for y = 8 in part c.