Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Calculus help Related Rates Problems!!?!?!?

1) An airplane is flying at an altitude of 7 miles and passes directly over a radar antenna. When the plane is 10 miles from the antenna (s=10), the radar detects that the distance s is changing at the rate of 300 mph. What is the speed of the airplane at that moment?

2) The mechanics at Lincoln Automotive are reboring a 6-in. deep cylinder to fit a new piston. The machine they are using increases the cylinder's radius one thousandth of an inch every 3 minutes. How rapidly is the cylinder volume increasing when the bore (diameter) is 3,800 in?

3) Water is flowing at the rate of 6 m cubed/min from a reservoir shaped like a hemispherical bowl of radius 13 m. Given that the volume of water in a hemispherical bowl of radius R is V= (pi/3)(y^2)(3R-y) when the water is y units deep....

a) At what rate is the water level changing when the water is 8 m deep?

b) What is the radius r of the water's surface when the water is y m deep?

c) At what rate is the radius r changing when the water is 8 m deep?

I have the answers to 1 & 2, but not 3. I don't know how to do these, so please show me the steps to do these problems.

The answer to 1 is about 420.08 mph (got that from 4000/sqrt51)

The answer to 2 is .0239 in cubed/min (got that from 19pi/2500)

I don't have the answer to 3, but if you can still help me, I'd appreciate it. Thanks!

1 Answer

  • 1 decade ago
    Best Answer


    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)(3R-y)

    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.

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