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Related Rates homework

  1. Jul 19, 2008 #1
    this assignment is on related rates. I believe questions 3/4 are on linear approximation and relate back to the last unit.

    My problem is that I am entirely unconfident on this work, and am going to be taking a test on the material soon. If someone could check my work, I'd be delighted.

    Homework Assignment
    1. Fabio stands atop his 16 foot ladder when he realizes that the ladder
    is slipping down the side of the building. He decides that the base of
    the ladder is moving away from the bottom of the building at a rate
    of 2 feet per second when it is 3 feet from the bottom of the building.
    How fast is Fabio falling at that instant?

    To do this problem I construct a triangle. Height is H Base is L and the distance between the two is 16.
    H^2 + L^2 = 256. Therefore (dh/dt)2h +(dl/dt)2l=0 At this instant, h=3, which means l = sqrt(256-9) = sqrt(247). I am going to divide both sides of my equation by 2 to obtain (dh/dt)h +(dl/dt)l = 0. plugging in gives (x)(3) +(-2)(sqrt(247))=0; x = -2/3 sqrt (247)

    2. A decaying nerf ball shrinks in such a way that its radius decreases
    by 1/6th inch per month. How fast is the volume changing when the
    radius is one quarter of an inch?

    dv/dt = (4)(pi)(r^2)(dr/dt)=4(pi)((1/4)^2)(-1/6)

    For the following two questions, suppose that y = 2x^2 − 3x + 1.
    3. Find and simplify a formula for the y increment, ∆y.

    ∆y = f(x+ ∆x) - f(x) = 2x^2 +4x(∆x)+2(∆x)^2-3x-3(∆x)+1-2x^2+3x-1= 4x(∆x)+2(∆x)^2 -3(∆x)

    4. Find a formula for the y differential dy.
    dy=f'(x)dx= (4x-3)dx

    5. When a stone is dropped in a pool, a circular wave moves out from
    the point of impact at a rate of six inches per second. How fast is the
    area enclosed by the wave increasing when the wave is two inches in

    dA/dt = dr/dt (2r)(pi) = (6)(2)(2)(pi) =24(pi) in^2/sec

    6. The electric resistance of espresso as a function of its temperature is
    given by
    R = 6.000 + 0.002T 2
    where R is measured in Ohms and T in degrees Celsius. If the tem-
    perature is decreasing at the rate of 0.2 degrees per second, find the
    rate of change of the resistance when T = 38 ◦ C.

    dR/dt = 0.004 T (dT/dt)= 0.004 x 38 x -0.2 = -.0304
    Last edited: Jul 19, 2008
  2. jcsd
  3. Jul 19, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Assuming that, for the last one, you meant R= 6+ 0.002T^2, they are all correct.
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