1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Related Rates (increasing cone radius question)

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I've worked through both parts of this question twice in what I assume is the correct manner, but I'm receiving an unexpected result from part B. The question is as follows:

    Sand is dumped such that the shape of the sandpile remains a cone with height equal to twice the radius.
    A) If the sand is dumped at the constant rate of 20ft^3/s, find the rate at which the radius is increasing when the height reaches 6 feet.
    B) Repeat for a sandpile for which the edge of the sandpile forms an angle of 45° with the horizontal.

    2. Relevant equations

    Volume of a cone [itex]V=(1/3)πr^2h[/itex]
    Where r = radius
    Where h = height

    3. The attempt at a solution

    [itex]dV/dt=d/dt((1/3)πr^2)*h + (1/3)πr^2*(d/dt)(h)[/itex]
    [itex]20ft^3/s=(2/3)πr(dr/dt)*h + (1/3)πr^2*(1)[/itex]

    If the height = 2*radius, and h=6, r=3

    [itex]20=(2/3)π(3)(dr/dt)*(6) + (1/3)π(3^2)[/itex]
    [itex]20=(36/3)π(dr/dt) + (9/3)π[/itex]
    [itex](20-3π)/12π = dr/dt[/itex]

    So when the height of the cone is 6ft, the radius is expanding at about (20-3π)/12π or .2805ft/s.

    B) If the edge of the sandpile forms a 45° angle with the horizontal, that means the peak of the cone is 90°, and the height is equal to the radius (since both the bisected angle at the peak and the angle at the edge of the base must be 45°)

    [itex]20=(2/3)π(6)(dr/dt)*(6) + (1/3)π(6^2)[/itex]
    [itex]20=(72/3)π(dr/dt) + (36/3)π[/itex]
    [itex](20-12π)/24π = dr/dt[/itex]

    So if the height is equal to the radius, and the height is 6ft, the radius seems to be expanding at about (20-12π)/24π or -.2347ft/s. Since the base of the cone can't be shrinking, I know I must have done something wrong, but I don't see my error.
  2. jcsd
  3. Nov 19, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    Why are you putting dh/dt=1? In the first case h=2r, so dh/dt=2dr/dt. I can't see any reason for setting it to be 1.
  4. Nov 19, 2013 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Also, factor the (pi/3) term out. All it does is clutter your calculation and make it hard to follow.

    I get a slightly different dr/dt for part 1.
  5. Nov 19, 2013 #4
    I set d/dt[h] = 1 because I understand the power rule to state that a variable with no exponent (or more correctly, an exponent of 1) to differentiate to 1 (h^1 = 1h^0 = 1(1) = 1). Are you suggesting I convert h to terms of r prior to differentiation?

    A) [itex]V = 1/3(π)(r^2)(h)[/itex]
    1. [itex]dV/dt = d/dt[1/3(π)(r^2)]2r + 1/3(π)(r^2)d/dt[2r][/itex]
    2. [itex]20 = 2π/3(r)dr/dt(2r) + π/3(r^2)2dr/dt[/itex]
    3. [itex]20 = 2π/3(3)dr/dt(6) + 2π/3(3^2)dr/dt[/itex]
    4. [itex]20 = 12πdr/dt + 6πdr/dt[/itex]
    5. [itex]20 = 6πdr/dt(2+1)[/itex]
    6. [itex]20/3 = 6πdr/dt[/itex]
    7. [itex]6.667/6π = dr/dt[/itex]
    Ans. [itex]0.3537 = dr/dt[/itex]

    B) [itex]V = 1/3(π)(r^2)(h)[/itex]
    1. [itex]dV/dt = d/dt[1/3(π)(r^2)]r + 1/3(π)(r^2)d/dt[r][/itex]
    2. [itex]20 = 2π/3(r)dr/dt(r) + π/3(r^2)dr/dt[/itex]
    3. [itex]20 = 2π/3(6)dr/dt(6) + π/3(6^2)dr/dt[/itex]
    4. [itex]20 = 24πdr/dt + 12πdr/dt[/itex]
    5. [itex]20 = 12πdr/dt(2 + 1)[/itex]
    6. [itex]20/3 = 12πdr/dt[/itex]
    7. [itex]6.667/12π = dr/dt[/itex]
    Ans. [itex]0.1768 = dr/dt[/itex]

    The answer for part B makes much more sense that way, thank you! I notice that the rate of the radius's expansion in part B is half that of part A, which makes sense considering the radius is twice that of A's. However, I'm not sure I understand why I would want to convert my independent variable into terms of the dependent variable? I see that in step 5 this allows me to factor out (Xπ(dr/dt)) and divide, rather than subtract (which is what led to a negative answer in my previous attempt), so does this mean I should always convert to dependent terms prior to differentiation?

    I'm sorry for the simple questions. I'm taking a self-taught college course while deployed, and there's no instructor to turn to.
    Last edited: Nov 19, 2013
  6. Nov 19, 2013 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Let's make this simple:

    The volume of a cone is V = [itex]\frac{π}{3}[/itex]r[itex]^{2}[/itex]h

    By implicit differentiation w.r.t. time and the chain rule,

    dV/dt = [itex]\frac{π}{3}[/itex][2rh dr/dt + r[itex]^{2}[/itex]dh/dt]

    This is the general equation for dV/dt, when no relationship between r and h is known.

    In your problem, two different relationships between r and h are given. In the first, h = 2r; in the second, r = h. Given these relationships, the equation for dV/dt may be adjusted:

    for h = 2r, then dh/dt = 2 dr/dt,

    dV/dt = [itex]\frac{π}{3}[/itex][2r(2r) dr/dt + r[itex]^{2}[/itex](2)dr/dt]

    dV/dt = [itex]\frac{π}{3}[/itex][4r[itex]^{2}[/itex] + 2r[itex]^{2}[/itex]](dr/dt)

    dV/dt = [itex]\frac{π}{3}[/itex][6r[itex]^{2}[/itex]](dr/dt)

    With this last equation, you can now substitute the given values for h and dV/dt and solve for dr/dt. (dV/dt = 20 ft[itex]^{3}[/itex]/s and h = 6 ft, which implies r = 3 ft.)

    When the relationship between r and h is altered, then the equations above can be changed according to the new relationship between variables.
  7. Nov 19, 2013 #6
    A variable to the first power's derivative is equal to one only when differentiating with respect to that variable. So, dh/dh=1, but dh/dt doesn't necessarily.

    You don't have to, but you would need to find dh/dt in terms of r and dr/dt and substitute that at some point. Personally I think it would be easier to get h in terms of r before differentiation though.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted