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!

Related Rates Problems

  1. Feb 9, 2013 #1
    I've tried these problems but I just can't wrap my head around them. Thanks to anyone that can help me out here. I have a couple more problems similar to these to solve, but if I can get to understand these then I'll have no problem with the others.
    Here are some diagrams I drew: http://gyazo.com/bf78da97179aa8d62d5e44b484ffcdc3 and http://gyazo.com/65689d4a0de1f23ceeae3c236c1c891e
    1. The problem statement, all variables and given/known data:
    A conical paper cup is 30cm tall with a radius of 10cm. The cup is being filled with water a rate of 16*pi/3 cm^3/sec. How fast is the water level rising when the water level is 6cm?
    dV/dt = 16*pi/3
    r=10
    h=30
    hw(height of water)=6
    dh/dt=?
    2. Relevant equations:
    V=1/3*pi*r^2*h


    3. The attempt at a solution:
    V=1/3*pi*r^2h
    dV/dt=1/3*pi*r^2*dh/dt
    16*pi/3=1/3*pi*10^2*dh/dt
    3/100*pi *16*pi/3=100*pi/3dh/dt *3/100*pi
    16*pi/100*pi=dh/dt
    16/100=dh/dt



    1. The problem statement, all variables and given/known data:
    A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 6/x ft/sec, where x is the distance from the person to the lamppost. Assume the scenario can be modeled with right triangles. At what rate is he length of the person's shadow changing when the person is 15 ft from the lamppost?
    y= height of lamppost, 20ft
    x= distance from lamppost to person, 15 ft
    dx/dt= 6/15
    2. Relevant equations:
    a^2 + b^2 = c^2


    3. The attempt at a solution:
    x^2 + y^2 = c^2
    2x*dx/dt + 2y*dy/dt = 0
    2*15*(6/15) + 2*20*dy/dt = 0
    12+40dy/dt=0
    dy/dt = -12/40
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2013 #2
    Radius of the cone filled with water varies as the water is poured in. At any instant, if the height of the water filled is h, the radius of the filled portion isn't r. You will have to relate r and h somehow. (Hint: Using similar triangles. :wink: )
     
  4. Feb 9, 2013 #3
    http://gyazo.com/62a49a2ec489bb6662afb35a72f981b9 . So how about r^2+h^2 = c^2? But if that is the case, I'm not sure what I would do with c^2. Or should I rearrange the equation in order to combine r^2+h^2=c^2 and V=1/3*pi*r^2*h?
     
  5. Feb 9, 2013 #4
    (see attachment)
    The angle θ always remains constant. You can find two values of tan θ from here and compare them to find a relation between r and h.
     

    Attached Files:

  6. Feb 9, 2013 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are confusing constants with variables. You can't say r=10 and then let r also be variable. Relabel some of them. Suppose R=10 is the radius of the cone and the height of the cone is H=30. Now suppose the height of the water is h. What is r, the radius of the surface at height h? As Pranav-Arora said use similar triangles.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Related Rates Problems
Loading...