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Homework Help: Applications of Derivatives using Pythagoras

  1. Mar 9, 2005 #1
    Hello. Could anyone help me with a few questions that I'm stuck on? Any suggestions would be much appreciatd.

    1) A man 2m tall walks away from a lampost whose light is 5m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow growing when he is 10m from the lampost?

    Using pythagoras and taking the derivative I got:

    [tex] \frac{dy}{dx}= \frac{-x}{y} \frac{dx}{dt}[/tex]

    The derivative of x with respect to t(time) is 1.5 m/s so I think what I'm trying to find is the derivative of y with respect to t. Not too sure what to do here now but the answer is suppose to be 1m/s.

    2) At 1:00 p.m. ship A was 80 km south of ship B. Ship A is sailing north at 30 km/h and ship B is sailing east at 40 km/h. How fast is the distance between them changing at 3:00 p.m.?

    I think it's the times that is messing me up. Tried using the derivative of pythagoras again but where does the 80 km fit in? Answer: [tex] \frac{130}{\sqrt{17}} = 31.5 km/h [/tex]

    3) A water trough is 10 m long and a cross section has the shape of an isosceles triangle that is 1m across at the top and is 50 cm high. The trough is being filled with water at a rate of 0.4 m^3/min. How fast will the water level rise when the water is 40 cm deep? Answer: 5 cm/min

    Would differiantiating the equation for the area of a triangle fit into this?

    Anyways, thank you in advance.
  2. jcsd
  3. Mar 10, 2005 #2
    I'll do this one for you. Yes, the answer is 1m/s. Make a diagram of the lampost and the man (let base of lampost be at origin (0,0) and base of man at (10,0). Horizontal distance from man to lampost is 10 m. Draw a line from top of lampost passing through top of man and extend it till it hits x-axis. It will hit x-axis at (16.67,0), so length of shadow = 16.67-10 = 6.67.

    Generalise: let x be the distance from base of post to base of man, let y be length of shadow. Then by properties of similar triangles:

    y/2 = (x+y)/5, so y = 2x/3, hence dy/dx = 2/3. So if x is growing by 1.5 m/s, then y will be increasing by 1.5*(2/3) = 1 m/s. QED.
  4. Mar 10, 2005 #3


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    Previous msg completed Problem #1. Solution to Prob #2 uses similar techniques to Prob #1. Here's Prob #3:
    {Length of Trough} = (10 m)
    {Base of Triangular Cross-section} = (1 m)
    {Height of Triangular Cross-section} = (50 cm) = (0.5 m)
    {Area of Triangular Cross-section} = (1/2)*{Base}*{Height} = (1/2)*(1)*(0.5) = (0.25 m^2)
    {Volume of Trough} = {Length}*{Area} = (10)*(0.25) = (2.5 m^3)

    {Height of Water in Trough} = H
    {Base of Water Triangular Cross-section} = (1 - 0)*(H - 0)/(0.5 - 0) = 2*H
    {Area of Water Triangular Cross-section} = (1/2)*{Base}*{Height} = (1/2)*(2*H)*(H) = H2
    {Volume of Water in Trough} = V = {Length}*{Area} = (10)*(H2) = 10*H2
    {Water Fill Rate} = (0.4 m^3/min)

    Water Volume time rate of change is given by (dV/dt). Thus, from the above data:
    (dV/dt) = (d/dt){10*H2} = (10)*(2)*(H)*(dH/dt) = (20)*H*(dH/dt)

    It's given that {Water Fill Rate}=(0.4 m^3/min), so we place this value into the above equation and solve for (dH/dt) when H=(40 cm)=(0.4 m):
    (0.4) = (20)*H*(dH/dt)
    ::: ⇒ (dH/dt) = (0.4)/{(20)*H} = (0.02)/H = (0.02)/(0.4)
    ::: ⇒ (dH/dt) = (0.05 m/min) = (5 cm/min)

    Last edited: Mar 10, 2005
  5. Mar 10, 2005 #4
    Ah, you have to remember the property of similiar triangles for number 1. I see. Any hints for question number two?
  6. Mar 10, 2005 #5
    Questions like #2 are solved as examples in most calculus texts. Please look them up. If you still can't get the solution by tomorrow evening, I'll do it for you.
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