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Homework Help: Rate of change in an Isosceles triangle

  1. Sep 30, 2010 #1
    An Isosceles triangle has two equal sides of length 10cm. Let x be the angle between the two equal sides.

    a. Express the area A of the triangle as a function of x in radians.

    b. Suppose that x is increasing at the rate of 10 degrees per minute. How fast is A changing at the instant x = pi/3? At what value of x will the triangle have a maximum area?

    I've set the triangle and drew a line to cut it in half. the angle is now x/2, and the base is b/2. The hypoteneuse of each triangle is 10.

    cos x/2 = adj/hyp = h/10
    sin x/2 = opp/hyp = b/2/10 = b/20

    A = 100sinx/2 cosx/2
    = 100 sinx/2

    would that be the equation for part a?

    any help for part b would be greatly appreciated.

    thanks
     
  2. jcsd
  3. Sep 30, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    How did you go from "100 sin(x/2)cos(x/2)" to just "100 sin(x/2)"? What happened to the "cos(x/2)"? Perhaps you were trying to use the fact that sin(2a)= 2sin(a)cos(a).


    Once you have the correct formula for A as a function of x, then dA/dt= (dA/dx)(dx/dt) and you are told that dx/dt= 10 degrees per minute.
     
  4. Sep 30, 2010 #3
     
  5. Sep 30, 2010 #4

    Mark44

    Staff: Mentor

     
  6. Jan 18, 2012 #5
    Hey guys, so I'd like to resurrect this thread since I'm also working on it.

    So for part A) area=50sinx by double angle formula

    B) My answer is:

    dx/dt=10=0.1745rad/min

    da/dt=50cosx(dx/dt)

    =50cos(pi/3)(0.1745rad)

    =4.36cm^2/min

    For max area, using optimization techniques:

    A'=50cosx >0 for 0<x<pi/2 and 3pi/2<x<2pi (since dimensions can't be negative)
    A' <0 for pi/2<x<3pi/2
    A'= 0 for x= pi/2 and 3pi/2
    Using closed interval method:
    A(pi/2)=50
    A(3pi/2)=-50, therefore A will be max when x=pi/2

    I have a feeling I did the first part of B) wrong, what do the pros think?
     
  7. Jan 19, 2012 #6
    Bump, any takers? Bribes? Love perhaps?
     
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