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Related Rates and Area of a Triangle

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data

    The length of the hypotenuse of a right triangle is 10 cm. One of the acute angles is decreasing at a rate of 5 degrees/s. how fast is the area decreasing when this angle is 30 degrees?


    2. Relevant equations




    3. The attempt at a solution

    I got the a and b using the cos and sin of the 30degrees. For a, I got 5[tex]\sqrt{3}[/tex] and for b, I got 5.

    I know that a will stay constant when the angle is decreasing so da/dt is 0. For the db/dt I got (-50[tex]\sqrt{3}[/tex])/2

    How am I supposed to find the area from this?

    Any help would be appreciated!
     
  2. jcsd
  3. Oct 21, 2007 #2

    G01

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    Try to start this problem by finding the Area of the triangle in terms of the angle that changing. After you do this, can you find the rate of change of the area in terms of the rate of change of the angle using some calculus?

    To sum up:

    [tex]\frac{d\phi}{dt}= 5 [/tex] where [tex]\phi[/tex] is the angle that is changing.

    Now, can you you find A in terms of phi, the angle?

    If so, you should be able to find [tex]\frac{dA}{dt}[/tex] in terms of [tex]\frac{d\phi}{dt}[/tex] using some calculus.

    Does this help?

    EDIT: I just realized you used "a" as a variable. Here, by "A," I mean the Area of the triangle. Just so you aren't confused.
     
    Last edited: Oct 21, 2007
  4. Oct 21, 2007 #3

    HallsofIvy

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    Don't start with what happens when [itex]\theta= 30 degrees[/itex]. This is a "dynamic" problem. What happens for any right triangle with hypotenuse 10 cm? Since you are asked, in particular, about the area, and area of a right triangle is "1/2 base times height, what is the height of a right triangl with hypotenuse 10 and angle [itex]\theta[/itex]? What is the base of a right triangle with hypotenuse 10 and angle [itex]\theta[/itex]? And so what is the area? Now differentiate both sides of that equation with respect to time.
     
  5. Oct 21, 2007 #4
    Thanks for the replies. I got the right answer by doing

    A = 1/2 b*h
    A = 1/2 (10cos [itex]\theta[/itex]* 10sin [itex]\theta[/itex])

    After that I differentiated to find dA/dt.
     
  6. Oct 21, 2007 #5

    G01

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    Nice Job!
     
  7. Oct 21, 2007 #6
    what is the answer to this problem, i want to check my work and answer! thanks.
     
  8. Oct 21, 2007 #7
    The answer to this is 25/36 [tex]\pi[/tex]
     
  9. Oct 22, 2007 #8
    mmph ... i did it the same method as you did and i am not getting that answer

    [tex]A=\frac{1}{2}xy[/tex]

    [tex]x=10\cos{30}[/tex]
    [tex]y=10\sin{30}[/tex]

    [tex]A=\frac{1}{2}10\sin{\theta}10\cos{\theta}[/tex]

    [tex]A=50\sin{\theta}\cos{\theta}[/tex]

    [tex]\frac{1}{2}\sin{2\theta}=\sin{\theta}\cos{\theta}[/tex]

    [tex]A=25\sin{2\theta}[/tex]

    [tex]\frac{dA}{dt}=50\cos{2\theta}\frac{d\theta}{dt}[/tex]

    [tex]\frac{dA}{dt}=-250\cos{60}\frac{cm}{s}[/tex]

    [tex]\frac{dA}{dt}=-125\frac{cm}{s}[/tex]

    argh!
     
    Last edited: Oct 22, 2007
  10. Oct 22, 2007 #9
    I got the same thing you got robo but my question is would it be negative? Because the rate is decreasing so shouldn't it be -5 degrees/sec?.
    My question to Bondgirl007 is what units is your answer in?
     
  11. Oct 22, 2007 #10
    yes it would be negative, my bad. but hmm ...
     
  12. Oct 22, 2007 #11
    Till here, I have the same answer as you but after this, I differentiated dA/dt and applied the Product Law to the sin[tex]\theta[/tex]cos[tex]\theta[/tex] and then I substituted pi/6 for [tex]\theta[/tex] and multiplied by that will give me a 25. And d[tex]\theta[/tex]/dt is [tex]\pi[/tex]/36 cm.
     
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