Derivatives of trig functions and isosceles triangles.

  1. The base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At what rate is the base angle increasing when the area is 100 cm2?

    I wasnt really sure where to start on this question so i tried my best at an answer. i'm sure i've gone wrong with this question so i appreciate any guidance.

    the isosceles triangle divides into two right triangles with bases of 10 cm and areas 50 cm2

    tan[tex]\theta[/tex] = [tex]\frac{h}{10}[/tex]

    [tex]\frac{d\theta}{dt}[/tex] sec[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{10}[/tex]

    [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex]. cos [tex]^{2}[/tex] [tex]\theta[/tex]

    A = [tex]\frac{1}{2}[/tex] b x h

    50 = [tex]\frac{1}{2}[/tex] (10) . h
    h = 10

    tan [tex]\theta[/tex] = [tex]\frac{10}{10}[/tex]
    tan [tex]\theta[/tex] = 1

    we know,

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

    (cos[tex]\theta[/tex] tan [tex]\theta[/tex]) [tex]^{2}[/tex] + cos [tex]^{2}[/tex] [tex]\theta[/tex] = 1

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


    [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex] . cos[tex]^{2}[/tex] [tex]\theta[/tex]

    = [tex]\frac{1}{10}[/tex] . cos[tex]^{2}[/tex][tex]\frac{1}{2}[/tex]
    = 0.077 rads / min

    or 4.6 rads /s


    Thanks in advance!
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,302
    Staff Emeritus
    Science Advisor

    Here's your error. "when the area is 100 cm2" refers to the original isosceles triangle and that had base 20, not 10.

     
  4. but if i plug in
    A = (1/2) b x h

    100 = (1/2) (20) h

    h is still equal to 10 cm. Is that the only error you see? because that does not effect the rest of the equation...:S
     
  5. is the rest of this equation anywhere near correct?? :(
     
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