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Math Help - Related Rates

  1. #1
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    Related Rates

    The altitude of a triangle is increasing at a rate of 1cm/min while the area of the triangle is in creasing at a rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100cm^2.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    The altitude of a triangle is increasing at a rate of 1cm/min while the area of the triangle is in creasing at a rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100cm^2.
    Let A be the area, b be the base and h be the height (or altitude)
    Let dh/dt be the rate at which the height is changing
    Let dhA/dt be the rate at which the area is changing
    Let db/dt be the rate at which the base is changing

    A = (1/2)bh

    when A = 100, h = 10
    100 = (1/2)(10)b
    => b = 20

    now, A = (1/2)bh
    => dA/dt = (1/2)h db/dt + (1/2)b dh/dt
    => dh/dt = [2dA/dt - h db/dt]/b
    now plug in all the values we know
    => dh/dt = [2(2) - 10(1)]/20
    => dh/dt = -3/10

    so the base is decreasing at a rate of -3/10 cm/min
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Let A be the area, b be the base and h be the height (or altitude)
    Let dh/dt be the rate at which the height is changing
    Let dhA/dt be the rate at which the area is changing
    Let db/dt be the rate at which the base is changing

    A = (1/2)bh

    when A = 100, h = 10
    100 = (1/2)(10)b
    => b = 20

    now, A = (1/2)bh
    => dA/dt = (1/2)h db/dt + (1/2)b dh/dt
    => dh/dt = [2dA/dt - h db/dt]/b
    now plug in all the values we know
    => dh/dt = [2(2) - 10(1)]/20
    => dh/dt = -3/10

    so the base is decreasing at a rate of -3/10 cm/min
    They are looking for db/dt and the answer is -1.6cm/min
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    They are looking for db/dt and the answer is -1.6cm/min
    sorry about that:

    Let A be the area, b be the base and h be the height (or altitude)
    Let dh/dt be the rate at which the height is changing
    Let dhA/dt be the rate at which the area is changing
    Let db/dt be the rate at which the base is changing

    A = (1/2)bh

    when A = 100, h = 10
    100 = (1/2)(10)b
    => b = 20

    now, A = (1/2)bh
    => dA/dt = (1/2)h db/dt + (1/2)b dh/dt
    => db/dt = [2dA/dt - b dh/dt]/h
    now plug in all the values we know
    => db/dt = [2(2) - 20(1)]/10
    => db/dt = -16/10 = -8/5
    => db/dt = -1.6 cm/min

    so the base is decreasing at a rate of -1.6 cm/min
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