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Math Help - Rate of Change

  1. #1
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    Rate of Change

    I have a homework problem that I am unable to figure out.

    Consider a triangle (not necessarily a right triangle!) with the length of the base increasing at a rate of 1/2 cm/s. At what rate must the height change so that the area remains constant at the instant the base is 10cm and the height is 18cm. Hint: What are the quantites that are changing?
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  2. #2
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    Re: Rate of Change

    the area of a triangle is calculated by the formula:

    A = 1/2(b)(h), the area remains constant so you can calculate it's area for b = 10 and h = 18; it would be the same bc it's the height that is changing,

    Next, having the Area and the rate of change of the base, db/dt = 1/2cm/s, take the derivative of A = 1/2(b)(h), and solve for dh/dt,

    dokrbb

    btw, can you say which sides of the triangle are changing? and in what direction (neg or pos), what that means?, cheers
    Last edited by dokrbb; April 21st 2013 at 04:48 PM.
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    Re: Rate of Change

    Thank you so very much! This is great and very helpful.

    I will think about those questions once I finish the problem. Thank you again for your help.
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    Re: Rate of Change

    As I am trying to solve it I end up with 90=1/2(1/2)dh/dt and I have no idea how to continue. Also if the area remains the same and the base is increasing, the height must be decreasing which would make it neg.
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  5. #5
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    Re: Rate of Change

    Quote Originally Posted by DreamingofWolves View Post
    I have a homework problem that I am unable to figure out.

    Consider a triangle (not necessarily a right triangle!) with the length of the base increasing at a rate of 1/2 cm/s. At what rate must the height change so that the area remains constant at the instant the base is 10cm and the height is 18cm. Hint: What are the quantites that are changing?
    Very important here is the 'per second' in the given 1/2 cm/s so you must differentiate the Aea function with respect to time ,

     A = \frac{1}{2}bh

     \frac{d}{dt}(A) = \frac{d}{dt}(\frac{1}{2}bh)

     \frac{dA}{dt} = \frac{1}{2} \frac{d}{dt}(bh)

    Now use the product rule on the right hand side of the equation ,

     \frac{dA}{dt} = \frac{1}{2}(b \frac{dh}{dt} + h \frac{db}{dt})

    Now you have...

     \frac{dA}{dt} = 0 because area remains constant

     \frac{db}{dt} = \frac{1}{2} because the base is increasing at that rate

     b = 10 \ h = 18

    so when you plug these in your only unknown is  \frac{dh}{dt}

    Can you solve it now?

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    Re: Rate of Change

    I got as far as plugging everything it, it is solving for dh/dt. Do I solve it as if it is another variable or am I going to have to take the derivative of something?
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    Re: Rate of Change

    As far as you have got this formula, already differentiated,

     \frac{dA}{dt} = \frac{1}{2}(b \frac{dh}{dt} + h \frac{db}{dt})

    you solve for  \frac{dh}{dt} as it asks you - at what rate is the height changing?

    dokrbb
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  8. #8
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    Re: Rate of Change

    Quote Originally Posted by DreamingofWolves View Post
    I got as far as plugging everything it, it is solving for dh/dt. Do I solve it as if it is another variable or am I going to have to take the derivative of something?
    You don't have to take any more derivatives, dh/dt is just like another variable now, isolate dh/dt.

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