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    cluelezz123's Avatar
    cluelezz123 Posts: 3, Reputation: 1
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    #1

    Oct 7, 2008, 09:51 PM
    aplications of differientiation
    Hi there,

    I am given a scenario stating that

    The sides of an equilateral triangle are decreasing at a rate of da/dt=-(root6) cm. Find the rate of change of:

    a) The area of the triangle
    b) The height

    I have drawn a diagram but I don't know how to go on about solving these problems
    cluelezz123's Avatar
    cluelezz123 Posts: 3, Reputation: 1
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    #2

    Oct 7, 2008, 10:11 PM

    So I Know that for a) I'm meant to find da/dt
    b) dh/dt

    But I don't know to do arrive at the answer... please help.
    cluelezz123's Avatar
    cluelezz123 Posts: 3, Reputation: 1
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    #3

    Oct 8, 2008, 03:55 AM
    For a) A=root(3)/4 * s^2
    So I'm trying to find Da/Dt
    So the related rate of change is

    dA/dt=DA/Da * Da/dt

    Diff A=root(30)/4 * s^2 would give root(3)/2 *s

    so this is what it would look like
    DA/dt=root(3)/2 * s * -root(6)
    So far is this right and how would I find that missing side length of s.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Oct 9, 2008, 12:30 AM

    You're using 'a', now 's'... i think that you're confusing yourself. If i understood well what you meant, then, this is an equilateral triangle and all the sides are equal so a=s.

    Use the formula:



    To find dA/dh, use the other formula for area of triangle,



    Now you need a formula linking a and h. When you find that, substitute a in the formula of the area of the triangle and differentiate to have dA/dh.

    Hope it helped.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #5

    Oct 10, 2008, 09:03 AM
    The area of an equilateral triangle is given by , where s is the length of a side.

    Differentiate:



    You are given ds/dt, plug it in.

    The height of an equilateral is given by

    Do the same thing as above.
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    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Oct 11, 2008, 02:12 AM

    In other words, that's neraly the same as what I said...
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #7

    Oct 11, 2008, 05:16 AM
    I'm sorry, I reckon you did. My apologies.

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