Serious calculus help needed.?

The sides of an equilateral triangle are increasing at the rate of 27 in/sec. How fast is the triangle's area increasing when the sides of the triangle are each 18 inches long?

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  • 1 decade ago
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    It's unclear whether each side is increasing at the rate of 27 in/sec or whether the perimeter is increasing at that rate. I am making the assumption that each side is increasing at that rate.

    The area of an equilateral triangle with side s is:

    A = (√3/4)s^2

    ds/dt = 27

    dA/ds = (√3/2)s

    dA/dt = (dA/ds)(ds/dt) = (√3/2)s*27 = (27√3/2)s

    = (27√3/2)(18) = 243√3 = 420.9 ft^2/sec

  • Puggy
    Lv 7
    1 decade ago

    Let s = the side of the triangle. We're given that

    ds/dt = 27

    We WANT to know what dA/dt is, where A is the area of the triangle.

    In order to solve this problem, we must relate A and s.

    We know that A = base times height, and our base is s but we need to know what height is.

    If you bisect an equilateral triangle, you'll notice that it forms two right angle triangles, with hypotenuse s and base s/2. That makes the height solveable by the Pythagoras Theorem.

    h^2 + (s/2)^2 = s^2, so

    h^2 = s^2 - (s/2)^2

    h^2 = s^2 - (s^2)/4, or

    h^2 = (3/4) s^2. Taking the square root of both sides, we get

    h = [sqrt(3)/2]s

    Since A = base times height, and base = s, height = [sqrt(3)/2]s:

    A = s[sqrt(3)/2]s, or

    A = [sqrt(3)/2]s^2.

    When solving related rates problem, we ALWAYS differentiate with respect to t. We have to solve this using implicit differentiation.

    dA/dt = 2[sqrt(3)/2]s (ds/dt)

    However, we know that ds/dt is equal to, which is given to be 27, so

    dA/dt = 2[sqrt(3)/2]s [27], or

    dA/dt = [27sqrt(3)] s

    In every related rates question, we have our "when" statement. It would be incorrect to plug in our numeric value (s = 18) early, and must be plugged in only AFTER the implicit differentiation and substitution of rates.

    WHEN THE SIDES OF THE TRIANGLE ARE EACH 18 INCHES LONG...

    Now, we plug in s = 18.

    dA/dt = [27sqrt(3)] [18]

    dA/dt = 486 sqrt(3)

    Now, your concluding statement would be:

    The triangle's area is increasing at a rate of 486 sqrt(3) square inches per second.

  • feanor
    Lv 7
    1 decade ago

    Area of equalateral triangle = s^2 * sqrt(3)/4

    dA/dt = s*sqrt(3)/2 ds/dt = 18 * sqrt(3)/2 * 27 = 243 * sqrt(3)

  • 4 years ago

    2(x^2+y^2)^2=25(x^2–y^2) differentiating implicitly 4(x^2 + y^2)(2x + 2yy ') = 25(2x - 2yy') divide by making use of two 4(x^2 + y^2) (x + yy') = 25(x - yy') 4(x^3 + xy^2 + x^2yy' + y^3y') = 25x - 25yy' y' (4x^2 y + 4y^3 + 25y) + x(4x^2 + 4y^2 - 25) = 0 y ' = x(25 - 4x^2 - 4y^2) /(4x^2 + 4y^3 + 25y) substitute x = 3 and y =a million y ' = 3(25 - 36 - 4) / (36 + 4 + 25) = -40 5 / sixty 5 = -9/13

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