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Thread: related rate problems: rectangle's diagonal / kite string

  1. #1

    related rate problems: rectangle's diagonal / kite string

    1) The length L of a rectangle is decreasing at 2 cm/s (centimeters per second), while the width is increasing at 2 cm/s. When L = 12 cm and w = 5 cm, find the rate of change of the length of a diagonal of the rectangle.

    Is there a formula for the length of a diagonal of a rectangle? I have no idea how to go about this problem.

    2) Terry flies a kite at a height of 300 ft. The wind carrying the kite horizontally away at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 ft away from her?

    I set up a triange and got h = 300, b = 400, and the hypotenuse equals 500. I don't know what the 25 is supposed to stand for though? h', b'? I don't really understand what the question is asking. Please help me!

    thank you in advance!

  2. #2
    Elite Member stapel's Avatar
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    1) Have you never heard of the Pythagorean Theorem...?

    Draw the length and the width. Draw the diagonal. Note that this forms a right triangle. For the given length and width, use the Pythagorean Theorem to find the hypotenuse. (You can find the Pythagorean Theorem online.)

    Pick variables for the length, the width, and the hypotenuse, such as "L", "w", and "h". Plug these into the Pythagorean Theorem, and differentiate implicitly with respect to time "t". Plug in the known values for L, w, h, dL/dt, and dw/dt. Solve for dh/dt.

    2) Draw the height "h" above ground, the horizontal distance "x" from Terry, and the hypotenuse marking the string length "s". Since this is a right triangle, the Pythagorean Theorem again applies. In this case, of course, dh/dt = 0, and dx/dt = 25. Solve in a manner similar to (1).

    If you get stuck, please reply showing how far you have gotten. Thank you.

    Eliz.

  3. #3
    ok, well, i did the first one, and i got half the answer. so it must be mostly right, i just divided by an extra two somewhere. (i cannot work in dy/dx format, i just call it y'...)

    so this is what i did:

    a^2 + b^2 = c^2
    25 + 144 = c^2
    c = 13

    then the deriv of that..
    (2a)(a') + (2b)(b') = (2c)(c')
    (10)(2) + (24)(-2) = (2c)(c')
    20-48 = (2c)(c')
    c' = -14/26

    the real answer for c' is -14/13. what did i do wrong??

  4. #4
    Elite Member
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    Quote Originally Posted by ihatecalc
    ok, well, i did the first one, and i got half the answer. so it must be mostly right, i just divided by an extra two somewhere. (i cannot work in dy/dx format, i just call it y'...)

    so this is what i did:

    a^2 + b^2 = c^2
    25 + 144 = c^2
    c = 13

    then the deriv of that..
    (2a)(a') + (2b)(b') = (2c)(c')
    (10)(2) + (24)(-2) = (2c)(c')
    20-48 = (2c)(c')
    c' = -14/26

    the real answer for c' is -14/13. what did i do wrong??
    You multiplied eveything by 2. See?.

    I will let l=length, w=width, D=diagonal

    [tex]\L\\l^{2}+w^{2}=D^{2}[/tex]

    Differentiate:

    [tex]\L\\2l\frac{dl}{dt}+2w\frac{dw}{dt}=2D\frac{dD}{dt }[/tex]

    Now, divide thorugh by 2


    The kite problem is a little misleading. "When the kite is 500 feet from her" can be interpreted as along the string or horizontally.

    Let's say it's along the length of the string.

    You have a triangle. Of course, ol' Pythagoras again.

    The kite is 300' high, ley y=300

    The kite is flying horizontally at 25 ft/min., dx/dt=25

    The kite is flying horizontal, therefore, dy/dt=0

    Using Pythagoras, we find x=400, when y=300 and D=500(as you did).

    Now, set it up and solve for dD/dt.

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