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Math Help - Driving me nuts!!! please help

  1. #1
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    Driving me nuts!!! please help

    A 9-ft long ladder is leaning against the side of a house, forming a right triangle with the ground and the side of the house. The base of the ladder begins to slide away from the house, which causes the top of the ladder to slide down the side of the house.
    The top of the ladder is falling at 4 ft/s.
    The top of the ladder 6 ft from the ground.

    1.How fast is the bottom of the ladder sliding away from the base of the house?
    2. What is the change in area for the triangle at this time?
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  2. #2
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    A 9-ft long ladder is leaning against the side of a house, forming a right triangle with the ground and the side of the house. The base of the ladder begins to slide away from the house, which causes the top of the ladder to slide down the side of the house.
    The top of the ladder is falling at 4 ft/s.
    The top of the ladder 6 ft from the ground.

    1.How fast is the bottom of the ladder sliding away from the base of the house?
    2. What is the change in area for the triangle at this time?

    1. Okay, for the first part, we will need to find the lengths of all sides of the triangle that is formed. We start by using the Pythagorean Theorem:

    Let's say that the height 6 ft. is y, and the ladder, which is 9 ft., represents the hypotenuse, and we'll call this z. Find x, or the horizontal distance on the ground:

    x^2 + y^2 = z^2
    x^2 + 6^2 = 9^2
    x^2 + 36 = 81
    x^2 = 45
    x = sqrt(45)
    x = sqrt(9)*sqrt(5)
    x = 3sqrt(5)

    Okay, now that we have all the distances we set up the related rate problem. Note, since the ladder, which is z, is a constant value in this process (the ladder's length doesn't change at all), we substitute it's constant value:

    x^2 + y^2 = 9^2
    Now, take derivatives with respect to time:
    2x*dx/dt + 2y*dy/dt = 0 Note, derivative of a constant value is zero
    x*dx/dt + y*dy/dt = 0 Divide both sides by 2

    Now, we substitute the necessary values to find dx/dt, which is the rate at which the ladder is sliding away from the base of the house

    Also, note that dy/dt is negative since it is falling downward, or in the negative direction:

    3sqrt(5)*dx/dt + 6*-4 = 0 Add 24 to both sides
    3sqrt(5)*dx/dt = 24 Divide both sides by 3sqrt(5)
    dx/dt = 8/sqrt(5) ft/s This value is positive because it is moving right along the x axis, which is defined to be positive.

    2. Now for part two, we have to find the change in area, or dA/dt. Start by writing the equation for the area of a triangle:

    A = xy/2
    A = (1/2) * (xy)

    Now take the derivative of both sides with respect to time:

    dA/dt = (x/2)*(dy/dt) + (y/2)*(dx/dt) This is gotten from the product rule.
    Now, just substitute the above values:

    dA/dt = (3sqrt(5))/2 * (-4) + (6/2) * 8/sqrt(5)
    dA/dt = -6sqrt(5) + 24/sqrt(5)
    dA/dt = -30/sqrt(5) + 24/sqrt (5)
    Multiplied the first term by sqrt(5)/sqrt(5)

    dA/dt = -6/sqrt(5) (ft^2)/s

    This should be it. Also, remember the units for the answers. Teachers sometimes get really picky and deduct points over stuff like this.

    Also, it should be noted that the area of the triangle will decrease until the ladder is resting completely on the ground, at which point the area will have decreased to zero units.
    Last edited by ajj86; November 7th 2008 at 09:18 AM. Reason: additional
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  3. #3
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    Quote Originally Posted by McDiesel View Post
    A 9-ft long ladder is leaning against the side of a house, forming a right triangle with the ground and the side of the house. The base of the ladder begins to slide away from the house, which causes the top of the ladder to slide down the side of the house.
    The top of the ladder is falling at 4 ft/s.
    The top of the ladder 6 ft from the ground.

    1.How fast is the bottom of the ladder sliding away from the base of the house?
    2. What is the change in area for the triangle at this time?
    1. The ladder forms a right triangle so the Pythagorean theorem applies. If x is the distance from the house to the foot of the ladder and y is the height of the top of the ladder up the house, then [tex]x^2+ y^2= 81[tex]. Now differentiate both sides of that equation with respect to t (using the chain rule of course) 2x dx/dt+ 2y dy/dt= 0. When the top of the ladder, y, is 6, x^2+ 6^2= 81 so x^2= 81- 36= 45. x= 3\sqrt{5} and you are told that dy/dt= 4. Put those numbers into the formula and solve for dx/dt.

    2. A= (1/2)xy so dA/dt= (1/2) (y dx/dt+ x dy/dt)
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