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Math Help - Homework for calculus thats hard URGENT

  1. #1
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    Exclamation Homework for calculus thats hard URGENT

    hey guys these 4 problems in my homework has been given me problems for the past two days. i just can't figure them out. please help me out i would greatly be thankful for it.

    1 (A) a cube of ice is melting a uniform rate of 3in cube per hr. Find the rate at which each side of the cube is decreasing when each side is 7 to the power of n.

    (B) a cube of ice is melting at a rate of 2 in cube /hr. find the rate at which the surface area is decreasing when the sides are 5 inches.

    2. If a ladder is 25 ft long and is sliding away from the bottom of a house at a rate of 3 ft/sec. Find the rate at which the top of the ladder is slideing down the side of the house when the top of the ladder is 20 ft above the ground.

    3. (A) A stone is dropped into a pond and makes concentric cirles expanding from the center. How fast is the area of the circle increasing when radius R=8ft and the radius is increasing at a rate of 3ft/sec

    4. A man 6ft tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. When he is 10ft from the base of the light at what rate is the tip of his shadow moving.

    Please help me out i got through the other 14 problems in my calculus packet and these are the last 4. PLEASE HELP.
    Last edited by uniquereason81; March 2nd 2008 at 08:30 PM.
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  2. #2
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    Oh man... word problems </3
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  3. #3
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    well, for the ice cube problem, I guess you have to use the formula for grow and decay, that's using e...
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by uniquereason81 View Post
    2. If a ladder is 25 ft long and is sliding away from the bottom of a house at a rate of 3 ft/sec. Find the rate at which the top of the ladder is slideing down the side of the house when the top of the ladder is 20 ft above the ground.
    Okay, if you draw the ground, and the edge of a building, they form a right angle. Then if you draw a ladder leaning against the building, you see you have a triangle with a right angle.

    And they give you the hypotenuse (ladder is 25 feet long) and the length of the opposite side (ladder is 20 ft above the ground)

    So we can use Pythagorean's Theorem of a^2 + b^2 = c^2

    Now, in this equation, a and b are variables since they each have the power to change. (as the height decreases, the length increases)

    We'll choose a to be our height, so we are trying to find our equation for the rate at which height changes. Now, the height changes as the width changes, as time changes. So we will differentiate with respect to the time.

    so a^2+b^2=c^2

    \frac{d}{dt}(a^2+b^2)=\frac{d}{dt}c^2

    2a~a\prime+2b~b\prime=0

    c is not a variable, it is a fixed size of 25, so it differentiates to zero.

    Now, a is the height of our ladder from the ground, a' is our rate at which the top of the ladder moves, b is the length from the side of the building, and b' is the rate at which the bottom of our ladder moves. We are trying to find a', so we solve for it.

    2a~a\prime=-2b~b\prime

    a\prime=-\frac{2b~b\prime}{2a}

    a\prime=-\frac{b~b\prime}{a}

    And now, a is given to us in the problem, the top of the ladder is 20 feet from the ground, so a = 20. b we can solve for with pythagorean's theorem (I won't do it, but the answer is 15), and b' is given to us in the problem, it is 3 feet per second.

    So we plug these values in:
    a\prime=-\frac{15*3}{20}

    a\prime=-\frac 94

    so the value of a' is decreasing at a rate of 9/4 feet per second. Which means that the height is decreasing, which means the top of the ladder is falling down at a rate of 9/4 feet per second.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by uniquereason81 View Post
    (A) A stone is dropped into a pond and makes concentric cirles expanding from the center. How fast is the area of the circle increasing when radius R=8ft and the radius is increasing at a rate of 3ft/sec
    Area of a circle is a=\pi~r^2

    and it changes with respect to time.

    So we differentiate with respect to time.

    \frac{da}{dt} = \frac{d}{dt}(\pi~r^2)

    a\prime = 2\pi~r~r\prime)

    Now plug in the values they gave to you to find the rate at which the area changes.
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