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Math Help - Homework Prob

  1. #1
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    Homework Prob

    Find the equation of f the tangent to the curve at the given point:

    y= (2x)/(x-1), point (2,4)


    Also A canister is dropped from a helicopter hovering 50 m above the ground Unfortuanately its parachute does not open. Its been designed to withstand an impact velovity of 100 m / s Will it burst or not ? Acceleration = 9.8m/s^2
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    Quote Originally Posted by tnkfub
    Find the equation of f the tangent to the curve at the given point:
    y= (2x)/(x-1), point (2,4)
    The slope of the tangent line is the first derivative. So:
    y=\frac{2x}{x-1}

    y'=\frac{2x}{x-1}+\frac{2x*-1}{(x-1)^2}
    (Note: If this doesn't look like the quotient rule, that's because it isn't. I rarely use the quotient rule because I can't ever remember the correct form. I always move the denominator up and use the product rule.)

    y'=\frac{2x(x-1)-2x}{(x-1)^2}

    y'=\frac{2x^2-4x}{(x-1)^2}.

    Now, to find the slope at the point (2,4) we plug in the x value:
    y'(2)=\frac{2*2^2-4*2}{(2-1)^2}=0.

    The equation for a line is y = mx + b. So we want the equation for the line with a slope of 0 that passes through the point (2,4). Thus:
    4=0*2+b
    b=4

    Thus your line is y = 4.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tnkfub
    Also A canister is dropped from a helicopter hovering 50 m above the ground Unfortuanately its parachute does not open. Its been designed to withstand an impact velovity of 100 m / s Will it burst or not ? Acceleration = 9.8m/s^2
    We know that motion under constant acceleration will follow the form:
    y=y_0+v_0t+(1/2)at^2

    In this case a = -9.8 m/s^2, y0= 50 m, and since it is dropped from a (presumably) stationary helicopter, v0 = 0 m/s. So:
    y=50-4.9t^2.

    How fast is it moving when it strikes the ground? Well, when it hits the ground, we know that y = 0 m. We need to know when this happens before we can calculate how fast. So:
    0=50-4.9t^2
    t=\sqrt{\frac{50}{4.9}}

    Let's leave that as it is for now.

    To find out how fast it is moving at that time, recall that v=y' so:
    y=50-4.9t^2
    v=y'=-9.8t (The negative sign simply means the object is moving downward.)

    Plugging in the value of t when the object strikes the ground:
    v=-9.8*\sqrt{\frac{50}{4.9}}

    That's the exact number. The approximate value of v at the ground is about -31.3 m/s. Since the canister can take up to 100 m/s, the canister will not burst.

    -Dan

    Note: If you want to use Physics , there is an equation that rarely comes up in a Math class that solves the problem immediately:
    v^2=v_0^2+2a(y-y_0). (For constant a only!)
    You can use this to verify your solution if you like.
    Last edited by topsquark; March 13th 2006 at 03:54 PM. Reason: addendum
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  4. #4
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    Thanks ALot for the help Dan
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    Forum Admin topsquark's Avatar
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    You are welcome!

    I live to serve...

    -Dan
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