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Math Help - Tricky Word Prob Decreasing Sphere

  1. #1
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    Tricky Word Prob Decreasing Sphere

    A sphere is losing volume at 6cm^3/minute

    How fast is the radius of the sphere decreasing when the surface are of the sphere is 24cm^2?

    "The volume of a sphere as a function of its radius is V=4pi/3*r^3. The surface area of a sphere as a function of its radius is SA=4pi*r^2."

    This is given as a problem in a calculus chapter containing derivatives. I can't get my initial equation started at all. The more I try to even think of it, I just complicate it more and more.

    We are talking radius decreasing so its not the total volume. Otherwise I'd just write 6cm^3/minute.

    Please help.
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  2. #2
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    Quote Originally Posted by Awesomeo View Post
    A sphere is losing volume at 6cm^3/minute

    How fast is the radius of the sphere decreasing when the surface are of the sphere is 24cm^2?

    "The volume of a sphere as a function of its radius is V=4pi/3*r^3. The surface area of a sphere as a function of its radius is SA=4pi*r^2."

    This is given as a problem in a calculus chapter containing derivatives. I can't get my initial equation started at all. The more I try to even think of it, I just complicate it more and more.

    We are talking radius decreasing so its not the total volume. Otherwise I'd just write 6cm^3/minute.

    Please help.
    \frac{dr}{dt} = \frac{dr}{dV} \cdot \frac{dV}{dt} = -6 \frac{dr}{dV} .... (1)

    Get \frac{dV}{dr} and hence \frac{dr}{dV} from V = \frac{4}{3} \pi r^3. Substitute into (1).

    Find the value of r when S = 24 by solving 24 = 4 \pi r^2. Substitute this value of r into (1) and take the magnitude of the result (why?)
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  3. #3
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    Thoroughly Confused

    So the wording there really confused me. I've gotten so far out of that, which looks like a fool proof plan. I just don't understand where to substitute.

    I can get the radius from the surface area. I get that. And the slope will be affected by the 6cm^3 at which its constantly losing per minute.

    I think I'm just not used to seeing those ways of talking about derivatives. If I was a calc pro, I would be done here, but I have to ask for clarifications.
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  4. #4
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    The radius for this is going to be sqrt 6 / sqrt pi... I'm still learning how to write it in math language, but now that I have the radius at this point in time, I believe I should take the derivative of that volume formula. I'm sorry but the more I look at that explanation the more I just want to chalk this one up as a loss.
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  5. #5
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    Quote Originally Posted by Awesomeo View Post
    I've gotten so far out of that, which looks like a fool proof plan. I just don't understand where to substitute.
    Yes, too far out! Just in case a picture helps...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to t, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which subject to the chain rule).

    Once you've differentiated with respect to r, you'll see you can just sub two values into these blanks, and virtually read off the value of dr/dt...




    __________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus: Gallery

    Balloon Calculus Drawing with LaTeX and Asymptote!
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    Wow, I mean, that is some nice stuff, the more I look at it the more I feel like I'm levitating! I feel close to it, but i don't see the step by step approach. I've found the radius. When I substitute the radius into the volume formula, I don't have a variable and I receive a constant value at my given radius of sqrt 6 / sqrt of pi .
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  7. #7
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    Quote Originally Posted by Awesomeo View Post
    When I substitute the radius into the volume formula...
    But you want to sub into a derivative - differentiate first. Then you'll notice something handy...
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    Are you saying I should differentiate the volume formula first? That would give me a variable. I've done that and I don't get something that looks like its part of my answer. It looks like a step, but I'm missing a good half of everything else I'm supposed to do.
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  9. #9
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    Quote Originally Posted by Awesomeo View Post
    Are you saying I should differentiate the volume formula first? That would give me a variable.
    Well, a function. What did you get?
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    4 pi r ^2 which is the same formula for the surface area... but that is dr/dV?
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  11. #11
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    Quote Originally Posted by Awesomeo View Post
    Are you saying I should differentiate the volume formula first? That would give me a variable. I've done that and I don't get something that looks like its part of my answer. It looks like a step, but I'm missing a good half of everything else I'm supposed to do.
    Have you been taught about related rates of change?
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  12. #12
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    Quote Originally Posted by Awesomeo View Post
    4 pi r ^2 which is the same formula for the surface area... but that is dr/dV?
    Yes it happens conveniently to also be the formula for SA... and no, it's dV/dr, but the point is yes it's the same as the formula for which you've been given a value. Fill those two blanks, then?
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    That is where we are at right now, I believe.
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  14. #14
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    I mean, don't you now have two numbers to fill the blank balloons...



    ?
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  15. #15
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    I'm really trying to get this balloon thing. So if I plug in the pieces of the derivative of that function into the ?'s such as

    4= ?1 and 2=?2 (the exponent)... Then I can use the chain rule with the formula for volume and differentiate that all together to get the dr/dt?

    Or am I just confusing myself with these balloons. I'm weak at implicit differentiation, and we are approaching related rates of change problems this week. Just so you know where I'm at with comprehension.
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