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Math Help - rate of change- calculus worded question

  1. #1
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    rate of change- calculus worded question

    Air is being pumped into a spherical ballon so that its volume increases at a rate of 80pi cm^3/s/ How fast is the radius, r, increasing when r=2? (you may find it useful to know the volume of a sphere of radius r is given by V=4/3 pi r^3)

    Been trying to figure this out for ages and no luck at all doing these worded calculus questions and I gotta figure out how to do them for tomorrow exam (first uni maths exam ever for me!)!!

    If anyone could help it would be fantastic!
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  2. #2
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    Hello, sooty1

    I bet you couldn't do worded problems in Algebra I, either.


    Air is being pumped into a spherical balloon so that its volume increases at a rate of 80\pi cm³/s.
    How fast is the radius r increasing when r=2 ?

    The volume of a sphere is given by: . V \;=\;\frac{4}{3}\pi r^3

    Differentiate with respect to time: . \frac{dV}{dt}\;=\;4\pi r^2\cdot\frac{dr}{dt}

    We are given: . \frac{dV}{dt} = 80\pi,\;\;r = 2

    Plug those in and solve for \frac{dr}{dt}

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  3. #3
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    Quote Originally Posted by Soroban
    Hello, sooty1

    I bet you couldn't do worded problems in Algebra I, either.



    The volume of a sphere is given by: . V \;=\;\frac{4}{3}\pi r^3

    Differentiate with respect to time: . \frac{dV}{dt}\;=\;4\pi r^2\cdot\frac{dr}{dt}

    We are given: . \frac{dV}{dt} = 80\pi,\;\;r = 2

    Plug those in and solve for \frac{dr}{dt}

    thanks dude. i understand it now.

    just wasn't too sure how to use the given variables.

    and I have no idea what Algebra I is! lol
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  4. #4
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    Quote Originally Posted by sooty
    Air is being pumped into a spherical ballon so that its volume increases at a rate of 80pi cm^3/s/ How fast is the radius, r, increasing when r=2? (you may find it useful to know the volume of a sphere of radius r is given by V=4/3 pi r^3)

    Been trying to figure this out for ages and no luck at all doing these worded calculus questions and I gotta figure out how to do them for tomorrow exam (first uni maths exam ever for me!)!!

    If anyone could help it would be fantastic!
    You are told the the volume is increasing at a rate of 80 \pi \mbox{ cm^3/s}, you are also told that volume:

    <br />
V=\frac{4}{3} \pi r^3<br />

    So differentiating this with respect to time will give the rate of increase of volume:

    <br />
\frac{dV}{dt}=\frac{d}{dt} \ \left[\frac{4}{3} \pi r^3 \right]=4 \pi r^2 \frac{dr}{dt}<br />

    Now when r=2:

    <br />
\frac{dV}{dt}=4 \pi 2^2 \frac{dr}{dt}=16 \pi \frac{dr}{dt}<br />

    but we are told that this rate of increase is 80 \pi \mbox{ cm^3/s}, so:

    <br />
\frac{dr}{dt}=5 \mbox{cm/s}<br />

    RonL
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  5. #5
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    Quote Originally Posted by Soroban
    Hello, sooty1

    I bet you couldn't do worded problems.


    okay, i've tried to solve a few more since figuring out that question. but i reckon im still a tad lost...

    i tried doing these but not much luck. i might sound silly puting these up but i think its better to have it out there and take time to learn it properly!

    1."The surface area of a cube is increasing at a rate of 20cm^2 per hour. Find the rate (in cm^3 per hour) of change in the volume of the cube and us it to compute the rate of change in the volume when the side of the cube is of length 10."

    2. "A spectator standing at a distance of 3000m from a lauch site, is observing a rocket launch. The rocket is launched vertically and is rising at a rate of 300m/sec.
    When its altitude is 4,000m how fast is the distance between the rocket and the spectator changing at that instant?"


    1. dSA/dt = 20cm^2 cm/hr

    V=l^3
    dV/dl = 3l^2

    so do we need to find dV/dt
    edit: found out how to do 1.


    2. dd/dt = 300m/s


    argh... i can find small bits from the worded equations but not the whole thing. its really annoying!!!


    What is everyone's method of attempting/solving these questions?
    Last edited by sooty; June 21st 2006 at 05:43 AM.
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  6. #6
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    Hello again, sooty!

    Advice: make a sketch . . .


    1. The surface area of a cube is increasing at a rate of 20 cm²/hr.
    Find the rate (in cm³/hr) of change in the volume of the cube
    and use it to compute the rate of change in the volume when the side of the cube is of length 10.

    I won't make a sketch for the cube; we know what one looks like, right?
    Let x = side of the cube.
    We are asked for the rate of change of the volume: . \frac{dV}{dt}

    The volume of a cube is: . V\;=\;x^3
    . . Differentiate with respect to time: . \frac{dV}{dt}\;=\;3x^2\cdot\frac{dx}{dt} [1]

    The left side is what we want. .If we know everything on the right side, we're golden.

    We are told that: x = 10 . . . but they didn't tell us \frac{dx}{dt}
    . . Instead they said the surface area is changing at 20 cm²/hr: . \frac{dA}{dt} = 20

    Okay, we know the area of a cube . . . six faces, each with an area of x^2
    . . So we have: . A\:=\:6x^2\quad\Rightarrow\quad\frac{dA}{dt}\,=\,1  2x\cdot\frac{dx}{dt}
    . . Then: . 12x\cdot\frac{dx}{dt}\:=\:20\quad\Rightarrow\quad \frac{dx}{dt}\,= \, \frac{5}{3x}

    Substitute into [1]: . \frac{dV}{dt}\;=\;3x^2\cdot\frac{5}{3x}\:=\:15x


    Therefore, when x = 10:\;\;\frac{dV}{dt}\;=\;15(10)\:=\:150 cm³/hr.



    2. A spectator standing at a distance of 3000 m from a lauch site, is observing a rocket launch.
    The rocket is launched vertically and is rising at a rate of 300 m/sec.
    When its altitude is 4,000 m, how fast is the distance between the rocket
    and the spectator changing at that instant?
    Code:
                                  * R
                              *   :
                     x    *       :
                      *           : y
                  *               :
              *                   :
        - * - - - - - - - - - - - * -
          S          3000         L

    The spectator is at S, the launch site is L, the rocket is at R.
    We are told that SL = 3000.
    Let y = RL, the height of the rocket.
    And let x = SR, the distance of the rocket from the spectator.
    . . And we want \frac{dx}{dt}.

    We need an equation that ties together all those factors.
    We have a right triangle . . . How about Pythagorus?

    Pythagorus says: . x^2 \:=\:y^2 + 3000^2

    Differentiate with respect to time: . 2x\cdot\frac{dx}{dt} \:= \:2y\cdot\frac{dy}{dt}

    So we have: . \frac{dx}{dt} \:= \:\frac{y}{x}\cdot\frac{dy}{dt} [2]


    The left side is what we want . . . Do we know everything on the right side?
    We know that, at that particular instant . . . the height is 4000 m: y = 4000
    . . and the rocket is rising at 300 m/sec: \frac{dy}{dt} = 300 . . . um, we don't know x.

    But we can find x . . . We have a right triangle with legs of 3000 and 4000.
    . . Hence: . x^2\:= \:3000^2 + 4000^2\:= \: 5000^2\quad\Rightarrow\quad x\,=\,5000

    Substitute into [2]: . \frac{dx}{dt} \:= \:\frac{4000}{5000}(300) \:= \: 240 m/sec.

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