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Math Help - So I'm having some trouble...

  1. #1
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    So I'm having some trouble...

    On a good number of questions. I realize I'm new here, but I hope to stick around. I have left a more sizable message after the problems, in case you're interested.

    Here goes:

    5. "The base of a solid object is the region in the xy-plane bounded by the graphs of y=x^2 and y=4. Every verticle cross section parallel to the y-axis is a rectangle with a height of 3. What is the volume of the object?"
    This is the part of an exorbitant amount of problems that I have been working on that I don't know how to do and couldn't find any help for on Wikipedia or by searching google for teacher's notes and the like.

    Anyways, even if you only want to tackle one problem, I would be greatly appreciative.

    I also wanted to say that I have been shadowing this forum for a while, and would like to start contributing.

    Thanks for any help you can provide.
    Last edited by Jon.Monreal; August 4th 2008 at 08:00 AM.
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  2. #2
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    Quote Originally Posted by Jon.Monreal View Post
    [snip]
    1. "For what values of k does the graph of y=(4/3)x^3+2kx^2+5x+3 have two tangent lines parallel to the x-axis."
    [snip]
    You want dy/dx = 0 to have two solutions. Hint: A quadratic equation has two solutions when \Delta > 0.

    Quote Originally Posted by Jon.Monreal View Post
    [snip]
    10. "Use differentials to estimate 4.1^4."
    [snip]
    You should know that f(x+h) \approx f(x) + h f'(x). Consider the function f(x) = x^{1/4}, x = 4 and h = 1/10.

    Quote Originally Posted by Jon.Monreal View Post
    [snip]
    "Find the volume of the solid formed when the region between y=tanx and the x-axis on the interval [0,(pi/4)] is revolved around the x-axis."
    [snip]
    Where exactly are you stuck here? Substitute into the usual volume of solid of revolution formula:

    Rotation around x-axis: V = \pi \int_a^b y^2 \, dx = \pi \int _{0}^{\pi/4} \tan^2 x \, dx.


    Most of the other questions are also basic application of formulae that should be in your class notes or textbook. It would help if you said where exactly you were stuck with each one.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You want dy/dx = 0 to have two solutions. Hint: A quadratic equation has two solutions when \Delta > 0.


    You should know that f(x+h) \approx f(x) + h f'(x). Consider the function f(x) = x^{1/4}, x = 4 and h = 1/10.


    Where exactly are you stuck here? Substitute into the usual volume of solid of revolution formula:

    Rotation around x-axis: V = \pi \int_a^b y^2 \, dx = \pi \int _{0}^{\pi/4} \tan^2 x \, dx.


    Most of the other questions are also basic application of formulae that should be in your class notes or textbook. It would help if you said where exactly you were stuck with each one.
    My problem is that I don't have a textbook, only a list of questions, and no resources at hand related to the problems. I'm pretty lost on these questions as to even setting up formulas.

    EDIT: I have solved #9.
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  4. #4
    Super Member 11rdc11's Avatar
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    the formula for finding volume using washers is

    pi intergal (outer radius)'2 - (inner radius')2

    That should help you solve prob 3,6,7

    If you still uncertain what to do ill work an example for you
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    the formula for finding volume using washers is

    pi intergal (outer radius)'2 - (inner radius')2

    That should help you solve prob 3,6,7

    If you still uncertain what to do ill work an example for you
    I'm just confused as to what the ' is for, and why it is on different sides of the perentheses.
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  6. #6
    Super Member 11rdc11's Avatar
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    I use the ' to indicate to the power of

    for example 2'2 is (2 to the power of 2 which equals 4)
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  7. #7
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    Quote Originally Posted by 11rdc11 View Post
    I use the ' to indicate to the power of

    for example 2'2 is (2 to the power of 2 which equals 4)
    Sounds good. But shouldn't this be a definite integral?

    Perhaps an example would help. In the meanwhile, I'm going to continue trying to work at these.
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  8. #8
    Super Member 11rdc11's Avatar
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    Example: Find the volume of the solid of revolution obtained by revolving the plane region R bounded by y = sqrtx, the x axis, and the line x=4 about the x-axis

    so the formula is

    V= pi intergal (outer radius)'2 - (inner radius)'2

    V= pi intergal (sqrtx)'2

    V= pi intergal x

    V=pi (1/2)x'2

    so now plug in your upper bound of 4 and lower bound of 0


    V= pi(16/2)
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  9. #9
    Super Member 11rdc11's Avatar
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    Yes it is a definite integral
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  10. #10
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    Quote Originally Posted by 11rdc11 View Post
    Example: Find the volume of the solid of revolution obtained by revolving the plane region R bounded by y = sqrtx, the x axis, and the line x=4 about the x-axis

    so the formula is

    V= pi intergal (outer radius)'2 - (inner radius)'2

    V= pi intergal (sqrtx)'2

    V= pi intergal x

    V=pi (1/2)x'2

    so now plug in your upper bound of 4 and lower bound of 0


    V= pi(16/2)
    So, just to make sure I'm doing this right, for #6 I would end up with:

    pi integral tan^2 x
    pi(tanx-x)
    pi(1-(pi/4)) which is about equal to 0.6741915533

    Right?
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  11. #11
    Super Member 11rdc11's Avatar
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    Yep
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    You should know that f(x+h) \approx f(x) + h f'(x). Consider the function f(x) = x^{1/4}, x = 4 and h = 1/10.
    My question with this is, how did you get x=4 and h=1/10? Or are you referring to my problem and not the example?

    I'm a bit confused.

    EDIT: Solved #6 then. Thanks for checking it. By the way, when all of this is done, I am going to edit my first post so those who come back in the future can reference this thread.
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  13. #13
    Super Member 11rdc11's Avatar
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    10. "Use differentials to estimate 4.1^4."

    x=4 and dx=.1

    y = x'4

    dy= 4x'3dx

    dy= 25.6

    thus

    x'4 +dy =

    256 + 25.6 = 281.6

    hope that helps
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  14. #14
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    Quote Originally Posted by 11rdc11 View Post
    10. "Use differentials to estimate 4.1^4."

    x=4 and dx=.1

    y = x'4

    dy= 4x'3dx

    dy= 25.6

    thus

    x'4 +dy =

    256 + 25.6 = 281.6

    hope that helps
    Okay, I understand now.

    I was using my calculator to calculate the actual value (282.5751), and my answer wasn't even coming close. But that's just because I was doing the problem completely wrong.

    That's it for #10 then. Right now I'm trying to work on #1. Anybody have anything?
    Last edited by Jon.Monreal; August 2nd 2008 at 06:18 PM.
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  15. #15
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jon.Monreal View Post
    On a good number of questions. I realize I'm new here, but I hope to stick around. I have left a more sizable message after the problems, in case you're interested.

    Here goes:

    :1. "For what values of k does the graph of y=(4/3)x^3+2kx^2+5x+3 have two tangent lines parallel to the x-axis."
    This is the part of an exorbitant amount of problems that I have been working on that I don't know how to do and couldn't find any help for on Wikipedia or by searching google for teacher's notes and the like.

    Anyways, even if you only want to tackle one problem, I would be greatly appreciative.

    I also wanted to say that I have been shadowing this forum for a while, and would like to start contributing.

    Thanks for any help you can provide.
    As mr fantastic suggested, you want to find where \frac{\,dy}{\,dx}=0

    Since y=\tfrac{4}{3}x^3+2kx^2+5x+3, \frac{\,dy}{\,dx}=4x^2+4kx+5

    To find where \frac{\,dy}{\,dx}=0, apply the quadratic formula:

    x=\frac{-4x\pm\sqrt{16k^2-80}}{8}

    However, to get 2 solutions, we need 16k^2-80>0

    Thus, k<-\sqrt{5} and k>\sqrt{5}

    Thus, k can be any value within these two intervals:

    (-\infty,-\sqrt{5}) and (\sqrt{5},\infty)

    Note the round brackets. This means k=\pm\sqrt{5} is not included in our answer set.

    Does this make sense?

    --Chris
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