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How do I do this?

  1. Apr 14, 2006 #1
    Hello,

    I have a few sums similar to this one:

    The following relation is given: dh/dx=e^3x + x^2
    Furthermore: h(0) =-9
    Calculate the value of h for x=3

    How would I solve this? Any hints?

    Hexa
     
  2. jcsd
  3. Apr 14, 2006 #2

    dav2008

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    You can rewrite the equation as [tex]dh=(e^{3x}+x^2)dx[/tex].

    Are you familiar with how to integrate an expression?
     
  4. Apr 14, 2006 #3

    matt grime

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    Please don't unnecessarily abuse notation and split dh/dx up. It is not a fraction.

    If dh/dx is what is given, what is h(x)?
     
  5. Apr 14, 2006 #4

    dav2008

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    Sorry, I guess I should have mentioned it as a disclaimer.

    I just find that it's easier to deal with problems such as this by pretending that it is a fraction. Same with the chain rule. You can pretend the denominator of one derivative cancels with another.

    I guess you're right and it's better to be rigorous in these situations to actually see what's going on and physically write out that you're integrating both sides with respect to x.
     
  6. Apr 14, 2006 #5

    matt grime

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    Well, the thing is you don't need to explicilty write down any integrals or do any manipulations of dh/dx. You can *solve it by inspection* (which is pretty much as unrigorous as it gets: write down the answer). Personally, if I wrote a question like this (say dy/dx=x^3 and y(0)=0) I would not expect anyone to use an integral sign to work out what y is, I would expect them to say, 'y(x)=x^4/4+k, and k is 0 since y(0)=0.' But , for any student reading, your teacher might have different ideas, so check.

    NB: I'd have no problem with going straight to the integrals, but I just don't like having dx's floating around at this level without an integral sign around.
     
  7. Apr 14, 2006 #6
    Some times I say multiply each side by dx and integrate. But some people have a problem with that act.
     
  8. Apr 15, 2006 #7
    Thanks a lot for your answers. I looked at the questions again and in only one of them was written that a pocket calculator is not allowed. That's the only instruction (I'm looking at old exams here). That's all.

    So how would I integrate it to get an answer? And Matt Grine: what are you exactly doing? It's just another example you're giving, right?

    Thanks,
    hexa
     
  9. Apr 15, 2006 #8
    Reply

    It is given that [tex] \frac{dh}{dx} = e^{3x} + x^2 [/tex]

    Since the expression on the right hand side is in terms of x, you can just integrate with respect to x on both sides of the equation!

    Also, remember to add an arbitrary constant once you have integrated!
     
  10. Apr 15, 2006 #9
    So that would then be (e^3*x^2)/2 + x^3/3 + K?

    but then there's still this left over:
    Furthermore: h(0) =-9
    Calculate the value of h for x=3

    and I don't see an h in the equation anymore :(

    Hexa
     
  11. Apr 15, 2006 #10
    When you integrate dh/dx you could say you are transforming h'(x) to h(x)... and you have an unknown constant in there such that when x = 0 h(x) = -9...what could possibly allow h(x) =-9 when all our terms in x add up to some number that isn't -9?...finally given that we know what allows h(0) = -9 what is h(x) when x = 3?

    *edit* My apologies if there is a more accurate way of saying the above
     
    Last edited: Apr 15, 2006
  12. Apr 15, 2006 #11

    dav2008

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    See what I mean, Matt ;)

    Also you might want to recheck your integration of [tex]e^{3x}[/tex].

    Here's an example of another problem if what Greg said was confusing...

    Given [itex]\frac{dy}{dx}=3x^2[/itex] and [itex]y(1)=9[/itex]
    [tex]\frac{dy}{dx}=3x^2 \\[/tex]

    [tex]\int \frac{dy}{dx} dx = \int 3x^2 dx\\[/tex]
    [tex]y(x)=x^3 + c\\[/tex]

    [tex]y(1)=9=(1)^3 + c\\[/tex]
    [tex]9=1+c\\[/tex]
    [tex]c=8\\[/tex]

    [tex]y(x)=x^3 + 8[/tex]
     
    Last edited: Apr 15, 2006
  13. Apr 15, 2006 #12

    matt grime

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    You integrated one side of an equality (e^{3x}+x^2}, so why didn't you integrate the other side of the equality (h) too?

    If x/3=2 and I wanted to find x what I'd presumably multiply by 3. if I multiply 2 by 3 I get 6, but now I've lost the x..... This is exactly what you've just said but in a different example.

    You have an equality, and you've forgotten to use it.
     
  14. Apr 15, 2006 #13
    ok, I'm still busy... so stay tuned ;)
    Next attempt for the integration:
    (E^(3*x) + x^3)/3
    1/3(x^3 + E^(3*x))

    I think this should be correct now. Sorry for the complicated display... I should finally learn to use latex for this, I know...

    Thus:
    y(x) = 1/3(x^3 + E^(3*x)) + c
    y(3) = -9 = 1/3(3^3 + E^(3*3)) +c
    -9 = 1/3(27 + E^9) +c
    -9 = 9 + 1/3 E^9 +c
    c = 1/3 E^9
    y(x) = 1/3(x^3 + E^(3*x)) + 1/3 E^9

    ?????
     
  15. Apr 15, 2006 #14
    Reply

    You are making progress, but here are 3 comments...

    1) The original function is h(x), so you should stick to it and avoid introducing y(x).

    2) Is h(3) = -9? :confused: Please check your first post...

    3) You wrote:
    [tex] -9 = 9 + \frac{1}{3}e^{\displaystyle9} + c [/tex]
    [tex]c = \frac{1}{3}e^\displaystyle9 [/tex]
    Something seems wrong here, apart from the incorrect substitution (see comment 2)
     
    Last edited: Apr 15, 2006
  16. Apr 16, 2006 #15
    Hello Pizzasky,

    1) you are right of course. Didn't pay attention to that.
    2) I know that h(0) =-9, and "calculate h for x=3"

    I simply don't know how to subsititude that into the equation, regardless if the integration is correct or not. What I did I just more of less copied from dav2008.

    Hexa
     
  17. Apr 16, 2006 #16

    Hootenanny

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    You need to find your constant of integration before substuting x =3 into your equation.

    You know that;

    [tex]h(0) = -9 \Rightarrow \frac{1}{3}e^{3\times 0} + \frac{1}{3} 0^3 + C = -9[/tex]

    Solve for C.

    Regards,
    ~Hoot
     
  18. Apr 16, 2006 #17
    ok, so
    C = -1/3e^(3*0) - 1/3*0^3 - 9
    So C = -9
    Then I put -9 into the euqaton before and subsititude for 3?
    1/3e^(3x) - 1/3x^3 + C
    1/3e^(3*3) - 1/3*3^3 - 9
    (getting desperate)

    Hexa
     
  19. Apr 16, 2006 #18

    Hootenanny

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    Not quite;

    [tex]e^0 \neq 0[/tex]

    [tex]e^0 = 1[/tex]
     
  20. Apr 16, 2006 #19
    err... right... *blush*

    thus
    C = -8
    and the rest of the process is correct?

    Hexa
     
  21. Apr 16, 2006 #20

    Hootenanny

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    Aren't you forgetting something? Take a look at the coefficent of the exponential function. :wink:
     
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