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Math Help - First order linear differential equations and an application

  1. #16
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    What is the answer to <br />
\frac{dt}{dv} = \frac{1}{kv - 9.8}<br />
?
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  2. #17
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    Quote Originally Posted by dark_knight_307 View Post
    Well I've never heard of that before, so....
    Well, you could make a shrewd guess that the integral might have the form (A \sin x + B \cos x) e^x. Differentiate this, compare with \cos x \, e^x and therefore get the value of A and B.
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  3. #18
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    Quote Originally Posted by dark_knight_307 View Post
    What is the answer to <br />
\frac{dt}{dv} = \frac{1}{kv - 9.8}<br />
?
    \int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln|ax + b| + C.

    So t = \frac{1}{k} \ln|kv - 9.8| + C.

    You should know this if studying differential equations. And you would not be asked to solve things like \int \cos x \, e^x \, dx if integration by parts was not pre-requisite to this topic.

    Look, the simple fact is that your ability to solve integrals is way below par for attempting to solve differential equations. You MUST go back and review all the basic integration techniques pre-requisite to a differential equations topic. Otherwise you're in for a world of pain.

    The trouble now is that from t = \frac{1}{k} \ln|kv - 9.8| + C you're probably going to want to know how to make v the subject.
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  4. #19
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    Quote Originally Posted by mr fantastic View Post
    \int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln|ax + b| + C.

    So t = \frac{1}{k} \ln|kv - 9.8| + C.

    You should know this if studying differential equations. And you would not be asked to solve things like \int \cos x \, e^x \, dx if integration by parts was not pre-requisite to this topic.

    Look, the simple fact is that your ability to solve integrals is way below par for attempting to solve differential equations. You MUST go back and review all the basic integration techniques pre-requisite to a differential equations topic. Otherwise you're in for a world of pain.

    The trouble now is that from t = \frac{1}{k} \ln|kv - 9.8| + C you're probably going to want to know how to make v the subject.
    I'm sorry, but our teacher is expecting way too much from us I guess. I can integrate to get that, I was just making sure that I wasn't making a mistake. It's just that in the next part of the question it asks us to take the limit of that function, when solved for v, as t approaches infinity to find the velocity, and I keep getting infinity (or negative infinity, I forget), and this seems impossible. Wouldn't the limit be a constant, the terminal velocity?

    Yeah, I know it seems like I have no idea what I'm doing (I mostly don't), but my teacher assigned this to us and it's a major grade. And yeah, that's really weird about not knowing the integration by parts thing. I flipped through my book and we're not learning that for a couple more chapters.... Ah well.
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  5. #20
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    Quote Originally Posted by dark_knight_307 View Post
    I'm sorry, but our teacher is expecting way too much from us I guess. I can integrate to get that, I was just making sure that I wasn't making a mistake.

    Mr F says: Then you should say so in the first place. And show the working you've done.

    It's just that in the next part of the question it asks us to take the limit of that function, when solved for v, as t approaches infinity to find the velocity, and I keep getting infinity (or negative infinity, I forget), and this seems impossible. Wouldn't the limit be a constant, the terminal velocity? Mr F says: Yes. And in one of my replies I mentioned how to get the value without using calculus (just solve 0 = kv - 0.98 for v).

    [snip]
    <br />
t = \frac{1}{k} \ln|kv - 9.8| + C \Rightarrow e^{k(t - C)} = kv - 9.8 \Rightarrow Ae^{kt} = kv - 9.8 \Rightarrow v = \frac{Ae^{kt} + 9.8}{k}.

    But you did realise that k < 0, yes?

    So that in the limit t \rightarrow +\infty, e^{kt} \rightarrow 0.

    So v \rightarrow \frac{9.8}{k}, as expected from the non-calculus argument.
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  6. #21
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    Quote Originally Posted by mr fantastic View Post
    <br />
t = \frac{1}{k} \ln|kv - 9.8| + C \Rightarrow e^{k(t - C)} = kv - 9.8 \Rightarrow Ae^{kt} = kv - 9.8 \Rightarrow v = \frac{Ae^{kt} + 9.8}{k}.

    But you did realise that k < 0, yes?

    So that in the limit t \rightarrow +\infty, e^{kt} \rightarrow 0.

    So v \rightarrow \frac{9.8}{k}, as expected from the non-calculus argument.
    And of course A can be got from the initial condition v(0) = v_0.
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  7. #22
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    Quote Originally Posted by mr fantastic View Post
    And of course A can be got from the initial condition v(0) = v_0.
    Which would be kv_0 - 9.8 ... right?
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  8. #23
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    Quote Originally Posted by dark_knight_307 View Post
    Which would be kv_0 - 9.8 ... right?
    v = \frac{Ae^{kt} + 9.8}{k}.

    Sub v(0) = v_0:

    v_0 = \frac{A + 9.8}{k} \Rightarrow kv_{0} = A + 9.8 .....

    The dark_knight strikes
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  9. #24
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    Quote Originally Posted by mr fantastic View Post
    v = \frac{Ae^{kt} + 9.8}{k}.

    Sub v(0) = v_0:

    v_0 = \frac{A + 9.8}{k} \Rightarrow kv_{0} = A + 9.8 .....

    The dark_knight strikes
    Hahaha... thanks.

    Lemme see if I can integrate that velocity function on my own....


    <br />
v = \frac{(kv_{0} - 9.8)e^{kt} + 9.8kt}{k^2} + C<br />
?
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  10. #25
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    Quote Originally Posted by dark_knight_307 View Post
    Hahaha... thanks.

    Lemme see if I can integrate that velocity function on my own....


    <br />
v = \frac{(kv_{0} - 9.8)e^{kt} + 9.8kt}{k^2} + C<br />
?
    Close .... It's s = ....., not v = ......
    Last edited by mr fantastic; February 10th 2008 at 02:43 PM. Reason: Removed a mistake - the dark knight's numerator >is< correct.
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  11. #26
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    Here's my work....

    <br />
v(t) = \frac {ds}{dt}<br />
    <br />
u = tk \implies du = k dt \implies \frac {du}{k} = dt<br />


    <br />
\int ds = \int  \frac {(kv_0- 9.8)e^{tk} + 9.8}{k}  dt<br />

    <br />
= \frac {kv_0 - 9.8}{k} \int e^{tk}dt  +  \int \frac {9.8}{k}dt<br />

    <br />
= \frac {kv_0 - 9.8}{k^2} \int e^{u}du + \frac {9.8t}{k} * \frac {k}{k}<br />

    <br />
s(t) = \frac {(kv_0 - 9.8) e^{kt} + 9.8kt}{k^2} + C<br />

    And is there any way to find C only with the information originally given?
    Last edited by dark_knight_307; February 10th 2008 at 11:35 AM.
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  12. #27
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    Quote Originally Posted by dark_knight_307 View Post


    Here's my work....

    <br />
v(t) = \frac {ds}{dt}<br />
    <br />
u = tk \implies du = k dt \implies \frac {du}{k} = dt<br />


    <br />
\int ds = \int  \frac {(kv_0- 9.8)e^{tk} + 9.8}{k}  dt<br />

    <br />
= \frac {kv_0 - 9.8}{k} \int e^{tk}dt  +  \int \frac {9.8}{k}dt<br />

    <br />
= \frac {kv_0 - 9.8}{k^2} \int e^{u}du + \frac {9.8t}{k} * \frac {k}{k}<br />

    <br />
s(t) = \frac {(kv_0 - 9.8) e^{kt} + 9.8kt}{k^2} + C<br />

    And is there any way to find C only with the information originally given?
    Whoops, my mistake. You were right. You got it over the common denominator. Well done.

    To get C, you need to know something about where the object is at a particular time. Typically, you'll be told its initial position.

    You could assume that at t = 0, s = 0. Note that since downwards is being taken as the negative direction (and k is less than zero) you will have s < 0 for t > 0 .....
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  13. #28
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    Quote Originally Posted by mr fantastic View Post
    Whoops, my mistake. You were right. You got it over the common denominator. Well done.

    To get C, you need to know something about where the object is at a particular time. Typically, you'll be told its initial position.

    You could assume that at t = 0, s = 0. Note that since downwards is being taken as the negative direction (and k is less than zero) you will have s < 0 for t > 0 .....
    And if you did make that assumption, you'd have:


    <br />
0 = \frac {kv_0 - 9.8}{k^2} + C \Rightarrow C = -\frac {(kv_0 - 9.8)}{k^2}<br />
.


    Then s(t) = \frac {(kv_0 - 9.8) e^{kt} + 9.8kt}{k^2} - \frac {(kv_0 - 9.8)}{k^2}


     = \frac {(kv_0 - 9.8) \left( e^{kt} - 1 \right) + 9.8kt}{k^2}
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