What is the answer to $\displaystyle

\frac{dt}{dv} = \frac{1}{kv - 9.8}

$?

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- Feb 9th 2008, 02:15 PMdark_knight_307
What is the answer to $\displaystyle

\frac{dt}{dv} = \frac{1}{kv - 9.8}

$? - Feb 9th 2008, 05:02 PMmr fantastic
- Feb 9th 2008, 05:13 PMmr fantastic
$\displaystyle \int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln|ax + b| + C$.

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

You should know this if studying differential equations. And you would*not*be asked to solve things like $\displaystyle \int \cos x \, e^x \, dx$ 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 $\displaystyle t = \frac{1}{k} \ln|kv - 9.8| + C$ you're probably going to want to know how to make v the subject. - Feb 9th 2008, 06:39 PMdark_knight_307
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. - Feb 9th 2008, 07:08 PMmr fantastic
$\displaystyle

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 $\displaystyle t \rightarrow +\infty$, $\displaystyle e^{kt} \rightarrow 0$.

So $\displaystyle v \rightarrow \frac{9.8}{k}$, as expected from the non-calculus argument. - Feb 9th 2008, 07:11 PMmr fantastic
- Feb 10th 2008, 10:03 AMdark_knight_307
- Feb 10th 2008, 10:47 AMmr fantastic
- Feb 10th 2008, 10:54 AMdark_knight_307
- Feb 10th 2008, 11:01 AMmr fantastic
- Feb 10th 2008, 11:22 AMdark_knight_307
:confused:

Here's my work....

$\displaystyle

v(t) = \frac {ds}{dt}

$

$\displaystyle

u = tk \implies du = k dt \implies \frac {du}{k} = dt

$

$\displaystyle

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

$

$\displaystyle

= \frac {kv_0 - 9.8}{k} \int e^{tk}dt + \int \frac {9.8}{k}dt

$

$\displaystyle

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

$

$\displaystyle

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

$

And is there any way to find C only with the information originally given? - Feb 10th 2008, 02:41 PMmr fantastic
Whoops, my mistake. (Headbang) 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 ..... - Feb 10th 2008, 05:49 PMmr fantastic
And if you did make that assumption, you'd have:

$\displaystyle

0 = \frac {kv_0 - 9.8}{k^2} + C \Rightarrow C = -\frac {(kv_0 - 9.8)}{k^2}

$.

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

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