# Math Help - differentiation then transposition..

1. ## differentiation then transposition..

Hi all, I am new to this site and I'm just starting to get back into education. I need to figure out acceleration from speed then transpose for t to find out when accel reaches 0.
I'm given the formula for speed

v=12te-(t/2) +5

so my acceleration, a= dv/dt = -6te^-(0.5t) - 6e-(0.5t) which after 4.5s a= -3.48 m/s 2

if this is correct so far, I have no idea. If by some miracle it is, I need to transpose for t, to find the time when acceleration reaches zero. Any help would be greatly appreciated.

steve

2. ## Re: differentiation then transposition..

The correct expression is $-6te^{-t/2} + 12e^{-t/2}$ (though I would have differentiated the product the other way round so as not to have a negative sign at the front).

There is no need to transpose for $t,$ you have $a = -6te^{-t/2} + 12e^{-t/2} = 6e^{-t/2}(2-t),$ and you simply have to ask yourself what value(s) of $t,$ will make this zero.

3. ## Re: differentiation then transposition..

thats great, thank you for the reply, I had used the product rule to differentiate, how should I have done it?

4. ## Re: differentiation then transposition..

You differentiated correctly, BobP is just saying that he would have written it as $12e^{-t}- 6t^{-t}= 6e^{-t}(2- t)$.

And to "transpose for t to find out when accel reaches 0" just means to solve $6e^{-t}(2- t)= 0$ for t.

(I have no idea why you calculated a when t= 4.5. Was there another part of the problem?)

5. ## Re: differentiation then transposition..

Suppose that you had to differentiate $x^{2}e^{-2x}$ (or $e^{-2x}x^{2})$ wrt $x.$

Using the product rule you could write the result as

$2xe^{-2x} - 2x^{2}e^{-2x},$

or

$-2x^{2}e^{-2x}+2xe^{-2x}.$

The results are identical but the first is preferable. It uses one fewer character, but more importantly, if you are writing this quickly on paper it's possible to lose the negative sign at the front of the expression, it can get tangled up with an equals sign.