Solving for a derivative - Question for an assignment

New to the forum, and ofcourse I type in Math help and up pops this great place!

I have an assignment due tomorrow and I cannot figure out this question... It is basic calc but not taking calc for over 10 years, I am rusty.

Could someone give me a hand with this question!!

Given the following equation for displacement:

S =S_{o }+ V_{o}T - 1/2(AT)^{2 }; Where S_{o}, V_{o }and A are constants.

Use the equation for the derivative to find V from the derivative of the displacement, and then A, from the derivative of velocity.

Any help with this would be great - If it can be broken down into steps so I can understand?

Thank you kindly....

-One frustrated student.

Jeff

Re: Solving for a derivative - Question for an assignment

I think the A should lie outside the parentheses (as in, $\displaystyle -\frac{1}{2}At^2$).

Find dS/dt:

$\displaystyle \frac{dS}{dt} = V - At \Rightarrow \frac{dS}{dt} + At = V$.

Differentiate again to find A.

Re: Solving for a derivative - Question for an assignment

To further this, I know you have to get the prime of S which will cancel out a bunch of the characters, but I am stuck with breaking that down.

To get the A... I am unsure. A friend mentioned to get the prime of the prime, but I am not sure if that is how to do it.

Re: Solving for a derivative - Question for an assignment

Quote:

Originally Posted by

**richard1234** I think the A should lie outside the parentheses (as in, $\displaystyle -\frac{1}{2}At^2$).

Find dS/dt:

$\displaystyle \frac{dS}{dt} = V - At \Rightarrow \frac{dS}{dt} + At = V$.

Differentiate again to find A.

THere is actually no brackets at all in the question as I placed there, but yes, you would be correct. I put them there to show the multiplication.

Can you please explain the above... I am looking forward to my calc tutor...

Re: Solving for a derivative - Question for an assignment

Basically what I did was differentiate both sides with respect to time (t). This is because, if X = Y, then dX/dt = dY/dt.

The LHS just becomes dS/dt. For the RHS, you can differentiate by finding the derivatives of each of the terms and then adding them up.

$\displaystyle \frac{dS}{dt} = \frac{d}{dt}S_0 + \frac{d}{dt}(V_0t) - \frac{d}{dt}(\frac{1}{2}At^2)$

The derivative of a constant ($\displaystyle S_0$) is zero. d/dt of any constant times t, is just that constant, so $\displaystyle \frac{d}{dt}(V_0t) = V_0$. The derivative of any expression in the form $\displaystyle At^n$ is just $\displaystyle Ant^{n-1}$ (look up "power rule").

Therefore, $\displaystyle \frac{dS}{dt} = 0 + V_0 - \frac{1}{2} (2At) = V_0 - At$

Re: Solving for a derivative - Question for an assignment

Quote:

Originally Posted by

**Hearzy** THere is actually no brackets at all in the question as I placed there, but yes, you would be correct. I put them there to show the multiplication.

Also, you should just skip the brackets. $\displaystyle (At)^2$ and $\displaystyle At^2$ mean entirely different things.

Re: Solving for a derivative - Question for an assignment

Richard,

Thank you kindly for your help - Next time I will drop the brackets... I see what I did there now. Thank you for questioning that...

I will be reviewing this in the morning but what you have wrote out does make sense. A bit of practice will go a long way.

Congrats on post 500 as well! Well deserved - Now how do I give a thumbs up? :)

-Jeff

Re: Solving for a derivative - Question for an assignment

Huh, didn't even notice my 500th post :)

You're welcome.