# Acceleration problem

• Aug 4th 2009, 01:55 PM
Acceleration problem
The position equation for the movement of a particle is given by $\displaystyle s(t) = \sqrt{t^3 +1}$ where s is measured in feet and t is measured in seconds. Find the acceleration when t = 2 seconds.

I know that I have to get y'', I got $\displaystyle {3t^2/2} -\sqrt{t^3+1}$, but when trying to get y'' I just made a huge mess and came out to a wrong answer. Someone want to walk me through the rest of this?
• Aug 4th 2009, 02:05 PM
slider142
Remember to keep the symbols straight; there is no y in the actual problem. (Wink) Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).
First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and $\displaystyle t^3 + 1$ is the inner function. To keep things clear, I like to use exponents in place of roots, so I would write it out:
$\displaystyle s(t) = (t^3 + 1)^\frac{1}{2}$
Explicitly, we can see that s(t) = f(g(t)) where $\displaystyle g(t) = t^3 + 1$ and $\displaystyle f(t) = t^{\frac{1}{2}}$. By the chain rule, s'(t) = f'(g(t))g'(t).
$\displaystyle s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)$
In order to get the second derivative you will have to use the product rule in addition to the chain rule. Can you find s''(t) from here?
• Aug 4th 2009, 02:10 PM
Quote:

Originally Posted by slider142
Remember to keep the symbols straight; there is no y in the actual problem. (Wink) Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).
First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and $\displaystyle t^3 + 1$ is the inner function. To keep things clear, I like to use exponents in place of square roots, so I would write it out:
$\displaystyle s(t) = (t^3 + 1)^\frac{1}{2}$
Explicitly, we can see that s(t) = f(g(t)) where $\displaystyle g(t) = t^3 + 1$ and $\displaystyle f(t) = t^{\frac{1}{2}}$. By the chain rule, s'(t) = f'(g(t))g'(t).
$\displaystyle s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)$
Explicitly, we can see that s(t) = f(g(t)) where g(t) = t^3 + 1 and f(t) = t^(1/2)
In order to get the second derivative you will have to use the product rule in addition to the chain rule. Did you find your problem here?

Yes sir, whenever I integrate the power rule and the chain rule together I usually have a difficult time. So can you explain this to me so I can understand the concept better?
• Aug 4th 2009, 02:20 PM
slider142
In the above, we have
$\displaystyle s'(t) = \frac{3}{2}(t^3 + 1)^{-\frac{1}{2}}(t^2)$
Since we do not know its derivative offhand, we note that it is a product of two functions, each of whose derivative we do know. Specifically
$\displaystyle f(t) = (t^3 + 1)^{-\frac{1}{2}}$
and
$\displaystyle g(t) = t^2$
so that we have
$\displaystyle s'(t) = \frac{3}{2}f(t)g(t)$
By the product rule, we know that
$\displaystyle s''(t) = \frac{3}{2}(f'(t)g(t) + f(t)g'(t))$
The only place you will need that chain rule here is in finding f'(t). Note how similar f(t) is to s(t). What do you propose s''(t) is?
• Aug 4th 2009, 02:34 PM
Okay I did it the way you asked me to and came out with $\displaystyle {-9t^4/2}+{3(t^3+1)^-3/2/2}+3t^4 +3t$
• Aug 4th 2009, 02:46 PM
slider142
It looks like you're missing some small steps. Don't skip any steps until you get more experience with these rules. The first step is to get the parts of the product rule that you do not have, f'(t) and g'(t). Do not worry about how they interact with the rest of the equation yet. What did you get for f'(t)? What did you get for g'(t)?
• Aug 4th 2009, 02:55 PM
for f' I got 6t and for g' I got $\displaystyle -1/2(-\sqrt{(t^3+1)}) (3t^2)$
• Aug 4th 2009, 03:12 PM
slider142
Quote:

for f' I got 6t and for g' I got $\displaystyle -1/2(-\sqrt{(t^3+1)}) (3t^2)$

Okay, so in your case, you switched them around from my post and you let
$\displaystyle f(t) = 3t^2$
and
$\displaystyle g(t) = (t^3 + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{t^3 + 1}}$.
Your f'(t) is correct, but your g'(t) is not. g'(t) should be
$\displaystyle -\frac{1}{2}(t^3 + 1)^{-\frac{1}{2} - 1}(3t^2) = -\frac{1}{2}(t^3 + 1)^{-\frac{3}{2}}(3t^2)$
The outer function gets taken down one unit to the -3/2 power by the power rule of differentiation.
• Aug 4th 2009, 03:25 PM
Quote:

Originally Posted by slider142
Okay, so in your case, you switched them around from my post and you let
$\displaystyle f(t) = 3t^2$
and
$\displaystyle g(t) = (t^3 + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{t^3 + 1}}$.
Your f'(t) is correct, but your g'(t) is not. g'(t) should be
$\displaystyle -\frac{1}{2}(t^3 + 1)^{-\frac{1}{2} - 1}(3t^2) = -\frac{1}{2}(t^3 + 1)^{-\frac{3}{2}}(3t^2)$
The outer function gets taken down one unit to the -3/2 power by the power rule of differentiation.

Just to make sure I'm doing this correctly
right now I have
$\displaystyle \frac{1}{2}[6t(t^3+1)^{-\frac{1}{2}}+{-\frac{9t^4}{2}}(t^3+1)^{-\frac{3}{2}}]$
• Aug 4th 2009, 03:31 PM
slider142
Quote:

Just to make sure I'm doing this correctly
right now I have
$\displaystyle \frac{1}{2}[6t(t^3+1)^{-\frac{1}{2}}+{-\frac{9t^4}{2}}(t^3+1)^{-\frac{3}{2}}]$

That is exactly correct. (Nod)
• Aug 4th 2009, 05:40 PM
Quote:

Originally Posted by slider142
That is exactly correct. (Nod)

Awesome, okay so the final answer is?

$\displaystyle 3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{2}$
• Aug 4th 2009, 06:27 PM
slider142
Quote:

Awesome, okay so the final answer is?

$\displaystyle 3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{2}$

Almost. The second term should also be multiplied by 1/2.
• Aug 4th 2009, 07:24 PM
$\displaystyle 3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{4}$
$\displaystyle 3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{4}$