1. ## Finding the Derivative

The function is L(t)=8.4sin((2pi/365)(t-80))+12.6

How would I go about finding the L'(t)? Should I simply multiply within the parenthesis and then multiply that with 8.4sin or is there another way to go about it?

-m

2. Originally Posted by blurain
The function is L(t)=8.4sin((2pi/365)(t-80))+12.6

How would I go about finding the L'(t)? Should I simply multiply within the parenthesis and then multiply that with 8.4sin or is there another way to go about it?

-m
it's just the chain rule:

$\displaystyle \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

thus, $\displaystyle L'(t) = 8.4 \cos \left( \frac {2 \pi}{365}(t - 80) \right) \cdot \left( \frac {2 \pi}{365}(t - 80) \right)'$

now find $\displaystyle \left( \frac {2 \pi}{365}(t - 80) \right)'$ and plug it in

3. So L'(t)=8.4sin((2pi/365)(t-80))x(2pi/365)?

And if so, the next part of the question asked that I use a graph of L(t) to obtain a rough estimate of $\displaystyle \int{L(t)}dt$ from 0 to 80. How should I go about estimating this problem?

4. Originally Posted by blurain
So L'(t)=8.4sin((2pi/365)(t-80))x(2pi/365)?
no, the derivative of sine is cosine

And if so, the next part of the question asked that I use a graph of L(t) to obtain a rough estimate of $\displaystyle \int{L(t)}dt$ from 0 to 80. How should I go about estimating this problem?
not sure about the estimate part. i would have probably directly computed the integral (it's actually not that hard), but i suppose that's not want you want. i'm trying to remember if there is any rule or theorem that allows us to estimate integrals, but nothing in this context is coming to mind... what do your notes/text say?

5. You can estimate the area using rectangle sums (if you have the graph). This is quite rough, but you can get better results as you add more rectangles.

6. Originally Posted by wingless
You can estimate the area using rectangle sums (if you have the graph). This is quite rough, but you can get better results as you add more rectangles.
ah yes, Riemann sums. i would think there was an easier way. that gets annoying, just doing the integral would be easier, which would make asking for an estimate useless...