Compute the necessary values/derivatives of
at
:






Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial)
at
by


###
Another way of doing this would be to solve for the coefficients
in

by expanding the right hand side and matching up terms with the same power of
.
I think it is 39a or whatever unit you're using
9514 1404 393
Answer:
109°
Step-by-step explanation:
You always start a problem by taking a careful look at the information given and how it relates to what is asked. Here, the key information is in the symbols marking the lines PQ and RS. They are parallel.
This means segments QR and PS are transversals. Marked angle 41° will be an "alternate interior angle" with angle TPQ, so angle TPQ is also 41°.
The desired angle, PTR, is an exterior angle to ΔQTP, so its measure is the sum of remote interior angles TQP (68°) and TPQ (41°). That is, ...
∠PTR = ∠TQP +∠TPQ = 68° +41°
∠PTR = 109°
Answer:
a) 
b) 
Step-by-step explanation:
Part a
Since the buoy oscillates in simple harmonic motion the equation to model this is given by:

For this case from the info given we know that:

"It returns to its high point every 10 seconds." That means period =10
, and the angular frequency can be founded like this:

Assuming that the value for the phase is
our model equation is given by:

Part b
From definition we can obtain the velocity with the derivate of the position function and if w calculate the derivate we got this:

