Can someone help me solve this problem?
Posted: 29 Apr 2015, 12:45Could someone help me out with how to solve this problem. Thank you in advance.
$mathjax$s(0) = 1$mathjax$
and $mathjax$s(0) = a \cos(0 \times \omega) = a$mathjax$
then we can find the value of $mathjax$a$mathjax$
We also know that the period is
$mathjax$P=1.1$mathjax$
that means that for all $mathjax$x,$mathjax$
$mathjax$s(x+P)= s(x)$mathjax$
. Therefore $mathjax$a \cos\left(\omega x + \omega \times P\right) = a\cos(\omega x)$mathjax$
.But the period of the cosine function is
$mathjax$2\Pi$mathjax$
. So $mathjax$\omega P = 2\Pi$mathjax$
. That will give you the value of $mathjax$\omega$mathjax$
b. You just have now to evaluate your function with the value of
$mathjax$a$mathjax$
and $mathjax$\omega$mathjax$
found at the previous question.c. Use your calculator to sketch the curve of your function and look around
$mathjax$t=1$mathjax$