We have that the displacement of the spring is given by
$$s = a e^{kt} \sin(2 \pi f t) $$
To find the velocity of the spring, we need to consider the derivative of the displacement.
$$ \frac{d s}{dt} = s'(t) =2 \pi a f e^{-k t} \cos (2 \pi f t)-a k e^{-k t} \sin (2 \pi f t). $$
Then, we need to replace the value of the constants given in the exercise:
$$ a = 6\ m = 600\ cm; \qquad k=0.55; \qquad f = 15\ Hz. $$
So that the derivative become,
$$s'(t) =18000 \pi e^{-0.55 t} \cos (30 \pi t)-330. e^{-0.55 t} \sin (30 \pi t).$$
Finally, we need to obtain the velocity at the time t =2.5 sec, so
$$ s'(2.5) = 18000 \pi e^{-0.55 \cdot 2.5} \cos (30 \pi\ 2.5)-330. e^{-0.55 \cdot 2.5} \sin (30 \pi\ 2.5) $$
$$ \blue{s'(2.5)= -14297.7\ \left[\frac{cm}{sec}\right]}.$$
b) $\displaystyle \int_0^{90} \cos^3 x \sin^3 x dx$;
First let us rewrite the integral in the following way,
$$ \int_0^{90} \cos^2 x \cos x \sin^3 x dx = \int_0^{90} (1-\sin^2 x) \cos x \sin^3 x dx $$
Then, let us made a change of variable,
$$ u = \sin x, \quad du = \cos x dx $$
$$ \Rightarrow u_0 = \sin 0 = 0 , \quad u_f = \sin 90 = 1 $$\
$$\int_{u_0 = 0}^{u_f = 1} (1-u^2) u^3 du = \int_{ 0}^{ 1} u^3-u^5 du = \frac{u^4}{4}-\frac{u^6}{6} \bigg|_0^1 = \blue{\frac{1}{12}}$$
c) $\int_1^3 \frac{5}{x^2(x+2)} dx $
Let's apply partial fractions,
$$ \frac{5}{x^2(x+2)}= \frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2} $$
$$ 5 = A (x+2) + B\ x(x+2) + C x^2$$
$$ if \ x =0 \Rightarrow A = \frac{5}{2},$$ $$ if \ x =-2 \Rightarrow C = \frac{5}{4},$$ $$ if \ x =2 \Rightarrow B = -\frac{5}{4},$$
Therefore, our integral becomes
$$\int_1^3 \frac{5}{x^2(x+2)} dx = \int_1^3 \frac{5}{2}\cdot \frac{1}{x^2} - \frac{5}{4}\cdot \frac{1}{x} +\frac{5}{4}\cdot \frac{1}{x+2} dx $$ $$ \int_1^3 \frac{5}{2}\cdot \frac{1}{x^2} - \frac{5}{4}\cdot \frac{1}{x} +\frac{5}{4}\cdot \frac{1}{x+2} dx = \frac{5}{4}\left[ \int_1^3 \frac{2}{x^2} - \frac{1}{x} + \frac{1}{x+2} dx \right]$$ $$ \frac{5}{4}\left[ \int_1^3 \frac{2}{x^2} - \frac{1}{x} + \frac{1}{x+2} dx \right] = \frac{5}{4} \left[ -\frac{1}{x} - \ln x + \ln (x+2) \right]_1^3$$ $$ \frac{5}{4} \left[ -\frac{1}{x} - \ln x + \ln (x+2) \right]_1^3 = \frac{5}{12} (4-6 \ln (3)+\ln (125)) \approx 0.9319 $$ $$\blue{\therefore \int_1^3 \frac{5}{x^2(x+2)} dx =\frac{5}{12} (4-6 \ln (3)+\ln (125)) \approx 0.9319 }$$
Anthonysrc
