Question

Hugo's bike tire has a piece of gum stuck to it. The distance G(t)G(t)G, left parenthesis, t, right parenthesis (in \text{cm}cmstart text, c, m, end text) between the gum and the sidewalk as a function of time ttt (in seconds) can be modeled by a sinusoidal expression of the form a\cdot\sin(b\cdot t)+da⋅sin(b⋅t)+da, dot, sine, left parenthesis, b, dot, t, right parenthesis, plus, d. At t=0t=0t, equals, 0, the gum is halfway between the ground and its maximum height, at 35\text{ cm}35 cm35, start text, space, c, m, end text. The gum reaches its maximum height of 70 \text{ cm}70 cm70, start text, space, c, m, end text from the ground \dfrac{\pi}{20} 20 π ​ start fraction, pi, divided by, 20, end fraction seconds later. Find G(t)G(t)G, left parenthesis, t, right parenthesis. \textit{t}tstart text, t, end text should be in radians.

Answer 1

G(t) = 35·sin(10t) +35

Explanation:

You are to find a, b, d to fill in the form ...

G(t) = a·sin(bt) +d

such that the maximum and minimum values of G(t) are 70 and 0, respectively, G(0) = 35 = d, and the quarter-period is π/20 seconds.

We know "a" is the difference between the maximum value and the starting value, so ...

a = 70 -35 = 35

The value of "b" is 2π divided by the period (which is four times the quarter-period). So, we have ...

b = 2π/(4·π/20) = 10

The desired function is ...

G(t) = 35·sin(10t) +35