Question

Kaliska is jumping rope. The vertical height of the center of her rope off the ground R(t) (in cm) as a function of time t (in seconds) can be modeled by a sinusoidal expression of the form a cos(b. t) + d. , At t = 0, when she starts jumping, her rope is 0 cm off the ground, which is the minimum. After $\frac{\pi }{12}$ seconds it reaches a height of 60 cm from the ground, which is half of its maximum height. Find R(t). t should be in radians.

Answer 1

Answer:

R (t) = 60 - 60 cos (6t)

Step-by-step explanation:

Given that:

R(t) = acos (bt) + d

at t= 0

R(0) = 0

0 = acos (0) + d

a + d = 0 ----- (1)

After $\dfrac{\pi}{12}$ seconds it reaches a height of 60 cm from the ground.

i.e

$R ( \dfrac{\pi}{12}) = 60$

$60 = acos (\dfrac{b \pi}{12}) +d --- (2)$

Recall from the question that:

At t = 0, R(0) = 0 which is the minimum

as such it is only  when a is  negative can acos (bt ) + d can get to minimum at t= 0

Similarly; 60 × 2 = maximum

R'(t) = -ab sin (bt) =0

bt = k π

here;

k  is the integer

making t the subject of the formula, we have:

$t = \dfrac{k \pi}{b}$

replacing the derived equation of k into R(t) = acos (bt) + d

$R (\dfrac{k \pi}{b}) = d+a cos (k \pi)$ $= \left \{ {{a+d \ for \ k \ odd} \atop {-a+d \ for k \ even}} \right.$

Since we known a < 0 (negative)

then d-a will be maximum

d-a = 60  × 2

d-a = 120 ----- (3)

Relating to equation (1) and (3)

a = -60 and d = 60

∴ R(t) = 60 - 60 cos (bt)

Similarly;

For $R ( \dfrac{\pi}{12})$

$R ( \dfrac{\pi}{12}) = 60 -60 \ cos (\dfrac{\pi b}{12}) =60$

where ;

$cos (\dfrac{\pi b}{12}) =0$

Then b = 6

∴

R (t) = 60 - 60 cos (6t)