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
`(\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

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

Explanation:

Given that:

R(t) = acos (bt) + d

at t= 0

R(0) = 0

0 = acos (0) + d

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

After
`(\pi)/(12)` seconds it reaches a height of 60 cm from the ground.

i.e

`R ( (\pi)/(12)) = 60`

`60 = acos ((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 = (k \pi)/(b)`

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

`R ((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 ( (\pi)/(12))`

`R ( (\pi)/(12)) = 60 -60 \ cos ((\pi b)/(12)) =60`

where ;

`cos ((\pi b)/(12)) =0`

Then b = 6

∴

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