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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 [tex]\frac{\pi }{12}[/tex] 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.

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ajeigbeibraheem

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 [tex]\dfrac{\pi}{12}[/tex] seconds it reaches a height of 60 cm from the ground.

i.e

[tex]R ( \dfrac{\pi}{12}) = 60[/tex]

[tex]60 = acos (\dfrac{b \pi}{12}) +d --- (2)[/tex]

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:

[tex]t = \dfrac{k \pi}{b}[/tex]

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

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

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 [tex]R ( \dfrac{\pi}{12})[/tex]

[tex]R ( \dfrac{\pi}{12}) = 60 -60 \ cos (\dfrac{\pi b}{12}) =60[/tex]

where ;

[tex]cos (\dfrac{\pi b}{12}) =0[/tex]

Then b = 6

∴

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