Question

Question #78bbb

Answer 1

a) Maximum height is `80` `"cm"`.
b) `0` `"seconds"`
c) Minimum height is `50` `"cm"`
d) `1.5` `"seconds"`
e)Time interval of `approx 1.2` `"seconds"`
f) `h(10) = 57.5` `"cm"`

Explanation:

Parts A through D To find the maximum or minimum height of the swing, find where the derivative of `h(x)`, `h'(x)` changes from positive to negative values or negative to positive, respectively.

`h(t)= 15cos((2pi)/3 t)+65`

Differentiate `h(x)`:

`h'(t) = -15((2pi)/3)sin((2pi)/3 t)`

`h'(t) = -10pi sin((2pi)/3 t)`

Set zero equal to `h'(x)`:
`0 = -10pi sin((2pi)/3 t)`

Use the zero product rule:
`0 = sin((2pi)/3 t)`

`(2pi)/3 t = 0, pi, 2pi, 3pi, ...`

`t = 0, 3/2 , 3, 9/2, ...`

Maxima: `t = 0, 3, 6, ...`
Minima: `t =3/2, 9/2, 15/2, ...`

Plug in some of these `t` values into the height function `h(x)`:
Max: `h(0) = 15cos0+65 = 15+65 = 80`
Min: `h(3/2) = 15cospi + 65 = -15+65 = 50`

Part E

After finding the points of intersection of the graphs of `h(t) = 15cos((2pi)/3) +65` and `y = 60` to be at `t approx 2.0877` and `t approx .9123`, I subtracted them to find the difference is a time length of `approx 1.1754` `"seconds"`

Part F
`h(10) = 15cos((20pi)/3) + 65`

`h(10) = 15(-1/2) + 65 = 65 - 7.5 = 57.5`

Answer 2

The given function representing the height `h` in cm of the person above the ground with time `t` is given by

`h=15cos((2pi)/3t)+65`

(a) h is a cosine function of t, So it will have maximum value when `cos((2pi)/3t)=1=cos0=>t=0` s
and the maximum height of the swing becomes

`h_(max)=15cos((2pi)/3xx0)+65=80cm`

(b) it takes `t=0`sec to reach maximum height after the person starts.

c) The minimum height of the swing will be achieved when `cos((2pi)/3t)=-1`

Minimum height

`h_(min)=15xx(-1)+65=50cm`

(d). Again `cos((2pi)/3t)=-1=cos(pi)=>t=3/2=1.5`s

Here `t=1.5`s represents the minimum time required to achieve the minimum height after start.

e) For `h=60` the equation becomes

`60=15cos((2pi)/3t)+65`

`=>60-65=15cos((2pi)/3t)`

`=>cos((2pi)/3t)=-1/3`

`=>t=[1/120(2nxx180pmcos^-1(-1/3))] sec" where " n in ZZ`

`=>t=(3npm0.912)`sec

when `n=0, t=0.912s` this is the minimum time to reach at height `h=60 cm` after start. just after that height `h=60 cm` will be achieved again for `n=1`

`=>t=(3xx1-0.912)=2.088`sec

So in `(2.088-0.912)=1.176s` sec (the minimum time difference) within one cycle the swing will be less than 60 cm above the ground.

f) The height of the swing at 10 s can be had by inserting `t=10`s in the given equation

`h_(10)=15cos((2pi)/3xx10)+65=57.5`cm