Question #78bbb

2 Answers
May 18, 2017

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
DesmosDesmos
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

May 18, 2017

drawndrawn

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=0sec 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=1.5s

Here t=1.5s 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)=2.088sec

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=10s in the given equation

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