Question #78bbb

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#

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