Re: trigonometric problem

On Nov 17, 7:10 pm, Temp <t...@xxxxxxxxxxxxxx> wrote:
On Sat, 17 Nov 2007 16:43:54 -0800 (PST), conrad <con...@xxxxxxxxxx>
wrote:





On Nov 17, 6:20 pm, Temp <t...@xxxxxxxxxxxxxx> wrote:
On Sat, 17 Nov 2007 15:37:42 -0800 (PST), conrad <con...@xxxxxxxxxx>
wrote:

h(t) = 64 -46cos(t*pi/5)

15. The function h above gives the height above
the ground, in feet, of a passenger on a Ferris
wheel t minutes after the ride begins. During
one revolution of the Ferris wheel, for how
many minutes is the passenger at least 100 feet
above the ground? Round your answer to the
nearest hundredth of a minute.

I approach as follows:

100 = 64 - 46cos(t*pi/5)
36/-46 = cos(t*pi/5)
acos(36/-46) = t*pi/5

So I can find t when it is
100 ft above ground.

But how would I find the duration
while it is above 100 feet?

The answer is 2.14 but I'm not
sure how to reach that answer.

Find the value of t such that h(t) is a maximum (when a passenger is
at the greatest height).

Wouldn't that involve calculus?
This is precalculus.

In this case, the value of t in question can be found by inspection.

You have h(t) = 64 - 46 cos(t*pi/5), and you know that the cosine
function has range [-1, 1]. If cos(t*pi/5) = -1, then h(t) = 64+46,
its largest possible value.

So, what value of t makes cos(t*pi/5) = -1?

I have 5.

I see now. After having found
where it equals 100, and then
finding the maximum we have
all the info we need to solve
it. The time elapsed from
height 100 to the max is
5 - 3.93 = 1.07
Of course, this is half
the time it is above height
100 ft. Therefore, 2*1.07 = 2.14
is our answer.

Thanks!

--
conrad

.

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