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Mathematics 17 Online
OpenStudy (anonymous):

I'm stuck on this one as well The function f(t) = 65 sin (pi over 5t) + 35 models the temperature of a periodic chemical reaction where t represents time in hours. What are the maximum and minimum temperatures of the reaction, and how long does the entire cycle take? maximum: 65°; minimum: 35°; period: pi over 5 hours maximum: 100°; minimum: −30°; period: pi over 5 hours maximum: 65°; minimum: 35°; period: 10 hours maximum: 100°; minimum: −30°; period: 10 hours

OpenStudy (anonymous):

I dont really want to download anything

OpenStudy (anonymous):

maximum of sine is when the input is \(\frac{\pi}{2}\) (you get 1 in that case, which is the largest sine can be)

OpenStudy (anonymous):

set \[\frac{\pi}{5}t=\frac{\pi}{2}\] and solve for \(t\)

OpenStudy (anonymous):

where would I start?

OpenStudy (anonymous):

ok now i have a question what moron wrote this problem? the question is posed with the measure in radians, but the answer is in degrees!!

OpenStudy (anonymous):

another copy and paste for my bad math folder hold on, we can answer in a second

OpenStudy (anonymous):

Okay

OpenStudy (anonymous):

oh wait, i am a moron, not the person who wrote the problem degrees is the output, it is temperature!!

OpenStudy (anonymous):

\[ f(t) = 65 \sin (\frac{\pi}{5}t) + 35 \] we need very little to find the maximum value

OpenStudy (anonymous):

I just saw that too

OpenStudy (anonymous):

the maximum value of sine is 1 the maximum value of 65 times sine is 65 the maximum value of 65 times sine plus 35 is 100

OpenStudy (anonymous):

i will let you find the minimum on your own, starting with the minimum value of sine is -1

OpenStudy (anonymous):

Where would I plug it in on the equation you gave me?

OpenStudy (anonymous):

as for the period, the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) put \(b=\frac{\pi}{5}\) so get your answer

OpenStudy (anonymous):

not sure what you are asking did you understand these lines: the maximum value of sine is 1 the maximum value of 65 times sine is 65 t he maximum value of 65 times sine plus 35 is 100

OpenStudy (anonymous):

I am so lost lol,sorry..math is my weak point

OpenStudy (anonymous):

ok lets go real slow

OpenStudy (anonymous):

the largest \(\sin(x)\) can ever be is \(1\) , it can never be bigger

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

if the largest \(\sin(x)\) can be is \(1\), what is the largest \(65\sin(x)\) can be?

OpenStudy (anonymous):

the largest would be 1 as well?

OpenStudy (anonymous):

let me rewrite it like this suppose \(\sin(x)=1\) then what is \[65\times \sin(x)\]?

OpenStudy (anonymous):

it would be 35?

OpenStudy (anonymous):

let me try again what is \[65\times 1\]?

OpenStudy (anonymous):

65

OpenStudy (anonymous):

k good

OpenStudy (anonymous):

then finally, what is \[65\times 1+35\]?

OpenStudy (anonymous):

100

OpenStudy (anonymous):

got it

OpenStudy (anonymous):

so 100 is the maximum

OpenStudy (anonymous):

now lets see what we know the biggest sine can be is \(1\) so the biggest \[f(t) = 65 \sin (\frac{\pi}{5}t) + 35\]can be is \[65\times1+35=100\]

OpenStudy (anonymous):

right

OpenStudy (anonymous):

now lets find the minimum, knowing that the minimum value of sine is \(-1\) what do you think the value of \[f(t) = 65 \sin (\frac{\pi}{5}t) + 35\] is if you replace sine by \(-1\) ?

OpenStudy (anonymous):

t could be -30?

OpenStudy (anonymous):

now we got it!

OpenStudy (anonymous):

So how would I apply that to the answer options I have?

OpenStudy (anonymous):

i forget what they were, but i hope one had "maximum is 100, minimum is -30"

OpenStudy (anonymous):

Yes I had two of those

OpenStudy (anonymous):

maximum: 100°; minimum: −30°; period: pi over 5 hours maximum: 100°; minimum: −30°; period: 10 hours

OpenStudy (anonymous):

ok so it is one of those now you need the period

OpenStudy (anonymous):

I think it would be the first one

OpenStudy (anonymous):

would you like to know how to find it?

OpenStudy (anonymous):

Yes please

OpenStudy (anonymous):

the period of \[\sin(bx)\] is \[\frac{2\pi}{b}\] it is NOT \(b\)

OpenStudy (anonymous):

therefore the period of \[\sin(\frac{\pi}{5}t)\] is \[2\pi\div \frac{\pi}{5}=2\pi\times \frac{5}{\pi}=10\]

OpenStudy (anonymous):

maximum: 65°; minimum: 35°; period: 10 hours

OpenStudy (anonymous):

hmm no

OpenStudy (anonymous):

we already found the max and min

OpenStudy (anonymous):

ssorry not that one

OpenStudy (anonymous):

maximum: 100°; minimum: −30°; period: 10 hours

OpenStudy (anonymous):

got it

OpenStudy (anonymous):

THANK YOU...you help me more than my online teacher

OpenStudy (anonymous):

more importantly i hope the steps were more or less clear on line teacher?

OpenStudy (anonymous):

It was more clear..Thank you again!!

OpenStudy (anonymous):

yw

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