Question

save me?
Final exam coming up need help on number 24 in this packet

Answer 1

number 24 need help please

Answer 2

and apparently there is no use of a calculator

Answer 3

Are you gonna fail ?

Answer 4

lol no why you say that?

Answer 5

Which letter are you stuck on?

Answer 6

just in general... how should i go about it? i dont want to just go plugging in numbers from 0-24 and waste all that time if there is a simpler way of doing it

Answer 7

Do you know how to find the max/min of a function?

Answer 8

care to remind me? lol

Answer 9

Take the derivative of the function and set it equal to zero

Answer 10

could you explain that by showing me how you'd find the answer to a?

Answer 11

Yeah first take the derivative of T(t) and let me know what you get

Answer 12

we havent dont derivatives yet... this is precalc lol

Answer 13

the sin function goes between -1 and 1
if you have
\[ A\sin(\omega t + \phi) \]
\(A\) is the amplitude, \(\omega\) is the angular frequency = \( 2 \pi f\) and ϕ is the phase, or in terms of period T
\[ A\sin(\frac{2\pi}{T} t + \phi) \]

Answer 14

The max value of your expression must occur when sin is at a max value of 1
Similarly, the min value occurs at sin() = -1

Answer 15

so how would i set up this equation to make it so that it gives me the max and min? sorry or being blunt but my final is in an hour lol

Answer 16

for part (d) plug in t= 12 and for (e) plug in t=18
for the period, match up
\[ \frac{\pi}{12}= \frac{2 \pi}{T} \] and solve for T
you get T= 24 hours

Answer 17

so how would i set up this equation to make it so that it gives me the max and min?

The equation is

13 sin(stuff) + 62
the only thing that changes is the value of sin(stuff). sin has a max of 1, so the max of the equation is
13*1+62= 75

the min of sin() is -1, so the min of the whole expression is
13*-1+62= 62-13= 49

Answer 18

lol you're a beast, got it =) and how did you get d and e? sorry

Answer 19

and c, when it tells us to plug in, how do we calculate that?

Answer 20

for (d) replace t with 12
\[ 13\sin \left(\frac{\pi}{12} \cdot 12 - \frac{2\pi}{3} \right) + 62 \]

They expect you to memorize sin, cos and tan of
0, 30, 45, 60 and 90 or in radians 0, pi/6, pi/4, pi/3, pi/2
and the sign according to which quadrant (for angles greater than 90)

Answer 21

if you have a calculator, you can use it to refresh your memory...

Answer 22

**(c) not (d). for (d) use t= 18 (6 hours past 12 noon)

Answer 23

so is it 13 sin (90-270)+62?

Answer 24

if want to use degrees
multiply by 180/pi
\[ 13\sin \left(\frac{\pi}{12} \cdot 12 - \frac{2\pi}{3} \right) + 62 \]
pi/12*12 becomes pi or just 180
2pi/3 becomes 2 pi /3 *180/pi = 120
180-120= 60 degrees
\[ 13\sin \left(60º \right) + 62 \]

Answer 25

and sin(60) is sqrt(3)/2

Answer 26

got it thanks a lot man you were great

Answer 27

good luck