Question

help me understand these 2 questions are driving me nuts!! >w<

3. Sandra is riding the ferris wheel, and her height can be modeled by the equation h(t)=25cos(pi/14t)+31, where H represents the height of the person above the ground in feet at T seconds.
Assuming Sandra begins the ride at the top, how far from the ground is the edge of the Ferris wheel, when Sandra's height above ground reaches a minimum?

Answer 1

4. Prove : sinθ-sinθ*cos^2θ=sin^3θ

Answer 2

Interesting...

so

\[\large h(t) = 25\cos(\frac{\pi}{14}t) + 31\]

Since we want the height at a MINIMUM, to me that screams take the derivative and set it equal to 0

So what is the derivative of h(t) ?

Answer 3

*

Answer 4

@perl @johnweldon1993
This is what I got so for number 3
h(t)=25cos(pi/14t)+31
h(t)=25cos+31
h(t)=25*-1+31 -1 = smallest cosine
h(t)=-25+31
h(t)=6ft

Answer 5

Exactly what I have as well...just took a longer path lol

Answer 6

really? thank you !!!~ I have a question though, i dont know why everything that is in the parenthesis disapears

Answer 7

well see thats were my step would have come into play

when you take the derivative and set it equal to 0...you get t = 14

and then when you plug in 14 into h(t)

you get

\[\large h(t) = 25cos(\frac{14\pi}{14}) + 31\]

Which is

\[\large h(t) = 25cos(\pi) + 31\]

And we know cos(pi) = -1 so that gives us our

\[\large h(t) = -25 + 31 = 6\]

Answer 8

Make sense now? :)

Answer 9

totally makes sense now!!~ thank you so so so much!!~ would you mind helping me out on number 4 as well?

Answer 10

Yeah no problem :D

This ones so easy all I'm gonna say is remember

\[\large cos^2(\theta) = (1 - sin^2(\theta))\]

And you should be all set :D but let me know if you're not!

Answer 11

this is what I got for number 4

sinθ-sinθ*cos^2θ=sin^3θ
0*cos^2θ=sin^3θ
0=sin^3θ

Answer 12

was this even right at all? lol

Answer 13

Nope lol...just look at it again quick

its sin - (sin*cos^2)

cant just subtract the sine's

Answer 14

ok gimme a second to make an attempt to solve this lol

Answer 15

ok idk how to multiply sinθ * cos^2θ in the parenthesis

Answer 16

is that equal to something?

Answer 17

because i know that there are 2 different identities

Answer 18

but i dont know if they apply here or not :/

Answer 19

Ignore the parenthesis...I put them there purely to say dont just subtract...

look a couple posts up about what I said about cos^2

Answer 20

ohhh so i plug it in like this?

sinθ-sinθ * (1-sin^2(θ))=sin^3θ

Answer 21

Perfect :)

now just distribute the sin into the parenthesis and simplify

Answer 22

do i convert them into 2 binomials to multiply them for distribution?

Answer 23

Just simple distribution

\[\large sin - sin(1 - sin^2) = sin^3\]

since -sin times 1 = -sin
and
-sin times -sin^2 = sin^3

this would become

\[\large sin - sin + sin^3 = sin^3\]

Answer 24

so would it take me here?

sinθ-sin+sin^3=sin^3θ
0+sin^3=sin^3θ
sin^3θ=sin^3θ

Answer 25

i think im not getting it. im still confused

Answer 26

@johnweldon1993

Answer 27

Okay so let me run through it (whole)

we have

\[\large sin(\theta) - sin(\theta)cos^2(\theta) = sin^3(\theta)\]

The important identity to remember is

\[\large sin^2 + cos^2 = 1\]
we can rewrite that as
\[\large cos^2 = 1 - sin^2\]

So...looking back to our equation...lets replace \(\large cos^2\) with \(\large 1 - sin^2\)

\[\large sin(\theta) - sin(\theta)cos^2(\theta) = sin^3(\theta)\]
becomes
\[\large sin(\theta) - sin(\theta)(1 - sin^2(\theta)) = sin^3(\theta)\]

Now we just need to multiply \(\large sin(\theta)\) into the parenthesis, so multiply it by each term inside the parenthesis

\[\large sin(\theta) - sin(\theta) + sin^3(\theta) = sin^3(\theta)\]

And obviously we see that sin - sin = 0 so we are left with the confirmation

\[\large sin^3(\theta) = sin^3(\theta)\]

Answer 28

IM CONFUSED ABOUT WHAT YOU DID HERE IN THIS STEP:

becomes
sin(θ)−sin(θ)(1−sin2(θ))=sin3(θ)

Now we just need to multiply
sin(θ) into the parenthesis, so multiply it by each term inside the parenthesis

sin(θ)−sin(θ)+sin3(θ)=sin3(θ)

Answer 29

when you multiply them int he parenthesis, well this is what I got

when I distribute them:

sinθ-sin^3θ-sinθ+sin^3=sin^3θ

after i simplify this, it also gives me 0=sin^3θ

Answer 30

wait i just saw the - sign between them

Answer 31

Why do you get that?

Oh I see what you did...no you're thinking again that there are parenthesis

\[\large sin(\theta) - sin(\theta)(1 - sin^2(\theta))\]

This is NOT the same as

\[\large (sin(\theta) - sin(\theta))(1 - sin^2(\theta))\]

Answer 32

sinθ-sinθ(1-sin^2θ)=sin^3θ
sinθ-sinθ+sin^3θ=sin^3θ
0+sin^3θ=sin^3θ
sin^3θ=sin^3θ

Answer 33

this is it right? hahaha

Answer 34

@johnweldon1993

Answer 35

That is perfect lol

Answer 36

thank you :)

Answer 37

Great job! :)