Question

Sandra is riding the Ferris wheel, and her height can be modeled by the equation H(t) = 25 cospi over 14t + 31, where H represents the height of the person above the ground in feet at t seconds.

Part 1: How far above the ground is Sandra before the ride begins?
Part 2: How long does the Ferris wheel take to make one complete revolution?
Part 3: 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 the ground reaches a minimum?

please help asap

Answer 1

part one
put \(t=0\) since that is when you start

Answer 2

i can't really read it
is it\[\large 25\cos(\frac{\pi}{14}t)+31\]?

Answer 3

yea

Answer 4

weird because that means she gets on a the highest point

Answer 5

i dont get this at all

Answer 6

put \(t=0\) and \(\cos(0)=1\) so you get
\[25\times 1+31=56\]

Answer 7

" H represents the height of the person above the ground in feet at t seconds."
that meas for example at time \(t=0\) which is when you start counting, replace \(t\) by \(0\) like i just did above

Answer 8

then after say \(7\) seconds the height would be
\[25\cos(\frac{7}{14}\pi)+31=25\cos(\frac{\pi}{2})+31=0+31=31\] since \(\cos(\frac{\pi}{2})=0\)

Answer 9

Is this still for the first part?

Answer 10

@satellite73

Answer 11

part one just asks for \(t=0\)

Answer 12

which is \(56\) since \(25+31=56\)

Answer 13

okay but for part 2 how do you find the complete revolution

Answer 14

that is the same as the period of the function

Answer 15

the period is always
\[\frac{2\pi}{b}\] and in your case \(b=\frac{\pi}{14}\)
how is your arithmetic?

Answer 16

so 28

Answer 17

that was a serious question, not a snark
many cannot do this
yes 28

Answer 18

and for part 3 if she starts at the top and ends at the bottom that would be 180°????????

Answer 19

well not since you are working in radians, no

Answer 20

what you really mean is when cosine is minus one (its minimum) so don't think too hard, just replace cosine by minus one

Answer 21

i understand you meant put \(t=180\) which won't work here since the measure is in radians
the idea is right, but really all you are asked is the minimum value of
\[25\cos(whatver)+31\] which is
\[-25+31\] since \(-1\) is the minimum value of cosine

Answer 22

H(t)=25(-1)(╥/14)t+31

Answer 23

and just solve

Answer 24

no no just replace cosine by minus one
forget the input

Answer 25

you are thinking way too hard
the maximum value of cosine is 1 so the max of
\[25\cos(x)+31\] is
\[25+31=56\] like in question one

the minumum of cosine is \(-1\) so the minimum of
\[25\cos(x)+31\] is \[-25+31=6\]

Answer 26

H(t)=6(╥/14)t ?

Answer 27

what about ╥/14

Answer 28

you are not asked "When" is it the minimum only "what is the minimum"

Answer 29

╥=pi

Answer 30

yeah i got that
if you were asked "when is it the minimum" then you would have to worry about that

Answer 31

oh okay can you help me with one more?

Answer 32

but you are only asked "what is the minimum" and that you can do in your head
it is \(-25+31=6\)

Answer 33

sure if i can

Answer 34

thanks Part 1: Using complete sentences, compare the key features and graphs of sine and cosine. What are their similarities and differences?

Part 2: Using these similarities and differences, how would you transform f(x) = 3 sin(4x - π) + 4 into a cosine function in the form f(x) = a cos(bx - c) + d?

Answer 35

oh damn the dreaded "complete sentences"

Answer 36

haha yea

Answer 37

if it was me, i would say the graphs are identical
they both have max an min 1 and -1
they are both periodic with period \(2\pi\)
the only difference is that say compared to the graph of sine, the graph of cosine is shifted \(\frac{\pi}{2}\) units to the left

Answer 38

okay i get that but what about the second part?

Answer 39

ok since
\[\sin(x+\frac{\pi}{2})=\cos(x)\] perhaps we could translate that way

Answer 40

the 3 out front will have to be the same, also the \(+4\) and the end will have to be the same so it will look like
\[3\cos(something)+4\] all we need is the "something"

Answer 41

so how do we find it

Answer 42

well i am sure there is a slicker way to do it, but i will do it the donkey way
lets see what makes \(4x-\pi=0\)

Answer 43

we get \(x=\frac{\pi}{4}\) pretty much right away

Answer 44

so when \(x=\frac{\pi}{4}\) we get 0 and the sine of 0 is 0
now the cosine of \(-\frac{\pi}{2}=0\) so we want
\[bx-c=-\frac{\pi}{4}\] when '
\[x=\frac{\pi}{4}\]

Answer 45

\[b\frac{\pi}{4}-c=-\frac{\pi}{4}\] we have lots of choices, since this is one equation and two unknowns so lets make \(b=4\) and get
\[\pi-c=-\frac{\pi}{4}\] making \(c=\frac{5\pi}{4}\) if my Lagrange is right

Answer 46

i am sure there is a slicker way to do it, but i think we end up with
\[3\cos(4x-\frac{5\pi}{4})+4\]

Answer 47

Thank you!!!!!!!!!!!!!!!

Answer 48

hold the phone i am wrong!!

Answer 49

i shifted incorrectly some how
give me a minute to figure out my mistake

Answer 50

okay thank you

Answer 51

somewere i made an algebra mistake
it should be
\[3\cos(4x+\frac{\pi}{2})+4\] or

Answer 52

jeez i screwed that up
\[bx-c=-\frac{\pi}{4}\] was a mistake, it should have been
\[bx-c=-\frac{\pi}{2}\]

Answer 53

should i replace all the pi/4 to pi/2

Answer 54

when \(x=\frac{\pi}{4}\) and we solve
\[\pi-c=-\frac{\pi}{2}\] and get \(c=\frac{3\pi}{2}\) making it
\[3\cos(4x-\frac{3\pi}{2}+4\]

Answer 55

So we find what makes 4x-╥=0, we get x=╥/4, so when x=╥/4 we get 0 and the sine of 0 is 0. Now the cosine of -╥/2=0 so we want bx-c=-╥/2 when x=╥/4. b*╥/4-c=-╥/4, lets make b = 4, ╥-c=-╥/4, making c=5╥/4, so we end up with 3cos(4x+3╥/2)+4.

Answer 56

the part where i solved
\[4x-\pi=0\] and got
\[x=\frac{\pi}{4}\] is still correct

Answer 57

is that okay?

Answer 58

╥=pi

Answer 59

So we find what makes 4x-╥=0, we get x=╥/4, so when x=╥/4 we get 0 and the sine of 0 is 0. Now the cosine of -╥/2=0 so we want bx-c=-╥/2 when x=╥/4. b*╥/4-c=-╥/2, lets make b = 4, ╥-c=-╥/2, making c=3╥/2, so we end up with 3cos(4x-3╥/2)+4.

Answer 60

there is the corrected version

Answer 61

i changed my mistaken \(\frac{\pi}{4} s\) to the correct \(\frac{\pi}{2}\)

Answer 62

i am sure there is a slicker way to do it ,but i can't think of it at the moment

Answer 63

its okay thank u very much

Answer 64

yw