Question

Need help with this question! Will attach pic!

A satellite is to be put into an elliptical orbit around a moon as shown below.The moon is a sphere with radius of 1000 km. Determine an equation for the ellipse if the distance of the satellite from the surface of the moon varies from 953 km to 466 km.

Answer 1

I know the denominators are 1953 and 1466, but i do not know which goes under x^2 and which goes under y^2

Answer 2

well your logic is absolutely correct , but the denominators will be 1953^2 and 1466^2
and since not much is mentioned in the question, you can assume x axes and y axes to be symmetrical
hence you can write 2 equations, both of which will be correct
one with 1953^2 under x^2 and other under y^2
and vice cersa for the second one

Answer 3

versa*

Answer 4

Or if the figure is to be taken into account , I mean if it is mentioned along with the question, you can see length along x axes will be minor axes and along y axes will be major axes.
hence your eqn will then be only
(x^2 /1466^2) + (y^2/1953^2) =1

Answer 5

oops yes I meant squared! So if you're not assuming it's symmetrical you'd go by the graph? Because when I graphed each of them the one you just mentioned seemed right! I just had to double check! Thank you so much!

Answer 6

Yes had graph not been mentioned, there'd have been 2 answers.
glad to help.

Answer 7

Can I ask you one more? It's about finding fifth roots

Answer 8

I can surely try to help

Answer 9

Find the fifth roots of 243(cos 240° + i sin 240°). I don't know where to start is the problem!

Answer 10

The euler formula, must have heard about it ?

Answer 11

cosx + isinx = e^(ix)

Answer 12

I actually don't recognize it! Could you maybe explain the e^ix?

Answer 13

hmm, let me show you the derivation

let us assume cosx + i sinx = y

=> -sinx + i cosx = dy/dx
=> i^2 sinx + i cosx = dy/dx
=> i( cosx + isinx) = dy/dx
=> iy = dy/dx
=>idx = dy/y

integrating both sides,
ix = ln(y) + C
y = e^(ix -C)

now recall that y = cosx + isinx

note that for x=0, y=1

we put that in out last eqn
1 = e^(i0 -C)
1 = e^(-C)

C=0

hence we have an identity
y= e^ix
or
cosx + isinx = e^(ix)

i'd recommend you to memorize this for future

Answer 14

are you getting it? :P

Answer 15

The d is confusing me :P are you referring to the derivative?

Answer 16

yes
perhaps you are not very comfortable with calculus yet ?

Answer 17

Well I'm in precalculus :)

Answer 18

hmm, well just memorize the result , thats what is important .

cosx + isinx = e^(ix)

Answer 19

How would I plug it in with the specific problem?

Answer 20

you can write
cos240 + isin240 = e^(i240) , right ?

Answer 21

yes :)

Answer 22

rest should be easier for you hmm ?

Answer 23

Side question: Could I use De Moivres?

Answer 24

thats exactly what we are using

Answer 25

Wow! I'm so sorry I'm supposed to be sleeping now! I'll have to review this in the morning when my brain can manage. I didn't realized how lost I was in this area! Thanks again for bearing with me! I apologize that I couldn't fully understand! I'll come back to this though after I review!

Answer 26

hmm..well take your time
this ain't that difficult, your main question was to find this :

[ 243 (cos240 + isin240) ]^(1/5)
after you express it as
[ 243 e^(i240) ] ^(1/5)

rest becomes easier .