Question

Find the curvature at the point (x, y) on the ellipse x^2/9+y^2/4=1.

Answer 1

@myininaya

Answer 2

@myininaya

Answer 3

so have you thought of doing using parametric equations?

Answer 4

Yes. Here's my work:
x^2/a^2+y^2/b^2=1
x(t)=acos(t)
y(t)=bsin(t)
-----------------
x(t)=3cos(t)
y(t)=2sin(t)
in this case
----------------
k=abs(x'y"-y'x")/[(x')^2+(y')^2]^(3/2)
x'(t)=-3sin(t)
y'(t)=2cos(t)
x"(t)=-3cos(t)
y"(t)=-2sin(t)
k=6/(9sin^2 t+4cos^2 t)^(3/2)
And I'm stucked.

Answer 5

\[k=\frac{6}{(9\sin^2(t)+4\cos^2(t))^\frac{3}{2}} \\ \text{ there isn't much you can do with this except } \\ k=\frac{6}{(5 \sin^2(t)+4\sin^2(t)+4 \cos^2(t))^\frac{3}{2}}=\frac{6}{(5 \sin^2(t)+4(1))^\frac{3}{2}} \\ k=\frac{6}{(5 \sin^2(t)+4)^\frac{3}{2}}\]

Answer 6

that looks a lil bit prettier

Answer 7

But that's not the answer. The answer in the book says \[\frac{ 162 }{ (81-5x^2)^{3/2} }\]

Answer 8

oh!
they want it back in terms of x and y (or I mean just x)
\[k=\frac{6}{(5(\frac{y}{2})^2+4)^\frac{3}{2}} \\ =\frac{6}{(\frac{5y^2}{4}+4)^\frac{3}{2}} \\ =\frac{6}{(5(1-\frac{x^2}{9})+4)^\frac{3}{2}}\]
so we are going to play with this

Answer 9

\[=\frac{6}{(5 -5 \frac{x^2}{9}+4)^\frac{3}{2}}=\frac{6}{(9-\frac{1}{9}5x^2 )^\frac{3}{2}}\]

Answer 10

now we have a compound fraction

Answer 11

so we need to find a way to get rid of that
we can do this...
\[=\frac{6}{(\frac{1}{9} \cdot (18-5x^2))^\frac{3}{2}}\]

Answer 12

guess what we can do by law of exponents?

Answer 13

oops 81 sorry I'm dyslexic sometimes

Answer 14

\[=\frac{6}{(\frac{1}{9} \cdot (81-5x^2))^\frac{3}{2}}\]

Answer 15

you know since 1/9*81=9
and 1/9 *5=5/9

Answer 16

but anyways by law of exponents we can say
\[=\frac{6}{(\frac{1}{9} \cdot (81-5x^2))^\frac{3}{2}}=\frac{6}{(\frac{1}{9})^\frac{3}{2}(81-5x^2)^\frac{3}{2}}\]

Answer 17

but you can simplify from here

Answer 18

I bet*

Answer 19

but if not let me know

Answer 20

But how did you get \[\frac{ 6 }{ (5(1-\frac{ x^2 }{ 9 })+4)^{3/2} }\]

Answer 21

\[\frac{x^2}{9}+\frac{y^2}{4}=1 \\ \frac{y^2}{4}=1-\frac{x^2}{9}\]

Answer 22

the ellispe equation

Answer 23

remember you called y=2 sin(t)
so sin(t)=y/2

Answer 24

Oh, and how did you get \[\frac{ 6 }{ (9-\frac{ 1 }{ 9 }5x^2)^{3/2} }\]

Answer 25

\[5 \sin^2(t)+4 \\ 5(\frac{y}{2})^2+4 \\ 5 \frac{y^2}{4}+4 \\ 5(1-\frac{x^2}{9})+4 \\ 5-5 \cdot \frac{x^2}{9}+4 \\ 9-5 \frac{x^2}{9} \\ \frac{1}{9}(9^2-5 x^2)\]

Answer 26

Oh, and how did you get \[\frac{ 6 }{ (9-\frac{ 1 }{ 9 }5x^2)^{3/2} }\]

Answer 27

Oh, and how did you get \[\frac{ 6 }{ (9-\frac{ 1 }{ 9 }5x^2)^{3/2} }\]

Answer 28

distribute and added like terms above

Answer 29

Thank you so much for the help, @myininaya !! I'm so sorry that in the morning, my computer was slow.