Question

determine all point on the circle x^2+y^2=25 where the slope is 3/4

Answer 1

First step: Implicitely differentiate the equation!

Answer 2

I know you can do this. I trust you.

Answer 3

hehe thanks!!!!
2x+2y=0

Answer 4

ooh nvm

Answer 5

\[2x+2y(dy/dx)=0\]

Answer 6

dy/dx=-x/y

Answer 7

i went on and found dy/dx

Answer 8

is that ok

Answer 9

Yes, that is fine.

Answer 10

so then do i do \[3/4=-x/y\]

Answer 11

then idk what to dooooo

Answer 12

You then need to solve for y in the original equation.

Answer 13

And plug that in for y for the derivative.

Answer 14

i got \[y=\sqrt(25-x^2)\]

Answer 15

It would be + or -

Answer 16

Now, plug that in to your derivative equation. And finally, solve!

Answer 17

\[dy/dx= -x/(\sqrt(25-x^2))\]

Answer 18

right

Answer 19

?

Answer 20

Yeah, and remember that dy/dx is 3/4

Answer 21

i got x=3 x=-3

Answer 22

That looks about right! Congratulations!!!

Answer 23

thats it?

Answer 24

\[x=-\frac{ 3 }{ 4 }y\]
\[x^2+y^2=25\]
\[\frac{ 9 }{ 16 }y^2+y^2=25\]
\[9y^2+16y^2=25\]
\[25 y^2=25\]
\[y^2=1,y=\pm 1\]
find x for each.

Answer 25

correction

Answer 26

@sshayer Are you sure you are trying to find the point when the tangent line is -3/4? I don't see where you are getting at.

Answer 27

\[9 y^2+16y^2=25*16=400,25 y^2=400,y^2=\frac{ 400 }{ 25 }=16,y=\pm 4\]
find x for each.

Answer 28

remember x and y are of different sign