Question

Help Please !!1

Answer 1

\[y\sin(3x)=x\cos(3y)\] right?

Answer 2

diff wrt x get \[y'\sin(3x)+3y\cos(3x)=\cos(3y)-3xy'\sin(3y)\]

Answer 3

btw your answer to 50 is a mistake

Answer 4

oh okay ty . i fixed it lol . but ummm

Answer 5

this is what i got for 51.

Answer 6

is it clear how to do this implicit diff?

Answer 7

thats the dy/dx right ?

Answer 8

yeah i mean how to find it

Answer 9

how to get \[y'\sin(3x)+3y\cos(3x)=\cos(3y)-3xy'\sin(3y)\]

Answer 10

yes thats gonna be this

Answer 11

then replace \(x\) by \(\frac{\pi}{3}\) and \(y\) by \(\frac{\pi}{6}\) and solve for \(y'\)

Answer 12

oh so then i plug it in directly?

Answer 13

but question do i have to move some stuff ?

Answer 14

you can use algebra to solve for \(y'\) first, but it is easier just to plug in the numbers, then solve

Answer 15

oh okay so then when its implicit its better to plug it in directly ?

Answer 16

\[y'\sin(3x)+3y\cos(3x)=\cos(3y)-3xy'\sin(3y)\]
\[y'\sin(\pi)+3\frac{\pi}{6}\cos(\pi)=\cos(\frac{\pi}{2})-3\frac{\pi}{3}\sin(\frac{\pi}{2})\]

Answer 17

i dropped a \(y'\) somewheres

Answer 18

that was on the left side ?

Answer 19

\[y'\sin(\pi)+3\frac{\pi}{6}\cos(\pi)=\cos(\frac{\pi}{2})-3\frac{\pi}{3}\sin(\frac{\pi}{2})y'\]

Answer 20

but now it is easy, since \(\sin(\pi)=0\) and \(\cos(\pi)=-1\)

Answer 21

oh okay so i got -1/2 ?

Answer 22

also \(\cos(\frac{\pi}{2}=0\) and \(\sin(\frac{\pi}{2})=1\)

Answer 23

\[-\frac{\pi}{2}=\frac{\pi}{2}y'\]

Answer 24

oops another mistake damn

Answer 25

\[-\frac{\pi}{2}=\pi y'\]

Answer 26

so yeah, \(-\frac{1}{2}\)

Answer 27

oh okay so then

Answer 28

then you are done, unless you need the equation for the line

Answer 29

yes the equation

Answer 30

you got the point \((\frac{\pi}{3},\frac{\pi}{6})\) and you got the slope \(-\frac{1}{2}\) use the point slope formula