Question

Find the slope of the tangent line to the curve 2(x^2+y^2)^2=25(x^2-y^2) at the point (-3,1).

Answer 1

the slope of the tangent line is given by the dy/dx. So this question is on implicit differentiation

Answer 2

yea i'm only having problems differentiatiog the right side of the equation

Answer 3

differentiating*

Answer 4

2(x^2+y^2)^2=25(x^2-y^2)

Differentiate both sides with respect to x:

4(x^2 + y^2)[2x + 2y(dy/dx)] = 25[2x - 2y(dy/dx)]

Answer 5

Do you know how to continue from there?

Answer 6

uhmm not quite

Answer 7

but for the right side isn't it 25 (2x-y^2 y' +x^2-2y)

Answer 8

The slope of the tangent line is given by the dy/dx, and now the question is to determine the slope at (-3,1). So substitute x= -3 and y = 1 into the equation and then rearrange it to get the dy/dx. Would you like to try it first?

Answer 9

Why is it 25 (2x-y^2 y' +x^2-2y)?

Answer 10

multiplying 25 by the derivative of x^2-y^2+x^2- the derivative of y^2

Answer 11

the right side is only 25(x^2 - y^2). Differentiate it you have

(d/dx)[25(x^2 - y^2)]
= 25 (d/dx)[x^2 - y^2]
= 25 [(d/dx)(x^2) - (d/dx)(y^2)]
= 25 [2x - 2y (dy/dx)]

Answer 12

Sorry, i still don't understand how did you get yours...

Answer 13

you're right

Answer 14

So are you able to continue it now? :)

Answer 15

still stuck :(

Answer 16

ok, we have x=-3 and y=1, so substituting in we get

4(x^2 + y^2)[2x + 2y(dy/dx)] = 25[2x - 2y(dy/dx)]
4(9 + 1)[(-6) + 2(dy/dx)] = 25[(-6) - 2(dy/dx)]
-240 + 80(dy/dx) = -150 - 50(dy/dx)
130 (dy/dx) = 90
dy/dx = 9/13

Answer 17

oh i thought i had to simplify until i find y' then i substitute in... oh well it makes sense though, thanks :)

Answer 18

you can also rearrange to find your y' then only you substitute. But I think in this case, it is easier to substitute to get your y'

Answer 19

you're welcome