Question

16(x^2+y^2)^2=400xy^2 ; (4,2) write an equation for the tangent line

Answer 1

I somewhat get it.. so do i distribute the 16 to x^2 and y^2 and get the dev. of that?

Answer 2

you cant distribute until you expand the binomial

there are two ways of going about it, you can expand the binomial then distribute the 16

or

you can use the chain rule

Answer 3

on the RHS you will need to use the product rule and chain rule
and remember this is implicit differentiation.

Answer 4

2(x^2 +y^2) * 2x + 2y *y'?

Answer 5

correct

Answer 6

well 2(x^2 +y^2) * (2x + 2y *y')

Answer 7

And then I move everything besides the y' to the right?

Answer 8

you need to implicite differentiate the right hand side

Answer 9

implicit*

Answer 10

y'= 2(x^2+y^2)/2x+2y like so?

Answer 11

you need to implicit differentiate the right hand side

Answer 12

\(400xy^2\implies400(y^2+x2yy')\)

Answer 13

now we have
\(2(x^2 +y^2) * (2x + 2y *y')=400(y^2+x2yy')\)

Answer 14

solve for y'

Answer 15

its not going to be fun.....

Answer 16

you with me still?

Answer 17

yup im trying to solve for y' now!

Answer 18

show me what you get, ill do it on paper:)

Answer 19

in this kind of problem you don't have to move x's and y's around. just plug in the numbers you have for x and y, then solve for y'

Answer 20

i'm not the best at this yet, so it might take some time! take you for helping me [:

Answer 21

np
i got
\(y'=\frac{400y^2-4x^3-4xy^2}{-800xy+4x^2y+4y^3}\)

Answer 22

@zzr0ck3r can I do that? instead of adding, subtracting, and dividing then plugging in the numbers can I just plug in the numbers for x and y and then work with only y' as a variable?

Answer 23

but its late.

after this you will plug in your point for (x,y) and that will give you the slope of the line
y=mx+b
then use the point again to find b
and we are done
\(\huge yay!!!!\)

Answer 24

sure, if its the one time thing

Answer 25

but sometimes it comes in handy to have the equation

Answer 26

isn't point-slope form easier than y=mx+b to get the equation?

Answer 27

thank you so much for the help! :D

Answer 28

if you know how to use it?

Answer 29

i.e if you have to evaluate more than one point, then find an explicit equation.

Answer 30

sure if its easier to you, there are both the same to me.. I always use that one because every knows it.

Answer 31

what i got for that was -.0979487 ... sorry i am so slow lol

Answer 32

y=-.0979487x+-1.91795?

Answer 33

16(x^2+y^2)^2=400xy^2
Differentiate both sides implicitly:
32(x^2 + y^2)(2x + 2yy') = 400(2xyy' + y^2)
To evaluate y' at (4,2), put x = 4 and y =2 right here and then find y'
32(16 + 4)(8 + 4y') = 400(16y' + 4)
32*20*4*(2 + y') = 400*4*(4y' + 1)
32(2+y') = 20(4y'+1)
8(2+y') = 5(4y'+1)
16+8y' = 20y' + 5
11 = 12y'
y' = 11/12 at the point (4,2)
y = mx + b
y = 11/12x + b
x = 4, y = 2
2 = 11/12 * 4 + b
b = 2 - 11/3 = (6-11)/3 = -5/3

y = 11x/12 - 5/3 is the equation of the tangent line.