I need someone to explain to me step by step the following problem:

- anonymous

I need someone to explain to me step by step the following problem:

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Find the equation of the tangent line to the curve
\[x^{2}y^{2}-1=3(x+y)\] at the points (1,-1)

- anonymous

I have the problem worked out in front of me (prof gave us the answers) but, his work is disorganized and I don't understand what it is he's doing.

- anonymous

I figure I have to do this implicitly.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- asnaseer

yes - you are correct - use implicit differentiation to get an expression for slope of the tangent line at any point (x,y).
then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1)
then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)

- anonymous

is the answer y=4-5x?

- anonymous

that's not the answer.... the answer is: y=-1/5x+4/5

- anonymous

its a plane....?

- anonymous

okay and the way my prof did his differentiation was really weird.
his first step looks like: x^{2}2y*y'+y^{2}(2x)=3(1+y')

- anonymous

a plane? huh?

- anonymous

that's not how i would start the implicit differentiation.

- anonymous

\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]

- asnaseer

what your prof did is correct - he is using the product rule, see here: http://www.1728.org/chainrul.htm

- asnaseer

how would you have done this?

- anonymous

haha no idea... i suck at math : (

- anonymous

so i use the product rule for x^2*y^2 ??

- asnaseer

correct

- anonymous

and for the 3*(x+y) ??

- asnaseer

that is just straight differentiation since it can be written as: 3x + 3y

- anonymous

so that will just equal 3y' ??

- anonymous

or 3+3y'

- asnaseer

so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces

- anonymous

so it's 3(1+y')

- asnaseer

yes - correct

- anonymous

okay so now i have the same first line as he does haha.

- anonymous

i ignored the 3 outside the brackets because it's a constant right?

- asnaseer

yes.
thats good - you are making progress - maybe you are a lot better at maths than you think. :)

- anonymous

haha maybe, maybe not...

- anonymous

so now i can take the y^2(2x) over to the other side?

- anonymous

and the right hand side over to the left?

- asnaseer

yes - what you need to do is to get an expression for y'.
so move all terms involving y' to the left and the rest to the right

- asnaseer

first expand the expression on the right

- anonymous

do i leave my 3 on the right hand side?

- asnaseer

snap! yes :)

- anonymous

okay so now i have: (x^2)(2y*y')-(1-y')=3(-2xy^2)

- anonymous

that look right to you?

- asnaseer

not quite :(

- anonymous

oh snappp

- asnaseer

you started with this:\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]so first expand the right to get:\[x^{2}(2y*y')+y^{2}(2x)=3+3y'\]then move terms involving y' to the left and the rest to the right to get:\[x^{2}(2y*y')-3y'=3-y^{2}(2x)\]

- asnaseer

then factorise y' on the left

- anonymous

\[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]

- asnaseer

perfect!

- anonymous

woooo. so now that part is done, right?

- asnaseer

yup - the rest should be straight forward. I need to go for dinner now. I'll check back afterwards to see if you had any further problems.

- asnaseer

the steps left are:
then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1)
then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)

- anonymous

okay thanks. : )

- asnaseer

yw :) and good luck!

- anonymous

i got the rest of it now ; ) thanks a bunch!

- asnaseer

thats great!
and you are more than welcome! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.