Question

Help please! (:

Consider the equation:
2(x^2+y^2)^2 = 25(x^2-y^2)

1.) verify that the point P(3,-1) lies on the curve given by the equation

2.) Use implicit Differentiation to solve for dy/dx

3.) Find the equations of the tangent line and normal line at point P.

Answer 1

let's do 1 first.
At the left-hand side of your equation, you have
\[\Large 2(x^2 + y^2)^2\]
Now, substitute the point P(3,-1) by replacing x with 3 and y with -1

and tell me what you get...

Answer 2

200?

Answer 3

@terenzreignz

Answer 4

Okay, now try it for the right-hand side
\[\Large 25(x^2 - y^2)\]

Answer 5

200? @terenzreignz

Answer 6

I hope you weren't guessing :)
But yes, it is 200 on both sides when x = 3 and y = -1

Thus verifying that the point P(3,-1) is indeed on the curve :3

Answer 7

haha no. I solved.
so was that the answer for 1.?
@terenzreignz

Answer 8

The 'answer' to number 1 is to just replace x and y with 3 and -1 respectively and show that the equation holds.

Answer 9

so it solving the two sides and having them equal each other? @terenzreignz

Answer 10

yup :)
Which they did, they both are 200...

Answer 11

okay(: so #2? @terenzreignz

Answer 12

There is no need to keep tagging me -.-

Just call me Terence, okay? :P

Now, you have to differentiate (both sides) of the equation...
Can you do that?

Answer 13

Sorry.... haha

the right side would be 4(x^2+y^2)*(2x+2y) right?

and the left side (100x-100y)?

Answer 14

I think you've got your right and left sides switched... :P

Answer 15

haha oh yeah. whoops.but did I derive them correctly?

Answer 16

Nope :)
Left one, almost... you just forgot that every time you differentiate the y bit, you affix a \(\large y'\) or a \[\Large \frac{dy}{dx}\] to it as per the chain rule...

Answer 17

but my right one was correct? how do i derive the left?

Answer 18

Left one is almost as you did, except...
\[\Large 4(x^2+y^2)\left(2x + 2y\color{blue}{\frac{dy}{dx}}\right)\]

Answer 19

i just needed to add dy/dx?

so where so I go from there?

Answer 20

Well, differentiate the right-side again, only this time, when you differentiate the y, include a dy/dx

Answer 21

so it would be (100x-100y(dy/dx))?

Answer 22

why 100?
Last I checked, 2x25 is 50 :P

Answer 23

omg. yes 25... so sorry my brain is not functioning well. Been doing math for 7 hrs now lol

soo. (50x-50y(dy/dx))?

Answer 24

Yup :)
Let's use y' instead... it looks prettier that way, see :P
\[\large 4(x^2+y^2)\left(2x + 2y \color{red}{y'}\right)= 50x +50y\color{red}{y'}\]

Answer 25

oh okay... so is that the answer? or? lol

Answer 26

No. Most unfortunately, the worst is yet to come...
You have to solve for y'.

What I mean is you have to get it alone on one side of the equation...
good luck :P

Answer 27

oh man... so both y' to one side?

Answer 28

Yup... and solve...

Answer 29

I'm stuck.. I dont know where to begin..

Answer 30

Begin by bringing all the terms with y' to the right (or left) side and those without, to the other side.

Answer 31

\[(4x^2+4y^2)(2x)-50x = 50y(dy/dx)-2y(dy/dx)\]

Answer 32

You forgot something....something you didn't distribute...

Answer 33

the 2x(4x^2+4y^2)?

Answer 34

6x^2+2x(4y)^2-50x = 50y(dy/dx)-27(dy/dx)

Answer 35

Hang on...
and for goodness' sake, use y' for now, not that menacing looking dy/dx :D

Answer 36

hah okay.
\[6x^3+(2x)4y^2-50x = 50yy'-2yy'\]

Answer 37

where do I go from here? hah

Answer 38

@terenzreignz

Answer 39

It was wrong... for one thing, 2x4 is not 6, it's 8.
Also, you forgot that this part also has a (x^2 + y^2)
\[\large 8x^3+(2x)4y^2-50x = 50y\color{red}{y'}-2y(x^2+y^2)\color{red}{y'}\]

Answer 40

But other than that, everything's fine...

Answer 41

Wow, I'm really bad at this.
okay. soo \[8x^3+(2x)4y^2-50x = 50yy'- (2y)x^2+2y^3(y')\]

Answer 42

Hold it there... let's start from here first...
\[\large 8x^3+(2x)4y^2-50x = 50y\color{red}{y'}-2y(x^2+y^2)\color{red}{y'}\]

Answer 43

And factor out the y' in the right-side
\[\large 8x^3+(2x)4y^2-50x = \color{red}{y'}\left[50y-2y(x^2+y^2)\right]\]

Answer 44

then divide right? to get y' on its own?

Answer 45

Yes.... So you get this really ugly expression..
\[\Large \color{red}{y' }= \frac{8x^3 + 8xy^2-50x}{50y-2y(x^2+y^2)}\]
Oh well... I suppose it could be worse :)

Answer 46

And there's your answer :P

Answer 47

wow. that was difficult lol I really appreciate the step by step process(:

Answer 48

no problem

Answer 49

So how do I go about finding the tangent and normal line at point P (3,-1)

Answer 50

You just need slopes.
to get the slope of the tangent line at the point P(3,-1), you... evaluate y' when x = 3 and

y = -1

get to it, champ :)

Answer 51

so just plug in x and y into y'?

Answer 52

Yup. Have fun.

Answer 53

-3 ?

Answer 54

so tangent line would be.
\[y=-3x-10\]

Answer 55

ehh?
y + 1 = -3(x - 3)
again please...

Answer 56

y+1=-3x-9

Answer 57

then you minus 1 from y. which makes the -9 a -10?

Answer 58

+9 not -9
sheesh what's negative times negative? :D

Answer 59

i meant y+1=-3x+9

so y=-3x+8 ?

Answer 60

Yes, much better.
Now for the normal line, the slope is just the negative of the reciprocal of -3

Answer 61

wait huh? negative of the reciprocal of -3? so 1/3?

Answer 62

yup :)
That slope, but passing through the point (3,-1)
that's the normal line :P

Answer 63

so y=1/3x-2 ?

Answer 64

good.
Well done :P

Answer 65

Yay! (: thank you so much Terence! I really appreciate all the help!

Answer 66

This is assuming you evaluated y' correctly...
but I'm sure you did :P

Answer 67

I hope so lol