Differentiation.

Christina. said:
The equation of the tangent to the curve (x − 2)2 + 2(y + 1)2 = 4 at (a, b) has y-intercept at 3[√(2/3)− 1] and is parallel to y = x. Find the point (a, b).

I'm not sure how to answer this question.
Please help.
Click to expand...

What have you tried so far? Have you solved for y' at least?
 
Christina. said:
Don't even know where to start? Sorry. :(

y' ?
Click to expand...

y' is just another notation for dy/dx, the derivative of y with respect to x.

Though you can solve explicitly for y and then find the derivative, it would be easier if you used implicit differentiation. Have you learned implicit differentiation?
 
Christina. said:
Yeah I know that :)

And yeah, I've learnt about implicit differentiation. Are we just interested in the curve equation? How does the y-intercept and information about being parallel to y=x come into it?

If we were to use implicit differentiation, how do we start?

(x-2)2 + 2(y+1) = 4
x2-4x+4 + 2(y2+y+1) = 4 ====> Should be 2y
x2-4x+4 + y2+2y+2 = 4

Thus
d(x2-4x+4)/dx = 2x-4
d (y2+2y+2)/dx = 2y * dy/dx + 2 * dy/dx

Thus
2x-4 + 2y * dy/dx + 2 *dy/dx = d(4)/dx

d(4)/dx = 0 (as 4 is a constant)
which means
2x-4 + 2y * dy/dx + 2 *dy/dx = 0

2x-4 + dy/dx (2y+2) = 0

dy/dx(2y+2) = -2x+4

dy/dx = -2x+4 / 2y+2 ...?????
Click to expand...

Try not to expand if you don't have to as it makes the math messy:

\(\displaystyle (x-2)^2+2(y+1)^2=4\)

Implicitly:

\(\displaystyle 2(x-2)+4(y+1)(\frac{dy}{dx})=0\)

\(\displaystyle \frac{dy}{dx}=\frac{-2(x-2)}{4(y+1)}\)

\(\displaystyle \frac{dy}{dx}=\frac{-(x-2)}{2(y+1)}\)

\(\displaystyle \frac{dy}{dx}=\frac{2-x}{2y+2}\)

It says that the line is parallel to y=x. Well, then the slope of the line must equal the slope of y=x, which is 1. Thus, plugging in (a,b) into the derivative you get:

\(\displaystyle \frac{2-a}{2b+2}=1\)

\(\displaystyle 2-a=2b+2\)

\(\displaystyle a+2b=0\) ====> This is your first equation with "a" and "b". Now you need one more equation with "a" and "b" in order to solve for each of them. What else do we know? We know the y-intercept. So we can use y = mx + b...we know b and we plug in "a" for "x" and "b" for "y" to get our second equation.

Try this and see if you can solve for "a" and "b".
 
Christina. said:
I'm not exactly sure

y=mx+b

b is the y-intercept ....3[sqrt(2/3) -1] ....(the square root doesn't include -1)

If yes, y=mx+ 3[sqrt(2/3) -1]

Sub in a for x

y=ma+ 3[sqrt(2/3) -1]

b=ma+ 3[sqrt(2/3) -1]

??

Not confident how to solve for a and b either....
Click to expand...

\(\displaystyle y = mx + b\)

\(\displaystyle m=1, b=3(\sqrt{\frac{2}{3}}-1)\)

\(\displaystyle b=(1)a+3(\sqrt{\frac{2}{3}}-1)\)

\(\displaystyle b-a=3(\sqrt{\frac{2}{3}}-1)\) [1]

From the previous post we have \(\displaystyle a+2b=0\)

Add this equation to equation [1] above and the a's subtract out leaving only b which we can solve for:

\(\displaystyle 3b=3(\sqrt{\frac{2}{3}}-1)\)

\(\displaystyle b=\sqrt{\frac{2}{3}}-1\)

Substitute this b back into [1] to solve for a. Cool beans?
 
Another side note- do NOT keep erasing your old posts. Other people, who could learn from this thread, will not be able to understand what is being said.
 
Last edited:
Christina,

Why are you trying to delete your post?

Are you trying hide from your instructor the fact that you sought and received help - or are you trying to hide these helpful hints from your class-mates?
 
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