consider the equation x^2-2xy+4y^2=64...the questions are below?

Question

a) Write an expression for the slope of the curve at any point (x,y)
b) find the equation of the tangent lines to the curve at the point x=2
c) find the second derivative at (0,4)

thanks!!!

Answer 1

1) You'd need to differentiate it explicitly:

d/dx (x² - 2xy + 4y²) = d/dx(64)
2x - 2(xy' + y) + 8yy' = 0
2x - 2xy' - 2y + 8yy' = 0
x - xy' - y + 4yy' = 0
y'(4y - x) = y - x
y' = (y - x)/(4y - x)

2) At x=2:

(2)² - 2(2)y + 4y² = 64
4 - 4y + 4y² = 64
y² - y - 15 = 0
y = (1 ± √61)/2

Messy...then you'd have to plug each of those y values into the derivative function above to get the slopes, then use the point-slope equation to get lines... Really messy... unless I miessed up somewhere...

3) Differentiate the first derivative:

d²y/dx² = d²/dx² (y - x)/(4y - x)
= [(4y - x)(y' - 1) - (y - x)(4y' - 1)]/(4y - x)²

And y'(0) = (4 - 0)/(16 - 0) = 1/4

Plugging in (0,4):

=[(16 - 0)(1/4 - 1) - (4 - 0)(1 - 1)]/(16 - 0)²
= -12/256
= -3/64

Since a_math_guy and I now agree, I think this is right. Really messy though....

Answer 2

a) (x-y)/(x-4y)
b) (-3/244*61^(1/2)+ 1/4)*x+32/61* 61^(1/2) and (3/244*61^(1/2)+ 1/4)*x-32/61* 61^(1/2)
c) -3/64

Jim forgot to d/dx x in part of the quotient rule and shasjing didnt use the product rule at a key point for his 2nd derivative. This could be an important lesson about using technology. Could be..... I think Michael is on track tho' hahahaha


WORK:

> fprime:=implicitdiff(x^2-2*x*y +4*y^2=64,y,x);
x - y
fprime := -------
x - 4 y
> solve(subs(x=2,x^2-2*x*y +4*y^2=64),y);
1/2 1/2
1/2 + 1/2 61 , 1/2 - 1/2 61
> collect(solve(y-(1/2+ 1/2*61^(1/2))=subs(x=2,y=1/2+ 1/2*61^(1/2),fprime) *(x-2),y),x);
> collect(solve(y-(1/2- 1/2*61^(1/2))=subs(x=2, y=1/2-1/2*61^(1/2),fprime)* (x-2),y),x);
1/2 32 1/2
(- 3/244 61 + 1/4) x + -- 61
61
1/2 32 1/2
(3/244 61 + 1/4) x - -- 61
61
> implicitdiff(x^2-2*x*y+ 4*y^2=64,y,x$2);
2 2
x - 2 x y + 4 y
3 ------------------------------
3 2 2 3
x - 12 x y + 48 x y - 64 y
> subs(x=0,y=4,3* (x^2-2*x*y+4*y^2)/ (x^3-12*x^2*y+48*x*y^2- 64*y^3));
-3
--
64

Answer 3

a) Take derivative,
2x - 2y - 2xy' + 8yy'= 0......(1)
y' = (x-y)/(x-4y)......(2)

b) Plugging x=2 into the original equation gives, y= (1±√61)/2
Therefore, you have two tangent lines. Can you finish this part now?

c) y'(0) = 1/4
Take derivative of (1),
2-2y'-2y'-2xy''+8y'^2+8yy'' = 0
Plugging in x, y and y' at (0,4) gives, y'' = -3/64

Answer 4

I considered it. And I don't like it. Besides, Algebra was a long time ago.