calculus help?

Question

Consider the curve given by the equation y^3-3xy=2
a)find dy/dx
b)write an equation for the line tangent to the curve at the point (1,2)
c)find d^2y/dx^2 at the point (1,2)

Answer 1

y^3-3xy=2
3y^2*dy/dx - (3y + 3x*dy/dx) =0
3y^2*dy/dx -3x*dy/dx= 3y
dy/dx (3y^2 - 3x) = 3y
dy/dx= 3y/(3y^2 - 3x)

At (1,2),
dy/dx= 3(2)/(3(2)^2 - 3(1))
dy/dx= 6/9 = 2/3
So y=mx+b
2=(2/3)(1) +b
b= 4/3
[Answer for part b: y=(2/3)x + (4/3)]

part c: Solve for derivative of dy/dx= 3y/(3y^2 + 3x)
d^2y/dx^2=(3*dy/dx)(3y^2 + 3x) - [3y(6y*dydx +3)]/ [(3y^2 + 3x)^2]
d^2y/dx^2= (3*(2/3))(3(2)^2 + 3(1)) - [3y(6(2)*(2/3) +3)]/ [(3(2)^2 + 3(1))^2]
d^2y/dx^2= -4/27

Answer 2

y^3 - 3 xy = 2

y^3 - 3 xy = 2
Let us represent dy/dx by y’ and d^2 (y) / dx by = y’’
(a) Solution-
Differentiating both the sides w.r.t x we get -
3 y^2 y’ - 3 [ y + ( x y’) ] = 0
=> 3 y^2 y’ = 3 [ y + ( x y’) ]
=> y^2 y’ = y + x y’
=> y’ = [ y divided by (y^2 - x) .......... (1) --------------- Answer
=> ( y’ ) at (1,2) = 2 / (4-3) = 2/3 .............. (2)
..............................................................................................
(b) Solution -
Tangent to the curve at (1,2) has slope = (2/3)
But its eqn may be written as -
y = mx + c where c = a constant.
Substituting m = 2/3, x= 1 and y = 2 we get c = 4/3
Hence the required eqn to the tangent is -
y = 2/3 x + 4/3
=> 3y = 2x + 4
=> 2x - 3y + 4 = 0 .......... Answer
.......................................................................................
(c) Solution
Differentiating eqn (1) with respect to x we get -
y’’ = [ (y’ (y^2 - x ) - y ( 2 y y’ - 1 ) ] divided by Squire of ( y^2 - x )
=> [ y’ ( y^2 - x - 2 y^2 ) + y ] divided by Squire of ( y^2 - x )
=> [ y’’] at (1,2) = [ 2 - (2/3) ( 1 + 4 ) ] divided by ( 4 - 1 )^2
=> [ y’’] at (1,2) = - ( 4/27) ................ Answer

Answer 3

1. Use implicit differentiation.
3y^2(y')-3y-3x(y')=0
The solve for y'
2. Use point slope after you figured out its derivative.
3. Use the point (1,2) and you've found its derivative.

Answer 4

a. 3y^2dy/dx-3xdy/dx-3y=0
dy/dx=y/y^2-x

b. y-2=2/3(x-1)
y=2/3x+4/3

c.d^2y/dx^2=(y^2-x)dy/dx-y(2ydy/dx-1)/(y^2-x)^2
d^2y/dx^2= (y^2-x)2/3-y(2y2/3-1)/(y^2-x)^2
d^2y/dx^2= -4/27