P3 Differentiation....Help!Watch

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#1
Hi....
I'm having trouble with q10, 15, reveiw exercise 1, book p3 so i neeed your help plz... :

10) Find the coordinates of the turning points on the curve with equation:

y^3+3xy^2-x^3 = 0

15) (b) Find the gradient of the curve with the equation

5x^2+5y^2-6xy = 13

at the point (1,2).

Thanks......
0
13 years ago
#2
10) Find the coordinates of the turning points on the curve with equation:

y^3+3xy^2-x^3 = 0
Differentiate with respect to x and set dy/dx=0. Remember that you differentiate y as you normally do with x, but multiply by dy/dx.
If necessary, solve simultaneously with the equation of the curve to give the coordinates.

15) (b) Find the gradient of the curve with the equation

5x^2+5y^2-6xy = 13

at the point (1,2).

Thanks......
Differentiate with respect to x, put in the values for the coordinates and hence find dy/dx.

You may need the product rule for the above two.
0
13 years ago
#3
(Original post by ossoss87)
10) Find the coordinates of the turning points on the curve with equation:

y^3+3xy^2-x^3 = 0
You can find the solution to this one here: http://www.thestudentroom.co.uk/t93948.html
0
13 years ago
#4
y³+3xy²-x³= 0

Differentiate both sides, with respect to x:

3y²(dy/dx) + 3y² + 2y3x(dy/dx) - 3x² = 0
(dy/dx)(3y²+6yx) = 3x²-3y²
dy/dx = (3x²-3y²)/(3y²+6yx)

turning points at dy/dx=0
0=3x²-3y²
x= ±y

so back to original equation
y³+3y³-y³=0

Hmm is your LHS of equation meant to equal 0?
0
13 years ago
#5
(Original post by ossoss87)
Hi....

15) (b) Find the gradient of the curve with the equation

5x^2+5y^2-6xy = 13

at the point (1,2).

Thanks......
5x² + 5y² - 6xy = 13

Differentiate wrt x:

10x + 10y(dy/dx) - 6y + 6x(dy/dx) = 0
dy/dx(10y - 6x) = 6y - 10x
dy/dx = (6y-10x)/(10y-6x)
dy/dx = (3y-5x)/(5y-3x)

When x=1, y=2

so at (1,2),
dy/dx = (3*2 - 5)/(5*2 -3)
dy/dx = 1/7
0
#6
(Original post by endeavour)
y³+3xy²-x³= 0

Differentiate both sides, with respect to x:

3y²(dy/dx) + 3y² + 2y3x(dy/dx) - 3x² = 0
(dy/dx)(3y²+6yx) = 3x²-3y²
dy/dx = (3x²-3y²)/(3y²+6yx)

turning points at dy/dx=0
0=3x²-3y²
x= ±y

so back to original equation
y³+3y³-y³=0

Hmm is your LHS of equation meant to equal 0?
it is equal to 3! sorrry.. my mistake !
0
#7
(Original post by endeavour)
5x² + 5y² - 6xy = 13

Differentiate wrt x:

10x + 10y(dy/dx) - 6y + 6x(dy/dx) = 0
dy/dx(10y - 6x) = 6y - 10x
dy/dx = (6y-10x)/(10y-6x)
dy/dx = (3y-5x)/(5y-3x)

When x=1, y=2

so at (1,2),
dy/dx = (3*2 - 5)/(5*2 -3)
dy/dx = 1/7
thanks alot
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