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A curve is described by the equation x^3 - 4y^2 = 12xy. a) Find the points on the curve where x = -8. b) Find the gradient at these points.

a) The first part of this question is asking us to find points, (x,y), such that        x = -8. By taking the equation and substituting in x = -8, we should find values, y, such that the points given are (-8,y).

(-8)3 - 4y2 = 12(-8)y 

-512 - 4y2= -96y

The coefficients of these terms are relatively large, so we can look to divide them through by some number to simplify the problem, luckily all the numbers are divisible by 4. They are also negative, so multiplying the whole equations by -1 will turn them positive, which makes it nicer to handle.

128 + y= 24y

Rearranging

y2 - 24y +128 = 0

Now here using the quadriatic formula: [-b ± √(b2 - 4ac)]/2

We get y = [24 ± √(242 - 4*1*128)]/2, = [24 ± √(576 - 512)]/2 = [24 ± √(64)]/2  

[24 ± 8]/2 = 12 ± 4.

So the two values of y for which x = -8 are 8 and 16, so the point (x,y) where x = -8 are

(-8,8) (-8,16)

b) This part of the question requires us to differentiate implicitly the formula given. The question asks for the gradient, in other words it wants us to find a value of dy/dx for the points we worked out in the first section. We will differentiate the equation term by term.

Starting with 12xy, we will need to use the product rule for differentiation as there are two variables multiplied together. In case you had forgotten, the product rules works by effectively differentiating one of the variables while leaving the other alone, taking the result of this and adding it to the result we get had we differentiated the other term whilst leaving the first term alone.

Ie. d/dx(12xy) = 12y + 12x(dy/dx)

The first term 12y comes from differentiating x and leaving y alone, d/dx(x) = 1

The second 12x(dy/dx) comes from differentiating y and leaving x alone, d/dx(y) = dy/dx 

The in both cases the 12 can be ignored as it is just a constant, d/dx(12xy) = 12*d/dx(xy)

Now moving to the 4y2 term. Here we can also use the product rule by considering this term as 4y*y. Giving:

d/dx(4y*y) = 4*d/dx(y*y) = 4*y*(dy/dx) + 4*(dy/dx)*y = 8y(dy/dx)

And then for the x3 term, it is easy to see using normal differentiation that d/dx(x3) = 3x2

This gives the whole equation as:

3x2 - 8y(dy/dx) = 12y + 12x(dy/dx)

Rearranging and factoring out the (dy/dx) gives:

3x2 - 12y = (dy/dx)(12x8y)

Dividing by (12x8y) gives us an expression for dy/dx:

(3x2 - 12y)/(12x8y) = (dy/dx)

Now substituting in the (x,y) values we got from the first part (-8,8) (-8,16)

For (-8,8), (3(-8)2 - 12(8))/(12(-8)8(8)) = (dy/dx)

(192 - 96)/(-96 + 64) = 96/-32 = -3 = dy/dx

For (-8,16), (3(-8)2 - 12(16))/(12(-8)8(16)) = (dy/dx)

(192 - 192)/(-96 + 128) = 0/32 = 0 = (dy/dx)

So at (-8,8), dy/dx = 3 and at (-8,16), dy/dx = 0

 
Ben C. GCSE Biology tutor, A Level Maths tutor, GCSE Spanish tutor, G...

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