Thread: differentiation past exam question

hello everybody!

1. if y= x(x^2 + 4) find dy/dx
2. then find the equation of the tangent to the curve where x=1 giving answer in the form y=mx+c
3. the normal to the curve y=x^3 + ax at coordinates (2,b) has gradient 1/2. find the values of the constants a and b.

1. dy/dx = 3x^2 + 4

2. I assume we would use y-y1 = m(m-m1) where the 1 is at the bottom of the y and m, but we only have one coordinate, which is x=1. So I am not sure about this

3. I haven't got a clue what the 'normal to the curve is'

2. Substitute into the equation . So your other co-ordinate is .

3. The normal to a curve at a given point P is the straight line perpendicular to the tangent at P and passing through P. If the tangent has gradient , the gradient of the normal is .

In your example, the normal to the curve at is ; hence at . You should be able to find a from here. Once you’ve found a, substitute into to find the y-coordinate, which is b.

Originally Posted by JaneBennet

In your example, the normal to the curve at is ; hence at .

I don't understand the above, if the normal is -1/m and m=1/2 then the normal is -2, right?

And what have you differentiated to give dy/dx = -2 ?

Here's what I got anyway,

for y=x^3 + ax

the normal = -1/m = -1/(1/2) = -2 = a

so, y= b = x^3 + 2x

b= 2^3 + (2*-2) = 4