# differentiation past exam question

• Apr 3rd 2008, 04:25 AM
Sweeties
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'
• Apr 3rd 2008, 06:30 AM
JaneBennet
2. Substitute $\displaystyle x=1$ into the equation $\displaystyle y=x(x^2+4)$. So your other co-ordinate is $\displaystyle y=5$.

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 $\displaystyle m\ne0$, the gradient of the normal is $\displaystyle -\,\frac{1}{m}$.

In your example, the normal to the curve $\displaystyle y=x^3+ax$ at $\displaystyle x=2$ is $\displaystyle \frac{1}{2}$; hence $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-2$ at $\displaystyle x=2$. You should be able to find a from here. Once you’ve found a, substitute $\displaystyle x=2$ into $\displaystyle y=x^3+ax$ to find the y-coordinate, which is b.
• Apr 3rd 2008, 07:47 AM
Sweeties
Quote:

Originally Posted by JaneBennet

In your example, the normal to the curve $\displaystyle y=x^3+ax$ at $\displaystyle x=2$ is $\displaystyle \frac{1}{2}$; hence $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-2$ at $\displaystyle x=2$.

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 ?
• Apr 3rd 2008, 08:02 AM
Sweeties
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