Question

How would I start this?

the line `x=a` intersects the curve `y= 1/3x^3+4x+2` at a point `P` and the curve `y=2x^2+x` at a point `Q`. for what value of `a` are the tangents to these curve at `P` and `Q` parallel

Answer 1

When two lines (in this case, the tangents) are parallel, it means they have the same slope.

The slope of a `y` at a point `p` is equal to the derivative of `y`, evaluated at the point `p`.

Define `f(x)=1/3 x^3 + 4x + 2` and `g(x)= 2x^2+x`.
Compute `f'(x)`, and `g'(x)`.

Since we want the slopes equal at the same x-coordinate `x=a`, set `f'(a)=g'(a)` and solve for `a`.

Answer 2

`a=1` or `a=3`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So if we differentiate the equations we have:

Consider the first curve

`y = 1/3x^3 + 4x + 2 => dy/dx = x^2 + 4`

This curve intersects the line `y=a` at the coordinate:

`P(a, 1/3a^3 + 4a + 2)`

So at `P`, the gradient of the tangent is given by:

`m_P = a^2+4`

Consider the second curve

`y = 2x^2+x => dy/dx = 4x+1`

Consider the second curve

`y = 2x^2+x => dy/dx = 4x+1`

This curve intersects the line `y=a` at the coordinate:

`Q(a, 2a^2+a)`

So at `Q`, the gradient of the tangent is given by:

`m_Q = 4a+1`

The Tangents to these curve at `P` and `Q` parallel provided:

`m_P = m_Q`
`=> a^2+4 = 4a+1`
`:. a^2 - 4a +3 = 0`
`:. (a-3)(a-1) = 0`
`:. a=1,3`