Question

How do you find the slope of a tangent line to the graph of the function`2xy^2+xy=y` at y=1?

Answer 1

`y = -9/2x + 5/2`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is `-1`).

We have:

`2xy^2+xy=y`

Differentiating implicitly wrt `x` we have:

`(2x)(d/dx y^2) + (d/dx 2x)(y^2) + (x)(d/dx y) + (d/dx x)(y) = dy/dx`

`:. (2x)(2ydy/dx) + (2)(y^2) + (x)(dy/dx) + (1)(y) = dy/dx`

`:. 4xydy/dx + 2y^2 + xdy/dx + y = dy/dx`

`:. (1 - x - 4xy)dy/dx = 2y^2 + y`

`:. dy/dx = (2y^2 + y)/(1 - x - 4xy)`

So when `y=1` we have;

`2x+x=1 => x=1/3`

And so:

`:. dy/dx = (2 + 1)/(1 - 1/3 - 4/3) = -9/2`

So the tangent passes through `(1/3,1)` and has gradient `m_T=-9/2`, and using the point/slope form `y-y_1=m(x-x_1)` the normal equation we seek is;

`y-1 = -9/2(x-1/3)`
`:. y-1 = -9/2x + 3/2`
`:. y = -9/2x + 5/2`

We can verify this graphically: