Question

A curve has a gradient function
(2x^3 - 9)/10
The point P is a point on the curve.
The tangent to the curve at the point P is perpendicular to the line
2x - 5y + 3 = 0
Work out the x-coordinate of P​

Answer 1

Answer:

The answer I came up with here brings to question how I solved it.  To double check the wording of the question though, the tangent to the curve is indeed perpendicular to that line and not parallel?

The answer I found is 5i√(1/6), which makes me think I made a mistake at one point.

Step-by-step explanation:

First we need to know the slope of the tangent at point p.  This is perpendicular to the slope of the given line.  So first, let's rearrange that line into the format y = sx + c:

$2x - 5y + 3 = 0\\5y = 2x + 3\\y = \frac{2}{5}x + \frac{3}{5}$

The line then has a slope of 2/5, meaning that a perpendicular slope would be -5/2.

Next, we can take the derivative of the curve:

$y = (2x^3 - 9) / 10\\= x^3/5 - 9/10\\\frac{dy}{dx} = 3x^2/5$

Finally, we match that perpendicular slope we found:

$-5/2 = 3x^2 / 5\\-25 / 6 = x^2\\x = \sqrt{-25/6}\\x = 5i\sqrt{1/6}$

The fact that I ended up with an imaginary number makes me think I hit an error there.  I suggest you double check this yourself.  I'm fairly certain that the logic here is correct.