Maths ‘A’ level: now for a tough one

From the same paper as the previous two questions, here is the hardest question (as measured by marks on offer – 12):

Sketch the curve given parametrically by x = t^2, y = t^3
Show that an equation of the normal to the curve at the point A(4,8) is x + 3y - 28 = 0
This normal meets the x-axis at the point N. Find the area of the region enclosed by the arc OA of the curve, the line segment AN and the x-axis.

While I can see the ideas behind each part of this question I am not at all sure where to begin with it.

For background, by the marking scheme you should take about 16 minutes to answer this – (and about 5 minutes per question for the earlier two). No pressure then.

8 responses to “Maths ‘A’ level: now for a tough one”

  1. Once you have the sketch everything else becomes much easier. You get the sketch by asking yourself “Which points (x, y) can be generated by varying t?”. You notice that x is always positive and that for each x there are two y values, one with positive and one with negative t. So draw two graphs: One for t positive and one for t negative. For each of those domains transform the two equations given into the form y = f(x). That makes drawing the two curves trivial and also allows to get the derivative later.

    1. Yes but it talks of one curve, not two. Or have I missed something?

      1. Not sure. It is one curve though because the two curves meet at (0, 0).

    2. OK, sketched the curve, it’s one curve with an inflection point at 0. But I make it f(x)=x^{\frac{3}{2}} and equation of normal as y=\frac{3}{2}\sqrt{x} + 2, so plainly I have got something wrong here

      1. OK so the equation for the normal is y = \frac{3}{2}\sqrt{x} + 5, and I can make that look like 3y - \frac{9}{4}x - 15 = 0 and -\frac{9}{4}x - 15 = x - 28 at A . But is there a cleaner way?

  2. ha ha. I now I realise I have completely misunderstood the question. U for me.

  3. Isn’t this basic calculus? The reason they call it a curve is because it’s not a function of a single variable. The only reason you can pretend it is because it only asks you about one piece of the curve.

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