Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 9

Table of contents
- a 9
  - Solution a
- b 9
  - Solution b

a

Solution

On

0 \leq x \leq 1/2

, the quadratic polynomial is positive. The required area is therefore its definite integral above the horizontal axis:

A = \int_{0}^{1/2} (x^{2} - 3 x + 2) d x = [\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + 2 x]_{0}^{1/2} = \frac{1}{24} - \frac{3}{8} + 1 = \frac{2}{3} .

The units are square coordinate units.

b

Solution

Real points require

x \geq 0

, and the two branches are

y = \pm x^{3/2}, x \geq 0.

This is a semicubical parabola, symmetric about the horizontal axis and with a cusp at the origin. The parametrization

x = t^{2}

,

y = t^{3}

traces both branches. Their limiting slopes at the origin are zero, so the common tangent direction is the positive horizontal axis, even though the curve has a cusp there.

Figure 1.
The two real branches of the semicubical parabola and their cusp at the origin
. Details.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version