Vertical and horizontal gaps to a parabola

C

Total marks: 12

Solution

(i)

Optimisation: vertical gap

$C:y=x^2$ and $L:y=\frac12x-2$ . $A$ on $C$ and $B$ on $L$ have the same $x$ -coordinate. Find $A$ and $B$ that give the shortest distance $AB$ .

Explore this idea

Hint

For the same $x$ , $AB=x^2-\left(\frac x2-2\right)$ .
Minimise this function of $x$ .

Answer · (i)

A=(1/4,1/16)

Explore the idea

The vertical gap AB against x

The gap is smallest where its gradient is zero: x = 1/4.

Position along the curvex = -1

Try an example

Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

f(x)

3.5

Gradient f′(x)

-2.5

negative

For x<1/4 the vertical gap shrinks as x increases.

Selected values are within the original drawing windows.

Vertical gap AB
Tangent
Gradient, and where the gap grows
Join P to Q; the dashed legs show Δx and Δy

Key idea

The gap is a quadratic in $x$ ; its vertex is at $x=\frac14$ .

Ideas to try

Find the

x

with the smallest gap.

(ii)

Optimisation: horizontal gap

$P$ on $C$ and $Q$ on $L$ have the same $y$ -coordinate. Find $P$ and $Q$ that give the shortest distance $PQ$ .

Explore this idea

Hint

Write $x$ in terms of $y$ for each: $P$ has $x=\sqrt y$ (the nearer branch) and $Q$ has $x=2(y+2)$ .
Minimise $PQ=2(y+2)-\sqrt y$ .

Answer · (ii)

P=(1/4,1/16)

Explore the idea

The horizontal gap PQ against y

As y increases from 0 the gap first shrinks, then grows. Its gradient is zero at y = 1/16.

Position along the curvey = 0.5

Try an example

Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

f(y)

4.29289

Gradient f′(y)

1.29289

positive

For y>1/16 the horizontal gap grows.

Selected values are within the original drawing windows.

Horizontal gap PQ
Tangent
Gradient, and where the gap grows
Join P to Q; the dashed legs show Δx and Δy

Key idea

The horizontal and vertical gaps are different functions, and here both are smallest at the same point $\left(\frac14,\frac1{16}\right)$ on $C$ .

Ideas to try

Is the minimum at the same

y

for both gaps?

(iii)

Area of a polygon

Find the exact area of the polygon formed by joining the points found in (i) and (ii).

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Hint

$A$ and $P$ are the same point, so the polygon is the triangle $PBQ$ .
The base $AB$ is vertical and the base $PQ$ is horizontal, so the angle at $A$ is a right angle.

Answer · (iii)

=961/256

Explore the idea

A right-angled triangle between the curve and the line

Switch on the points and the two gaps.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

Ideas to try

Why is the area

\frac12\times AB\times PQ

(iv)

Related rates with an integral

A point $(s,s^2)$ starts at $O$ and moves along $C$ with $s$ increasing at 2 units/s. Find the rate of change of the area bounded by the curve, the $y$ -axis and the line $y=s^2$ when $s=\sqrt2$ .

Hint

The area is $\int_0^{s^2}\sqrt y\,dy$ .
Differentiate with respect to $s$ , then multiply by $\frac{ds}{dt}=2$ .

Answer · (iv)

8 square units/s.

Vertical and horizontal gaps to a parabola

Solution

Optimisation: vertical gap

The gap

Differentiate

Check

The points

The vertical gap AB against x

Optimisation: horizontal gap

The gap

Differentiate

Check

The points

The horizontal gap PQ against y

Area of a polygon

The legs

Area

A right-angled triangle between the curve and the line

Related rates with an integral

The area

Chain rule

Evaluate