AI HL · 12.10
0 of 79 skills done
Grazing goat problems: sectors round a shed
12.10.14
Screenshot and video to come
Screenshot and video to come
A goat is tied to a 12 m by 8 m shed, seen from above. Change the length of the rope. The grass it can reach is made of sectors, and the area is their sum.
On the long wall, drag the hook X along the wall, or click it and use the arrow keys. Angles are in radians.
Move the rope slider to change the grass the goat can reach.
Set the radius with the slider and drag the point to set the angle at the centre. The arc length and the area are worked out with the numbers.
Drag the point to change the angle.
Two fences meet at a corner post at an angle of \(\frac{\pi}{4}\) radians. A pony is tied to the post by an 8 m rope and is outside the angle between the fences. Both fences are longer than 8 m. Find the area of grass the pony can graze.
Have a go first. Then open the working one line at a time.
\(\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\)
The fences take up π/4. The rest of the full turn, 2π, is open to the pony. This is a reflex angle.
\(A = \frac12 r^2\theta = \frac12 \times 8^2 \times \frac{7\pi}{4}\)
The formula needs the angle in radians.
\(A = 56\pi = 176\text{ m}^2\)
Give 3 significant figures, or leave the exact answer 56π.
A goat is tied by a 15 m rope to the corner X of a shed. The shed is a rectangle 12 m by 8 m. The goat can graze all the open ground it can reach. Find the area of grass the goat can graze.
Have a go first. Then open the working one line at a time.
\(\frac34 \times \pi \times 15^2 = 168.75\pi\)
The shed blocks a right angle at X, so the goat sweeps 3π/2 with the full 15 m.
\(\text{rope left} = 15 - 12 = 3\text{ m},\quad \frac14 \pi \times 3^2 = 2.25\pi\)
The rope runs along the 12 m wall and wraps at the far corner. A quarter circle of radius 3.
\(\text{rope left} = 15 - 8 = 7\text{ m},\quad \frac14 \pi \times 7^2 = 12.25\pi\)
Same idea along the other wall. Check: 3 + 7 < 12 + 8, so the two quarter circles do not overlap.
\(A = 168.75\pi + 2.25\pi + 12.25\pi = 183.25\pi = 576\text{ m}^2\)
Three sectors, one total.
A shed is a rectangle 12 m by 9 m. A goat is tied by a 10 m rope to a hook X on a 12 m wall, 3 m from the left-hand corner. Find the area of grass the goat can graze.
Have a go first. Then open the working one line at a time.
\(\frac12 \times \pi \times 10^2 = 50\pi\)
The wall is straight at X, so the goat sweeps a half turn with the full rope.
\(\text{rope left} = 10 - 3 = 7\text{ m},\quad \frac14\pi\times 7^2 = 12.25\pi\)
X is 3 m from the left corner. 7 m of rope are left to swing along the 9 m side wall.
\(\text{rope left} = 10 - (12-3) = 1\text{ m},\quad \frac14\pi\times 1^2 = 0.25\pi\)
X is 12 − 3 = 9 m from the right corner.
\(A = 50\pi + 12.25\pi + 0.25\pi = 62.5\pi = 196.3\text{ m}^2\)
Both left-over lengths are shorter than the 9 m side wall, so the rope does not wrap again.
A shed is 10 m wide and 8 m deep. A goat is tied by a 9 m rope to a hook X on the front wall, x m from the left-hand corner, where \(1 \le x \le 9\). Write the grazed area A in terms of x. Then find A when x = 4.
Have a go first. Then open the working one line at a time.
\(\frac12\pi\times 9^2 = \frac{81\pi}{2}\)
The rope is 9 m for the half turn in front of the wall.
\(\text{radius } 9 - x:\quad \frac{\pi}{4}(9-x)^2\)
The hook is x m from the left corner, so 9 − x m of rope are left.
\(\text{radius } 9 - (10-x) = x - 1:\quad \frac{\pi}{4}(x-1)^2\)
The hook is 10 − x m from the right corner. The domain 1 ≤ x ≤ 9 keeps both radii from 0 to 8, so neither wraps a second time.
\(A = \frac{81\pi}{2} + \frac{\pi}{4}(9-x)^2 + \frac{\pi}{4}(x-1)^2 = \frac{\pi}{2}(x^2 - 10x + 122)\)
Expand: (9 − x)² + (x − 1)² = 2x² − 20x + 82. Then 81π/2 = 162π/4 gives the 122.
\(A = \frac{\pi}{2}(16 - 40 + 122) = 49\pi = 154\text{ m}^2\)
Check with the radii: 5 and 3 for the quarter circles.
For the shed in the last example, the goat grazed 160 m² of grass. Find the possible distances x of the hook from the left-hand corner.
Have a go first. Then open the working one line at a time.
\(\frac{\pi}{2}(x^2 - 10x + 122) = 160\)
Use the area function from the last example.
Solve for x on 1 ≤ x ≤ 9: x = 2.80 or x = 7.20
A quadratic can have two solutions. Both lie between 1 and 9, so both are possible.
\(2.80 + 7.20 = 10\)
The two hooks are mirror images in the middle of the 10 m wall (x = 5). They give the same area.
For the same shed and rope, where should the hook be, for \(1 \le x \le 9\), so the goat grazes (a) the most grass, (b) the least grass?
Have a go first. Then open the working one line at a time.
\(y = \frac{\pi}{2}(x^2 - 10x + 122)\) on 1 ≤ x ≤ 9
Set the window to match the domain. The graph is a U-shaped parabola.
Minimum: x = 5, A = 152 m²
In the middle both quarter circles are small. The lowest point of a parabola is at the vertex: x = 5.
At x = 1 or x = 9: A = 177 m²
On a closed domain the highest point is at an end. At x = 1 the right quarter circle vanishes and the left one has radius 8: the rope is nearly all spent on one side.