✨ Math Magic: Exercise 5.5 Complete Solutions ✨

1
Parabolic Bridge Arch

A bridge has a parabolic arch that is 10m high in the centre and 30m wide at the bottom. Find the height of the arch 6m from the centre, on either sides.

1 Understand the problem: We have a parabola opening downward with vertex at (0,10).
2 Set up coordinates: Let's place the vertex at the top (0,10). The arch is 30m wide at bottom, so it touches ground at (-15,0) and (15,0).
3 Equation of parabola: Standard form is y = a(x-h)² + k. With vertex at (0,10):
y = ax² + 10
4 Find 'a': Using point (15,0):
0 = a(15)² + 10 → 225a = -10 → a = -10/225 = -2/45
5 Final equation:
y = (-2/45)x² + 10
6 Find height at x=6m:
y = (-2/45)(6)² + 10 = (-2/45)(36) + 10 = -1.6 + 10 = 8.4m
The height of the arch 6m from the center is 8.4 meters.
Did you know? The Gateway Arch in St. Louis is also a parabola (more precisely, a catenary curve) standing 192m tall!
2
Elliptical Tunnel Opening

A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16m, and the height at the edge of the road must be sufficient for a truck 4m high to clear if the highest point of the opening is to be 5m approximately. How wide must the opening be?

1 Understand the ellipse: The tunnel opening is an ellipse with vertical major axis (since height > width).
2 Standard form: Equation of vertical ellipse:
x²/b² + y²/a² = 1
where 2a = major axis (height), 2b = minor axis (width)
3 Given parameters:
  • Highest point (vertex) is 5m → a = 5
  • At edge of road (x=8m since highway is 16m wide), height must be 4m
4 Find b: Using point (8,4):
8²/b² + 4²/5² = 1 → 64/b² + 16/25 = 1 → 64/b² = 9/25 → b² = 64×25/9 → b = 40/3 ≈ 13.33
5 Opening width: Total width is 2b:
2b ≈ 2 × 13.33 ≈ 26.67m
The elliptical opening must be approximately 26.67 meters wide.
Fun fact: Elliptical shapes are used in tunnels because they distribute stress more evenly than circular shapes!
3
Water Fountain Parabola

At a water fountain, water attains a maximum height of 4m at horizontal distance of 0.5m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75m from the point of origin.

1 Understand the parabola: Water originates at (0,0), reaches max height at (0.5,4).
2 Equation form: Using vertex form y = a(x-h)² + k with vertex at (0.5,4):
y = a(x-0.5)² + 4
3 Find 'a': Using origin point (0,0):
0 = a(0-0.5)² + 4 → 0.25a = -4 → a = -16
4 Final equation:
y = -16(x-0.5)² + 4
5 Find height at x=0.75m:
y = -16(0.75-0.5)² + 4 = -16(0.25)² + 4 = -1 + 4 = 3m
The height of water at 0.75m from origin is 3 meters.
Cool fact: The water's path is a parabola because gravity accelerates all water particles equally downward!
4
Satellite Dish Design

An engineer designs a satellite dish with a parabolic cross section. The dish is 5m wide at the opening, and the focus is placed 1.2m from the vertex.

(a) Position a coordinate system with the origin at the vertex and the x-axis on the parabola's axis of symmetry and find an equation of the parabola.

(b) Find the depth of the satellite dish at the vertex.

1 Part (a): Find equation of parabola
2 Set up coordinate system: Vertex at (0,0), opening upwards.
3 Standard form: For vertical parabola with vertex at origin:
x² = 4py
where p is distance from vertex to focus (1.2m)
4 Equation:
x² = 4(1.2)y → x² = 4.8y
5 Part (b): Find depth at vertex
6 Find y when x=2.5m (half of 5m width):
(2.5)² = 4.8y → 6.25 = 4.8y → y = 6.25/4.8 ≈ 1.302m
(a) Equation of parabola: x² = 4.8y
(b) Depth at vertex: ≈1.302 meters
Tech fact: Satellite dishes are parabolic because this shape reflects all incoming signals to the focus where the receiver is placed!
5
Suspension Bridge Cables

Parabolic cable of a 60m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical Cables are to be spaced every 6m along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.

1 Understand the problem: Assuming the parabola has vertex at the center (lowest point) and spans 60m (from x=-30 to x=30).
2 Standard form: y = ax² (vertex at origin)
3 Find 'a': Need more information. Assuming the ends are at height 10m (typical bridge example):
10 = a(30)² → a = 10/900 = 1/90
4 Equation:
y = (1/90)x²
5 First cable at x=6m:
y = (1/90)(6)² = 36/90 = 0.4m
6 Second cable at x=12m:
y = (1/90)(12)² = 144/90 = 1.6m
Lengths of first two vertical cables:
First cable (6m from vertex): 0.4 meters
Second cable (12m from vertex): 1.6 meters
Engineering fact: The Golden Gate Bridge's cables form a parabola because this shape evenly distributes the bridge's weight!
6
Nuclear Cooling Tower

Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation \(\frac{x^2}{30^2} - \frac{y^2}{44^2} = 1\). The tower is 150m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.

1 Understand the hyperbola: Given equation:
x²/900 - y²/1936 = 1
2 Tower dimensions: Total height = 150m. Let distance from center to top = d, then to base = 2d.
d + 2d = 150 → 3d = 150 → d = 50m
So top is at y = +50m, base at y = -100m
3 Find x at y=50m (top):
x²/900 - (50)²/1936 = 1 → x²/900 = 1 + 2500/1936 ≈ 2.291
x² ≈ 900 × 2.291 ≈ 2062 → x ≈ ±45.4m
Diameter = 2 × 45.4 ≈ 90.8m
4 Find x at y=-100m (base):
x²/900 - (-100)²/1936 = 1 → x²/900 = 1 + 10000/1936 ≈ 6.166
x² ≈ 900 × 6.166 ≈ 5549.4 → x ≈ ±74.5m
Diameter = 2 × 74.5 ≈ 149m
Diameter at top: ≈90.8 meters
Diameter at base: ≈149 meters
Science fact: Cooling towers are hyperbolic because this shape provides structural strength and optimal air flow!
7
Elliptical Locus Problem

A rod of length 1.2m moves with its ends always touching the coordinate axes. The locus of a point P on the rod, which is 0.3m from the end in contact with x-axis is an ellipse. Find the eccentricity.

1 Understand the problem: Rod slides between axes, point P traces an ellipse.
2 Let end on x-axis be A, on y-axis be B: AB = 1.2m, AP = 0.3m, PB = 0.9m
3 Parametric equations: Let angle between rod and x-axis be θ.
A = (a,0), B = (0,b), where a² + b² = 1.2² = 1.44
4 Point P coordinates:
P = (0.9cosθ, 0.3sinθ) (since AP/PB = 0.3/0.9 = 1/3)
5 Ellipse equation:
(x/0.9)² + (y/0.3)² = cos²θ + sin²θ = 1
So semi-major axis a=0.9, semi-minor axis b=0.3
6 Eccentricity:
e = √(1 - b²/a²) = √(1 - 0.09/0.81) = √(1 - 1/9) = √(8/9) = (2√2)/3
The eccentricity of the ellipse is \(\frac{2\sqrt{2}}{3}\) ≈ 0.9428
Geometry fact: This is an example of how ellipses appear in mechanical systems - the same principle is used in elliptical gears!
8
Water Projectile Problem

Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5m below the line of the pipe, the flow of water has curved outward 3m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground?

1 Set up coordinates: Let pipe end be at (0,7.5) - vertex of parabola.
2 Equation form: Since vertex is at (0,7.5) and opens downward:
y = -ax² + 7.5
3 Find 'a': Using point (3,5) [2.5m below pipe]:
5 = -a(3)² + 7.5 → -9a = -2.5 → a = 2.5/9 ≈ 0.2778
4 Final equation:
y = -0.2778x² + 7.5
5 Find x when y=0 (ground):
0 = -0.2778x² + 7.5 → x² = 7.5/0.2778 ≈ 27 → x ≈ ±5.196m
The water will strike the ground approximately 5.2 meters beyond the vertical line.
Physics fact: This parabolic trajectory is the same principle that governs how water fountains and even Olympic diving work!
9
Rocket Projectile Angle

On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4m when it is 6m away from the point of projection. Finally it reaches the ground 12m away from the starting point. Find the angle of projection.

1 Understand the trajectory: Vertex at (6,4), roots at (0,0) and (12,0).
2 Equation form: Using roots form y = a(x)(x-12):
At vertex (6,4): 4 = a(6)(-6) → 4 = -36a → a = -1/9
3 Final equation:
y = (-1/9)x(x-12)
4 Find initial velocity components: From projectile motion equations:
Time to reach max height: t = v₀sinθ/g
Max height: h = (v₀sinθ)²/(2g) = 4
Horizontal distance at max height: R/2 = v₀cosθ × t = 6
5 Range equation: Total range R = 12m
R = (v₀²sin2θ)/g → 12 = (v₀²sin2θ)/g
6 Solve for θ: From max height:
4 = (v₀sinθ)²/(2g) → v₀sinθ = √(8g)
From horizontal distance: 6 = v₀cosθ × (v₀sinθ/g)
6 = √(8g) × v₀cosθ/g → v₀cosθ = 6g/√(8g) = 3√(g/2)
7 Find tanθ:
tanθ = (v₀sinθ)/(v₀cosθ) = √(8g)/(3√(g/2)) = √(8)/√(1/2)/3 = (2√2)/(√2/2)/3 = (2√2 × 2/√2)/3 = 4/3
8 Calculate θ:
θ = tan⁻¹(4/3) ≈ 53.13°
The angle of projection is approximately 53.13°.
Projectile fact: The optimal angle for maximum range (when landing at the same elevation) is 45° - here it's higher because the rocket reaches significant height!
10
Explosion Location Problem

Points A and B are 10km apart and it is determined from the sound of an explosion heard at those points at different times that the location of the explosion is 6 km closer to A than B. Show that the location of the explosion is restricted to a particular curve and find an equation of it.

1 Understand the problem: This describes a hyperbola where the difference in distances to two fixed points is constant.
2 Set up coordinates: Let A be at (-5,0) and B at (5,0) (10km apart).
3 Distance condition: For any point P on the curve:
PB - PA = 6km
4 Hyperbola properties: This is a hyperbola with foci at A and B.
For hyperbola: |PF₂ - PF₁| = 2a = 6 → a = 3
Distance between foci = 2c = 10 → c = 5
5 Find b: For hyperbola, c² = a² + b²:
25 = 9 + b² → b² = 16 → b = 4
6 Equation: Standard form for horizontal hyperbola centered at origin:
x²/a² - y²/b² = 1 → x²/9 - y²/16 = 1
The explosion location lies on the hyperbola: \(\frac{x^2}{9} - \frac{y^2}{16} = 1\)
Real-world application: This is how GPS systems and earthquake detection works - using time differences to locate positions on hyperbolic curves!