Question 64 of 70intermediate🔧 ApplyLong Answer4 marks

An arch is in the form of a parabola with its axis vertical. The arch is 1010 m high and 55 m wide at the base. How wide is it 22 m from the vertex of the parabola?

Correct Answer

The arch is approximately 2.242.24 m (or 5\sqrt{5} m) wide at a point 22 m from the vertex.

Exercise: Miscellaneous Exercise on Chapter 10 | Q: 2 | (Page 29)
For More Understanding

Explanation

This problem involves finding the equation of a parabola given its dimensions and then using it to find the width at a specific height. Since the arch has its axis vertical and opens downward, we use the standard form x2=−4ayx^2 = -4ay with vertex at origin. The base points give us the value of aa, and then we substitute y=−2y = -2 to find the width at 22 m from the vertex.

Solution Steps

  1. Step 1: Set up coordinate system with vertex at origin (0,0)(0, 0). Since the axis is vertical and the arch opens downward, the equation is:

    x2=−4ayx^2 = -4ay

  2. Step 2: At the base: Height =10= 10 m, so y=−10y = -10. Width =5= 5 m, so x=±2.5x = \pm 2.5 m.

  3. Step 3: Substitute base point (2.5,−10)(2.5, -10) into the equation:

    (2.5)2=−4a(−10)(2.5)^2 = -4a(-10)

    This gives 6.25=40a6.25 = 40a, so:

    a=6.2540=0.15625 ma = \frac{6.25}{40} = 0.15625 \text{ m}

  4. Step 4: The equation of the parabola is:

    x2=−4(0.15625)y=−0.625yx^2 = -4(0.15625)y = -0.625y

    or

    x2=−5y8x^2 = -\frac{5y}{8}

  5. Step 5: At 22 m from vertex (y=−2y = -2):

    x2=−5(−2)8=108=54x^2 = -\frac{5(-2)}{8} = \frac{10}{8} = \frac{5}{4}

    So:

    x=±54=±52x = \pm\sqrt{\frac{5}{4}} = \pm\frac{\sqrt{5}}{2}

  6. Step 6: Width =2×52=5≈2.236 m≈= 2 \times \frac{\sqrt{5}}{2} = \sqrt{5} \approx 2.236 \text{ m} \approx 2.242.24 m.

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