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Fiona finds the volumes of five different cylinders - Junior Cycle Mathematics - Question 11 - 2016

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Fiona finds the volumes of five different cylinders. Each of them has a height of K centimetres. (a) Complete the table below to show the volume of each of the five... show full transcript

Worked Solution & Example Answer:Fiona finds the volumes of five different cylinders - Junior Cycle Mathematics - Question 11 - 2016

Step 1

Complete the table for the volume of each cylinder

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Answer

To find the volume of a cylinder, we use the formula:

V=πr2hV = \pi r^2 h

where:

  • VV is the volume,
  • rr is the radius, and
  • hh is the height.

Using this formula, we can fill in the missing volumes for the cylinders:

  1. For radius = 1 cm:

    • Volume = π(12)(K)=πK\pi (1^2)(K) = \pi K

  2. For radius = 2 cm:

    • Volume = π(22)(K)=4πK\pi (2^2)(K) = 4\pi K

  3. For radius = 3 cm (given):

    • Volume = 9πK9\pi K

  4. For radius = 4 cm:

    • Volume = π(42)(K)=16πK\pi (4^2)(K) = 16\pi K

  5. For radius = 5 cm:

    • Volume = π(52)(K)=25πK\pi (5^2)(K) = 25\pi K

Thus, the completed table is:

Radius of cylinder (cm)Height of cylinder (cm)Volume of cylinder (cm³)
1KπK
2K4πK
3K9πK
4K16πK
5K25πK

Step 2

Is the sequence of volumes linear, quadratic, exponential, or none?

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Answer

The sequence of volumes is:

  • πK
  • 4πK
  • 9πK
  • 16πK
  • 25πK

By observing the coefficients: 1, 4, 9, 16, 25, we can see that these correspond to the square of the integers:

  • 121^2, 222^2, 323^2, 424^2, 525^2.

The first difference is:

  • 4πK−πK=3πK4\pi K - \pi K = 3\pi K
  • 9πK−4πK=5πK9\pi K - 4\pi K = 5\pi K
  • 16πK−9πK=7πK16\pi K - 9\pi K = 7\pi K
  • 25πK−16πK=9πK25\pi K - 16\pi K = 9\pi K

And the second difference is constant:

  • 5πK−3πK=2πK5\pi K - 3\pi K = 2\pi K
  • 7πK−5πK=2πK7\pi K - 5\pi K = 2\pi K
  • 9πK−7πK=2πK9\pi K - 7\pi K = 2\pi K

Since the second difference is constant, the sequence is quadratic.

Answer: The sequence is quadratic.

Justification: The volumes follow a pattern where each term is the square of the radius multiplied by πK\pi K, indicating a quadratic relationship.

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