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Joseph is doing a training session - Leaving Cert Mathematics - Question 7 - 2022

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Joseph is doing a training session. During the session, his heart-rate, h(x), is measured in beats per minute (BPM). For part of the session, h(x) can be modelled us... show full transcript

Worked Solution & Example Answer:Joseph is doing a training session - Leaving Cert Mathematics - Question 7 - 2022

Step 1

e) Work out how many times, in total, the smart watch beeps during the session, including the first and last beep.

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Answer

To find the total number of beeps, we need to calculate the time interval from 2:55 p.m. to 3:23 p.m.

First, calculate the total time in minutes:

  • From 2:55 p.m. to 3:00 p.m. is 5 minutes.
  • From 3:00 p.m. to 3:23 p.m. is 23 minutes.
  • Total time is 5 + 23 = 28 minutes.

Since the watch beeps every 15 seconds, we convert 28 minutes into seconds:

  • 28 minutes = 28 × 60 = 1680 seconds.

Next, we calculate the number of 15-second intervals in 1680 seconds:

  • Number of intervals = 1680 seconds / 15 seconds = 112.

Including the first beep at 2:55 p.m. and the last beep at 3:23 p.m., the total number of beeps is:

  • 112 + 1 = 113 times.

Step 2

f) Solve the equation h'(x) = -1.14x^2 + 5.2x - 0.13 = 0

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Answer

To solve for x, we need to identify the roots of the derivative:

  1. Identify a, b, and c from the equation:

    • a = -1.14
    • b = 5.2
    • c = -0.13

  2. Use the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    • Substituting in our values: x=−5.2±(5.2)2−4×(−1.14)×(−0.13)2×(−1.14)x = \frac{-5.2 \pm \sqrt{(5.2)^2 - 4\times(-1.14)\times(-0.13)}}{2\times(-1.14)}
    • Calculate the discriminant: =27.04−0.5916=26.4484≈5.1447= \sqrt{27.04 - 0.5916} = \sqrt{26.4484} \approx 5.1447

  3. Now, substituting back into the quadratic formula: x=−5.2±5.1447−2.28x = \frac{-5.2 \pm 5.1447}{-2.28}

  4. Calculate both solutions:

    • Positive solution: x=0.0553−2.28≈0.0243x = \frac{0.0553}{-2.28} \approx 0.0243
    • Negative solution (discarded since x must be positive): x=−10.3447−2.28≈4.539x = \frac{-10.3447}{-2.28} \approx 4.539

Thus, the maximum heart-rate occurs at approximately 4.54 minutes after the session starts.

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