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Hannah is doing a training session - Leaving Cert Mathematics - Question 7 - 2022 Question 7
View full question Hannah is doing a training session. During this session, her heart-rate, h(x), is measured in beats per minute (BPM), where x is the time in minutes from the start o... show full transcript
View marking scheme Worked Solution & Example Answer:Hannah is doing a training session - Leaving Cert Mathematics - Question 7 - 2022
Work out Hannah's heart-rate 4 minutes after the start of the session. Only available for registered users.
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To find Hannah's heart-rate at 4 minutes, substitute x = 4 into h(x):
h ( 4 ) = 2 ( 4 ) 3 − 28 ( 4 ) 2 + 105 ( 4 ) + 70 h(4) = 2(4)^3 - 28(4)^2 + 105(4) + 70 h ( 4 ) = 2 ( 4 ) 3 − 28 ( 4 ) 2 + 105 ( 4 ) + 70
Calculating this gives:
h ( 4 ) = 2 ( 64 ) − 28 ( 16 ) + 420 + 70 h(4) = 2(64) - 28(16) + 420 + 70 h ( 4 ) = 2 ( 64 ) − 28 ( 16 ) + 420 + 70 h ( 4 ) = 128 − 448 + 420 + 70 h(4) = 128 - 448 + 420 + 70 h ( 4 ) = 128 − 448 + 420 + 70 h ( 4 ) = 128 − 448 + 490 = 170 h(4) = 128 - 448 + 490 = 170 h ( 4 ) = 128 − 448 + 490 = 170
Thus, Hannah's heart-rate at 4 minutes is 170 BPM.
Find h'(x). Only available for registered users.
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To find h'(x), we differentiate h(x):
h ( x ) = 2 x 3 − 28 x 2 + 105 x + 70 h(x) = 2x^3 - 28x^2 + 105x + 70 h ( x ) = 2 x 3 − 28 x 2 + 105 x + 70
The derivative is:
h ′ ( x ) = 6 x 2 − 56 x + 105 h'(x) = 6x^2 - 56x + 105 h ′ ( x ) = 6 x 2 − 56 x + 105
Find h'(2), and explain what this value means in the context of Hannah's heart-rate. Only available for registered users.
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First, we evaluate h'(2):
h ′ ( 2 ) = 6 ( 2 ) 2 − 56 ( 2 ) + 105 h'(2) = 6(2)^2 - 56(2) + 105 h ′ ( 2 ) = 6 ( 2 ) 2 − 56 ( 2 ) + 105 h ′ ( 2 ) = 6 ( 4 ) − 112 + 105 h'(2) = 6(4) - 112 + 105 h ′ ( 2 ) = 6 ( 4 ) − 112 + 105 h ′ ( 2 ) = 24 − 112 + 105 = 17 h'(2) = 24 - 112 + 105 = 17 h ′ ( 2 ) = 24 − 112 + 105 = 17
This means that at 2 minutes, Hannah's heart-rate is increasing at a rate of 17 BPM per minute.
Find the least value and the greatest value of h(x), for 0 ≤ x ≤ 8, x ∈ ℝ. Only available for registered users.
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To find the maximum and minimum values of h(x), we evaluate h(x) and its critical points.
Firstly, set h'(x) = 0:
6 x 2 − 56 x + 105 = 0 6x^2 - 56x + 105 = 0 6 x 2 − 56 x + 105 = 0
Using the quadratic formula:
⇒ x = 56 ± ( − 56 ) 2 − 4 ( 6 ) ( 105 ) 2 ( 6 ) ⇒ x = 56 ± 3136 − 2520 12 ⇒ x = 56 ± 616 12 ⇒ x = 56 ± 24.8 12 ⇒ x ≈ 5.67 , 2.67. E v a l u a t e h ( x ) a t t h e s e p o i n t s a n d e n d p o i n t s 0 a n d 8 t o f i n d t h e l e a s t a n d g r e a t e s t . A f t e r c a l c u l a t i n g : − h ( 0 ) = 70 − h ( 2.67 ) = 203.36 ( m a x i m u m ) − h ( 5.67 ) = 186.96 − h ( 8 ) = 70 T h e l e a s t v a l u e i s ∗ ∗ 70 ∗ ∗ , a n d t h e g r e a t e s t v a l u e i s ∗ ∗ 203.36. ∗ ∗ \Rightarrow x = \frac{56 \pm \sqrt{(-56)^2 - 4(6)(105)}}{2(6)}\
\Rightarrow x = \frac{56 \pm \sqrt{3136 - 2520}}{12}\
\Rightarrow x = \frac{56 \pm \sqrt{616}}{12}\
\Rightarrow x = \frac{56 \pm 24.8}{12}\
\Rightarrow x \approx 5.67, 2.67.\
Evaluate h(x) at these points and endpoints 0 and 8 to find the least and greatest.
After calculating:
- h(0) = 70
- h(2.67) = 203.36 (maximum)
- h(5.67) = 186.96
- h(8) = 70
The least value is **70**, and the greatest value is **203.36.** ⇒ x = 2 ( 6 ) 56 ± ( − 56 ) 2 − 4 ( 6 ) ( 105 ) ⇒ x = 12 56 ± 3136 − 2520 ⇒ x = 12 56 ± 616 ⇒ x = 12 56 ± 24.8 ⇒ x ≈ 5.67 , 2.67. E v a l u a t e h ( x ) a tt h ese p o in t s an d e n d p o in t s 0 an d 8 t o f in d t h e l e a s t an d g re a t es t . A f t erc a l c u l a t in g : − h ( 0 ) = 70 − h ( 2.67 ) = 203.36 ( ma x im u m ) − h ( 5.67 ) = 186.96 − h ( 8 ) = 70 T h e l e a s t v a l u e i s ∗ ∗ 70 ∗ ∗ , an d t h e g re a t es t v a l u e i s ∗ ∗ 203.36. ∗ ∗
How long after the start of the session is Hannah's heart-rate decreasing most quickly, within the first 8 minutes? Only available for registered users.
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To find when the heart-rate is decreasing most quickly, we need to find where h''(x) is 0, as this indicates a local maximum or minimum.
We find h''(x):
h ′ ′ ( x ) = 12 x − 56 h''(x) = 12x - 56 h ′′ ( x ) = 12 x − 56
12 x − 56 = 0 ⇒ x = 56 12 = 4.67 12x - 56 = 0 \Rightarrow x = \frac{56}{12} = 4.67 12 x − 56 = 0 ⇒ x = 12 56 = 4.67
Calculating the time in minutes and seconds, it is approximately 4 minutes and 40 seconds.
Use this information to write b'(x) in terms of h'(x), where 0 ≤ x ≤ 8, x ∈ ℝ. Only available for registered users.
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Since Bruno's heart-rate is always 15 BPM more than Hannah's:
b ( x ) = h ( x ) + 15 b(x) = h(x) + 15 b ( x ) = h ( x ) + 15 b ′ ( x ) = h ′ ( x ) . b'(x) = h'(x). b ′ ( x ) = h ′ ( x ) .
Use this information to write k'(x) in terms of h'(x), where 0 ≤ x ≤ 8, x ∈ ℝ. Only available for registered users.
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Karen's heart-rate is always 10% less than Hannah's:
k ( x ) = 0.9 h ( x ) k(x) = 0.9h(x) k ( x ) = 0.9 h ( x ) k ′ ( x ) = 0.9 h ′ ( x ) . k'(x) = 0.9h'(x). k ′ ( x ) = 0.9 h ′ ( x ) .
Use h(x) to write m(x) in the form m(x) = ax^3 + bx^2 + cx + d where a, b, c, and d ∈ ℝ, for 0 ≤ x ≤ 10. Only available for registered users.
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We know:
= 1.2(2x^3 - 28x^2 + 105x + 70) \
= 2.4x^3 - 33.6x^2 + 126x + 84$$
Thus, the coefficients are:
a = 2.4, b = -33.6, c = 126, d = 84. Join the Leaving Cert students using SimpleStudy...97% of StudentsReport Improved Results
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