Logarithms and Exponentials

Question 1

Question 1: Power output (6 marks)
Power output is an important metric that cyclists use to determine their endurance during the Tour de France. To better understand how a cyclist’s power output decreases over time, the following function was derived:

𝑃(𝑡)=P₀ 𝑒 ^{−𝑘𝑡</sub>}

Where:
P(t) is the cyclist’s power output in watts at a given time in hours.

𝑃₀ is a constant that represents the cyclist’s initial power output in watts.

𝑘 is a positive constant that represents the power decay rate.

𝑡 is time measured in hours from the start of the course.

The first stage of the Tour de France has cyclists completing a course that starts at 4:00 AM. Two checkpoints during this stage recorded the power output of the cyclist at 8:00 AM and 1:00 PM. While passing the first checkpoint, the power output was recorded as 324.47 W at 8:00 AM. At the second checkpoint, the power output was recorded as 248.22 W at 1:00 PM.

(a) Find the rule of a function that models the decay in power output of the cyclist with respect to time,
𝑃(𝑡). Round your constants to two decimal places.

Answer 1

P(4) = 324.47
P(9) = 248.22
solve on CAS for Po and k</sub>

P^{o</sub> = 402.02
and k = 0.05}

P(t) = 402.02e_0.05t

Question 2

Hence, how long did it take from the start of the race for the cyclist to reach a power output of 212 W? Express your answer in hours and minutes. Round to the nearest minute.

Answer 2

Let P(t) = 212, solve for t on cas.
t= 12.7983
answer = 12 hours and 48 minutes

Question 3

Another two cyclists were also monitored for their power output as they completed the same first stage of the Tour de France. Recordings were taken and used to create two functions that model the power output of Cyclist 2 and Cyclist 3:

𝑃₂(𝑡) = 407.70𝑒²(−1/24𝑡)

𝑃₃(𝑡) = 407.70𝑒²(−1/36𝑡)
c. Determine the ratio of the power decay rate (𝑘) of Cyclist 2 to Cyclist 3.
Express your answer in the form a/b, where a and b are positive integers.

Answer 3

cyclist 2/cyclist 3 = 407.70𝑒²(−1/24𝑡)/407.70𝑒²(−1/36𝑡)
=-1/24 / -1/36 = 3/2

OR
cyclist 2/cyclist 3 =
2/3

Question 4

At what time do both Cyclist 2 and Cyclist 3 share the same power output?

Answer 4

t = 0, 4 am

Question 5

Often, the greatest challenge faced by cyclists is the change in altitude. Race teams often compile data regarding changes in altitude to build difficulty models so that measures can be put in place to aid cyclists and ensure success.

For this Tour de France, two stages are being compared for their difficulty with respect to altitude and will be referred to as Stage 1 and Stage 2.

Race teams will be able to choose between the two stages that are best for their cyclists.

Both Stage 1 and Stage 2 have cyclists start at their respective initial altitudes and finish the stage at their respective peak altitudes.

Stage 1 has a peak altitude of 3450 meters, and Stage 2 has a peak altitude of 3980 meters.

The following generalized formula models the difficulty of Stage 1 at any given altitude:
D₁(h)=D_bln(1+h/h₀)
D₁ (h) is the difficulty of stage one at any given altitude (in meters).
ℎ is the altitude in meters above the lowest point in the stage.
D_b is the baseline difficulty of the climb.
ℎ₀ is the initial altitude of Stage 1.
Stage 1 was found to have a baseline difficulty of e³ points and an initial altitude of 1000 meters.
a. Use the difficulty formula above to find a rule and domain for
𝐷₁(ℎ) which determines the difficulty of Stage 1 with respect to altitude.

Answer 5

D₁h=e¹3 log_e (1+h/1000) where h is an element [1000, 3450]

Question 6

Use the difficulty formula above to find the rule for
𝐷₁h which determines the altitude of Stage 1 with respect to the difficulty of Stage 1.

Answer 6

D₁= e³log_e(1+h/1000)

D₁/ e³ = log_e(1+h/1000)

e(^{D₁/ e3})=1+h/1000

e(^{D₁/ e3})-1=h/1000

hD₁=1000(e(^{D₁/ e3})-1)

Question 7

For Stage 2, a modified formula is used to represent difficulty with respect to altitude:
D₂(h)=log₁₀(Σ)+168/5log₁₀(1+h/850)
Where: Σ represents a positive constant for the difficulty required for any given stage in the Tour de France.

Owing to the surface being of lesser quality for Stage 2 compared to Stage 1, the difficulty constant applied to Stage 2 is
Σ=10^5

Answer 7

Since Σ = 10^5
D₂(h)=log₁₀(10^5)+168/5log₁₀(1+h/850)

=5+168/5log₁₀(1+h/850)

Question 8

State the sequence of transformations that could transform
D₁(h) to the function D₂(h). State any transformation values to two decimal places.

Answer 8

D₁: y = e³log_e(1 + x/1000)
D₂: y’ = 5 + (168/5)log₁₀(1 + x’/850)

y = e³log_e(1 + x/1000)
e³log_e(1 + x/1000) = 5(y’ - 5)

5(y’ - 5)/168 = log₁₀(1 + x’/850)

(log_e(10)/168)(5y’ - 25) = log_e(1 + x’/850)

y/e³ = (log_e(10)/168)(5y’ - 25)

1 + x/1000 = 1 + x’/850

x/1000 = x’/850

x’ = (17x)/20

x’ = 0.85x

y/e³ = (log_e(10)/168)(5y’ - 25)

168y/5log_e(10) = y’ - 5

y’ = (168y)/(5log_e(10)) + 5

y’ = 0.73y + 5

(h, D₁) → (0.85h, 0.7D₂p + 5)

Dilation of a factor 0.85 from the D axis

Dilation of a factor 0.73 from the h axis

Translation of 5 units up

Question 9

Determine a realistic domain for D₂h

Answer 9

h element [850, 3980] CAS

Question 10

Question 4: Altitude versus time (11 marks)**

For any new stage to be approved, a mathematical model of the altitude changes with respect to time (in hours) is needed. A sigmoid function was chosen to best represent this scenario. The basic form of a sigmoid function is as follows:

h(t) = h_m / (1 + e^{-a(t - b)}) + c

Where:

• h(t) is the altitude of a cyclist in meters above sea level.
• t is time measured in hours.
• h_m is the difference in altitude from the lowest point to the highest point.
• a is a constant that represents the change in the gradient of the new stage.
• b is a constant that represents at what time in hours the new stage has the steepest gradient.
• c is a constant that represents the lowest point of the new stage in altitude.

An example of a sigmoid function is graphed below with a h_m value of 4. Note that asymptotes exist at h = 1 and h = 5. Constants have been left as the following: a = 1, b = 0, and c = 1.

For this example, at t = 0, the gradient is the steepest for this sigmoid function. The sigmoid function that models the altitude of the new stage will need to have the following characteristics:

The lowest point of the new stage has an altitude of 1000 meters.

The highest point of the new stage has an altitude of 2384 meters.

The new stage is steepest gradient at t = 34 hours.

At t = 30 hours, the altitude of the new stage is
1384 / (e^2√2 + 1) + 1000 meters.

a. Show that the rule for the sigmoid function modelling the new stage is:

h(t) = 1384 / (1 + e^-√2/2(t - 34)) + 1000

Answer 10

h_m = 2384 - 1000 = 1384 (M)

Since steepest gradient occurs at t = 34, b = 34. (M)

Since lowest point is 1000m, c = 1000. (M)

h(t) = 1384 / (1 + e^{-a(t - 34)}) + 1000

Let h(30) = 1384 / (e^a√2 + 1), solve for a

∴ a = √2/2

Question 11

b. Sketch the sigmoid function, h(t), based on the above characteristics. Clearly label any asymptotes, point of inflection and the coordinates of one other point. (4 marks)

Answer 11

asymptotes
(inflection point) (M)
(coordinate) (M)
CAS

Question 12

Identify a sequence of transformations from the basic rule determined in part a. (3 marks)

Answer 12

y = 4 / (1 + e^x) + 1
y - 1 = 4 / (1 + e^x)
(y - 1) / 4 = 1 / (1 + e^x)
y’ = 1384 / (1 + e^{(√2 / 2) (x - 34)}) + 1000
y’ - 1000 = 1384 / (1 + e^{(√2 / 2) (x - 34)})
(y’ - 1000) / 1384 = 1 / (1 + e^{(√2 / 2) (x - 34)})

Dilation by a factor 346 from the x-axis.

Dilation by a factor √2 from the y-axis.

Translation 34 units right and 654 units up.

Question 13

12MAM SAC1B 2023

<b>Question 5: Cyclist performance (5 marks)</b>

A cyclist’s performance is affected by the cyclist’s power output, and the difficulty of the stage due to changing altitude. To study this phenomenon a cyclist participating in the race on stage 13 was chosen at random.

The cyclist’s power output with respect to time in hours while completing stage 13 is modelled as follows:

P₁₃: [0,60] → ℝ, P₁₃(t) = 300e^(-t/50)

The difficulty of the race due to the altitude of stage 13 is modelled as follows:

D₁₃: [724,3230] → ℝ, D₁₃(h) = 20 log_e(1 + h/1000)

The change in altitude of the cyclist with respect to time in hours is modelled as follows:

h₁₃: [0,60] → ℝ, h₁₃(t) = 2000 / (1 + e^{(-1/20 (t - 10))})

A composite of these functions allows for a function that models a cyclist’s performance with respect to time:

C₁₃(t) = P₁₃(t) * D₁₃(h₁₃(t))

Where:
C₁₃(t) represents the cyclist’s performance (in points).
t represents time (in hours) from the start of stage 13.

a. Show that the composite function C₁₃(t) exists.

Answer 13

ran h₁₃ ⊆ dom D₁₃

[755.08,1848.28] ⊆ [724,3230]

∴ D₁₃(h₁₃(t)) exists

ran D₁₃ ∘ h₁₃ ⊆ dom P₁₃

[11.25, 20.93] ⊆ [0,60]

∴ C₁₃(t) exists.

Question 14

b. Define a function, C₁₃(t), that models the cyclist’s overall performance with respect to time during stage 13.

Answer 14

C₁₃: [0,60] → ℝ, C₁₃(t) =

300 (e^x/20 + e^1/2)^2/5
—————————————————
(3e^x/20 + e^1/2)^2/5

Question 15

If the cyclist has been racing for 40 hours, what is their performance level at the 40 hour mark?
Round your answer to the nearest whole number.

Answer 15

C₁₃(40)= 203.61
204 points