The depth of water, 𝑑, in metres, varies with the tides throughout the day and can be modelled by the equation 𝐷(𝑡) = 2.5𝑐𝑜𝑠(0.675(𝑡 − 3.5)) + 4.3, where 𝑡 is the number of hours past 12𝑎𝑚.

Question

The depth of water, 𝑑, in metres, varies with the tides throughout the day and can be modelled by the equation 𝐷(𝑡) = 2.5𝑐𝑜𝑠(0.675(𝑡 − 3.5)) + 4.3, where 𝑡 is the number of hours past 12𝑎𝑚.

Abithan claims that at exactly 10𝑎𝑚, the depth is decreasing by 0.5𝑚 per hour. Is Abithan’s claim, correct? Describe and explain the steps you would take to check Abithan’s claim. Use terminology and formulae learned in this unit. Provide calculations as needed.

Answer 1

t = 10 hours past midnight

dD/dt = -2.5 (0.675) sin (0.675(𝑡 − 3.5))

Answer 2

To check Abithan's claim that at exactly 10am, the depth is decreasing by 0.5m per hour, we can find the derivative of the depth function with respect to time. If the derivative is equal to -0.5 at t = 10, then Abithan's claim is correct.

Step 1: Find the derivative of the depth function with respect to time.

The depth function 𝐷(𝑡) = 2.5𝑐𝑜𝑠(0.675(𝑡 − 3.5)) + 4.3 can be simplified using the chain rule of differentiation:
d𝐷(𝑡)/dt = d(2.5𝑐𝑜𝑠(0.675(𝑡 − 3.5)) + 4.3)/dt

Applying the chain rule, we get:
d𝐷(𝑡)/dt = 2.5 *(-sin(0.675(𝑡 − 3.5))) * (0.675)

Simplifying further, we have:
d𝐷(𝑡)/dt = -1.688sin(0.675(𝑡 − 3.5))

Step 2: Evaluate the derivative at t = 10.

Substitute t = 10 into the derivative equation:
d𝐷(𝑡)/dt = -1.688sin(0.675(10 − 3.5))

Simplifying further, we have:
d𝐷(𝑡)/dt = -1.688sin(0.675(6.5))

Calculating sin(0.675(6.5)), we get approximately sin(4.3875) ≈ -0.999979

Substituting into the equation:
d𝐷(𝑡)/dt = -1.688 * (-0.999979)
≈ 1.688

Step 3: Compare the calculated derivative at t=10 with -0.5.

Since the calculated derivative is approximately 1.688 and not equal to -0.5, Abithan's claim that the depth is decreasing by 0.5m per hour at exactly 10am is not correct.

Answer 3

To check Abithan's claim, we need to find the rate of change of the depth of water at exactly 10 am and determine if it is indeed decreasing by 0.5 m per hour.

Step 1: Convert the time of 10 am to the number of hours past 12 am.
Since 10 am is 10 hours after 12 am, we have t = 10.

Step 2: Calculate the derivative of the depth function with respect to time (t).
The derivative will give us the rate of change of the depth of water. In this case, the derivative will be the rate of change in meters per hour.
Differentiating the equation 𝐷(𝑡) = 2.5𝑐𝑜𝑠(0.675(𝑡 − 3.5)) + 4.3 gives:

𝐷'(𝑡) = 2.5 × −sin(0.675(𝑡 − 3.5)) × 0.675

Step 3: Substitute the value of t = 10 into the derivative function.
𝐷'(10) = 2.5 × −sin(0.675(10 − 3.5)) × 0.675

Step 4: Calculate the numerical value of the derivative at t = 10.
Using a scientific calculator or software, we can evaluate the above expression to find 𝐷'(10).

Step 5: Compare the calculated derivative value with -0.5 m/hr.
If the calculated value of 𝐷'(10) is approximately -0.5 m/hr, then Abithan's claim is correct. Otherwise, the claim is false.

Note: The value of 𝐷'(10) may not be exactly -0.5 m/hr due to rounding errors, but if close enough, we can consider the claim to be correct.

By following these steps, we can check Abithan's claim and determine if the depth of water at exactly 10 am is decreasing by 0.5 m per hour.

Answer 4

2.5cos(0.675(t − 3.5)) + 4.3

D' = -1.6875sin(0.675(t − 3.5))
10am is 10 hours past midnight, so just find D'(10)

please do not blindly regurgitate specific instructions on how to proceed. I am not responsible for following your teacher's directions, and I certainly have no idea about the "terminology and formulae learned in this unit."