#### To determine

**To find:** The velocity of the rise or fall of the tide for the timing 3:00am.

#### Answer

The velocity of the tide rises at 3 AM by 2.391 m/hour.

#### Explanation

**Given:**

The equation of time is given as below.

D(t)=7+5cos[0.503(t−6.75)] (1)

**Calculation:**

Calculate the velocity of rise or fall of the tide at 3 AM.

Differentiate equation (1) with respect to t.

D'(t)=5×(−sin)[0.503(t−6.75)](0.503)

D'(t)=−2.52sin(0.503t−3.4) (2)

Substitute 3 for t in equation (2).

D'(3)=−2.52×sin(0.503(3)−3.4)=−2.52×sin(1.509−3.4)=−2.52×sin(−1.891)

=−2.52×(−0.949)D'(3)=2.391 m/hour

The velocity of the tide rises at 3 AM by 2.391 m/hour.

#### To determine

**To find:** The velocity of the rise or fall of the tide for the timing 6:00am.

#### Answer

The velocity of the tide rises at 6 AM by 0.939 m/hour.

#### Explanation

Calculate the velocity of rise or fall of the tide at 6 AM.

Substitute 6 for t in equation (2).

D'(6)=−2.52×sin(0.503(6)−3.4)=−2.52×sin(3.018−3.4)=−2.52×sin(−0.382)

D'(6)=0.939 m/hour

The velocity of the tide rises at 6 AM by 0.939 m/hour.

#### To determine

**To find:** The velocity of the rise or fall of the tide for the timing 9:00am.

#### Answer

The velocity of the tide falls at 9 AM by 2.275 m/hour.

#### Explanation

Calculate the velocity of rise or fall of the tide at 9 AM.

Substitute 9 for t in equation (2).

D'(9)=−2.52×sin(0.503(9)−3.4)=−2.52×sin(3.018−3.4)=−2.52×sin(1.127)

=−2.52×0.903D'(9)=−2.275 m/hour

The velocity of the tide falls at 9 AM by 2.275 m/hour.

#### To determine

**To find:** The velocity of the rise or fall of the tide for the Noon.

#### Answer

The velocity of the tide falls at 12 AM by 1.22 m/hour.

#### Explanation

Calculate the velocity of rise or fall of the tide at 12 PM or Noon.

Substitute 12 for t in equation (2).

D'(12)=−2.52×sin(0.503(12)−3.4)=−2.52×sin(6.036−3.4)=−2.52×sin(2.636)

=−2.52×0.484D'(12)=−1.22 m/hour

The velocity of the tide falls at 12 AM by 1.22 m/hour.