Solving WWII Submarine Strike with Pythagoras and Differentiation

[rastakitty]

1. A WWII sub sailing north spots a Japanese carrier directly above it sailing due east at 10 knots. If the submarine at the time of sighting the carrier is 5 miles away, has torpedoes that travel 40 knots/hr, it takes 3 mins to load and fire torpedo.
• At what angle with respect to original position of sighting must the submarine fire torpedo to strike the carrier. ?
• What is the travel time and distance traveled of the torpedo?

Also for the same data above show that taking the derivative of a trigonometric function can be used to yield the same results.

I started using Pythagoras, but got stuck,, this seems to be challenging to most of us who battled for hours.. if you can help I would be grateful.. this is a practice question to help me prepare for an exam..
Thank you


2. pythagorus and differentiation



3. I converted the time east (3mins) to distance, used pythag to get hyp,but that is only the distance before shooting the torpedo. not sure if i set the prob up correctly. but it is a real challenge. thanks if you can help...

 

[Valkarie]

Answer:1. First, we need to calculate the angle of the submarine relative to the carrier. We can use the Pythagorean theorem to solve this. The right triangle formed between the submarine and the carrier has sides of length 5 miles and 10 knots for 3 minutes. Thus, the hypotenuse of the triangle will be the distance traveled by the torpedo in 3 minutes, which is 12.5 miles. If we divide this distance by 5 miles, we get the ratio of 2.5. This means that the angle θ formed between the two points is the inverse tangent of 2.5, which is 63.4 degrees. Therefore, the angle with respect to the original position of sighting at which the submarine must fire the torpedo is 63.4 degrees. The travel time and distance traveled of the torpedo is 3 minutes and 12.5 miles, respectively.2. Taking the derivative of a trigonometric function can also be used to calculate the angle of the submarine relative to the carrier. Taking the derivative of the inverse tangent (tan-1) of the ratio of the distances will yield the angle of the triangle formed. In this case, the ratio of the distances is 2.5, so taking the derivative of tan-1(2.5) will result in an angle of 63.4 degrees, which is the same angle calculated using the Pythagorean theorem. Therefore, both methods yield the same result of 63.4 degrees as the angle with respect to the original position of sighting at which the submarine must fire the torpedo.

 

[piercas]

Thank you for sharing your practice question with me. I am always excited to see how different mathematical concepts can be applied to real-life scenarios. The problem you have presented is indeed challenging, but I am happy to help you work through it.

Let's first tackle the first part of the problem. We have a submarine sailing north and a Japanese carrier sailing east. We know the distance between them is 5 miles and the carrier's speed is 10 knots, which means it travels 10 nautical miles in one hour. The submarine has torpedoes that travel at 40 knots/hr.

To determine the angle at which the submarine must fire the torpedo, we can use the Pythagorean theorem. Since the submarine is sailing north and the carrier is sailing east, we can draw a right triangle with the submarine's starting position as one of the vertices. The hypotenuse of this triangle will represent the distance traveled by the torpedo before hitting the carrier. Using the Pythagorean theorem, we can write the equation as:

(5)^2 + (10t)^2 = (40t)^2

Where t is the time it takes for the torpedo to reach the carrier. Solving for t, we get t = 0.1875 hours or 11.25 minutes.

Now, to determine the angle, we can use trigonometric functions. We know that the tangent of an angle is equal to the opposite side (5 miles) over the adjacent side (10t miles). So, we can write the equation as:

tanθ = 5/10t

Substituting the value of t, we get tanθ = 5/10(0.1875) = 0.2667. Taking the inverse tangent, we get θ = 14.477°. Therefore, the submarine must fire the torpedo at an angle of approximately 14.477° to hit the carrier.

Next, let's look at the travel time and distance of the torpedo. We already calculated the time it takes for the torpedo to reach the carrier, which is 11.25 minutes. To find the distance traveled, we can use the formula d = rt, where d is the distance, r is the rate or speed, and t is the time. Substituting the values, we get d = 40(0.1875) = 7.5 nautical miles.