# Related Rates, Boats Moving Around

• QuarkCharmer
In summary, the problem involves finding the distance between two ships, A and B, at different times and determining how fast the distance is changing. Using the distance formula and implicit differentiation, the distance between the ships is found to be D(t)=5\sqrt{74t^2-420t+900}. Taking the derivative, D'(t)=\frac{5}{2}(74t^2-420t+900)^{\frac{-1}{2}}(150t-420). When t=4, D'(4)=21.3932995895, which is the rate at which the distance between the ships is changing at 4pm.
QuarkCharmer

## Homework Statement

Stewart Calculus 6E, 3.8 #14
At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h, and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4pm?

## Homework Equations

Distance Formula I think.

## The Attempt at a Solution

Again, I am having trouble phrasing the question. It seems to me, that I need to find an equation in terms of distance and then implicitly differentiate. If the coordinate points of the ships are A(0,0) and B(150,0) at time 12, I thought that I could solve for the distance like this.

$$D = \sqrt{\Delta X^2+\Delta Y^2}$$

Since, I am looking for D', I would differentiate that to be:
$$D' = 1/2(\Delta X^2+\Delta Y^2)^{\frac{-1}{2}}\frac{d}{dt}(\Delta X^2+\Delta Y^2)$$

$$D' = 1/2(\Delta X^2+\Delta Y^2)^{\frac{-1}{2}}(2 \Delta X \Delta X'+2 \Delta Y \Delta Y')$$

No, first you need to find the general position of the boats at time t.

Let's begin with boat A.
What is the position of the boat at time 12?
What is the position one hour after?
Two hourds after?
t hours after?

Boat A
What is the position of the boat at time 12?
(0,0)
What is the position one hour after?
(35,0)
Two hours after?
(70,0)
t hours after?
(35t,0)

Indeed, now do the same for boat B!

Boat B
What is the position of the boat at time 12?
(150,0)
What is the position one hour after?
(150,25)
Two hours after?
(150,50)
t hours after?
(150,25t)

Ok, so at t, boat A is at (35t,0) and boat B is at (150,25t). What is the distance between the boats?

micromass said:
Ok, so at t, boat A is at (35t,0) and boat B is at (150,25t). What is the distance between the boats?

$$D(t)=\sqrt{(35t-150)^{2}+(0-25t)^{2}}$$
$$D(t)=\sqrt{(35t-150)^{2}+(-25t)^{2}}$$
?

Correct!

Now, how do you find how fast the ship changes?

micromass said:
Correct!

Now, how do you find how fast the ship changes?

You mean the distance right?
Find D'(t)
I'm working that out now...

QuarkCharmer said:
You mean the distance right?
Find D'(t)
I'm working that out now...

Yes,sorry!

You might want to simplify D first...

$$D(t)=\sqrt{1225t^2-10500t+22500+625t^2}$$
$$D(t)=\sqrt{1850t^2-10500t+22500}$$
$$D(t)=\sqrt{50(37t^2-210t+450)}$$
$$D(t)=5\sqrt{2(37t^2-210t+450)}$$
$$D(t)=5\sqrt{74t^2-420t+900}$$

$$D'(t)=5\frac{d}{dt}(74t^2-420t+900)^{\frac{1}{2}}$$
$$D'(t)=5\frac{1}{2}(74t^2-420t+900)^{\frac{-1}{2}}\frac{d}{dt}(74t^2-420t+900)$$
$$D'(t)=\frac{5}{2}(74t^2-420t+900)^{\frac{-1}{2}}(150t-420)$$

Then for when t=4;
$$D'(4)=\frac{5}{2}(74(4)^2-420(4)+900)^{\frac{-1}{2}}(150(4)-420)$$
$$D'(4)=\frac{5}{2}(86836)^{\frac{-1}{2}}(180)$$
$$D'(4)=\frac{5}{2}(86836)^{\frac{-1}{2}}(180)$$
$$D'(4)=132605.769105$$

?? That can't be right! (It should be 21.3932995895, so says my calculator)

Last edited:
QuarkCharmer said:
$$D(t)=\sqrt{1225t^2-10500t+22500+625t^2}$$
$$D(t)=\sqrt{1850t^2-10500t+22500}$$
$$D(t)=\sqrt{50(37t^2-210t+450)}$$
$$D(t)=5\sqrt{2(37t^2-210t+450)}$$
$$D(t)=5\sqrt{74t^2-420t+900}$$

$$D'(t)=5\frac{d}{dt}(74t^2-420t+900)^{\frac{1}{2}}$$
$$D'(t)=5\frac{1}{2}(74t^2-420t+900)^{\frac{-1}{2}}\frac{d}{dt}(74t^2-420t+900)$$
$$D'(t)=\frac{5}{2}(74t^2-420t+900)^{\frac{-1}{2}}(150t-420)$$

In the last term I get 148t-420

Then for when t=4;
$$D'(4)=\frac{5}{2}(74(4)^2-420(4)+900)^{\frac{-1}{2}}(150(4)-420)$$
$$D'(4)=\frac{5}{2}(86836)^{\frac{-1}{2}}(180)$$

Wow, this step can't be correct! The 86836 is way to big! Can you calculate this again, I get something very different.

Haha, yeah, I figured it out. I was just unsure how to set up the problem. Thanks for the help.

## 1. How do you determine related rates in a scenario involving boats moving around?

In order to determine related rates in a scenario involving boats moving around, you will need to identify the variables involved and their rates of change. These variables can include the position, velocity, and acceleration of the boats. Once you have identified the variables, you can use the chain rule to find the rate of change for each variable in relation to the others.

## 2. What is the importance of using the chain rule in related rates problems?

The chain rule is essential in related rates problems because it allows us to find the rate of change of one variable with respect to another variable. In scenarios involving boats moving around, there are often multiple variables that are changing simultaneously, and the chain rule helps us to find the relationship between these variables and their rates of change.

## 3. Can related rates problems involving boats moving around be solved using basic algebra?

No, related rates problems involving boats moving around typically require the use of calculus to solve. This is because they involve finding the derivative of a function, which is a fundamental concept in calculus. However, basic algebra can be used to simplify the problem and make it easier to apply the chain rule.

## 4. What is the relationship between the position, velocity, and acceleration of a boat in a related rates problem?

The position of a boat is determined by its velocity, and the velocity of a boat is determined by its acceleration. In a related rates problem involving boats moving around, the position, velocity, and acceleration are all changing simultaneously, and the chain rule allows us to find the rate of change of each variable in relation to the others.

## 5. How can related rates problems involving boats moving around be applied in real-life situations?

Related rates problems involving boats moving around can be applied in many real-life situations, such as determining the speed and direction of a boat on a river or the rate at which two boats are approaching each other. These problems can also be used to calculate the optimal route for a boat to take in order to reach a destination in the shortest amount of time.

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