# Math Help - [SOLVED] Two boats leave a dock at the same time

1. ## [SOLVED] Two boats leave a dock at the same time

Am I working this problem correctly?
Two boats leave a dock at the same time. One boat is headed directly east at a constant speed of 35 knots(nautical miles per hour) and the other is headed directly south at a constand speed of 22 knots. Express the distance d between the boats as a funtion of the time t.

(35,0)(0,22)
f(t)=sqrt(35^2 + 22^2)
=sqrt(1709)
=41.34

Is that right?

2. Originally Posted by ninobrn99
Am I working this problem correctly?
Two boats leave a dock at the same time. One boat is headed directly east at a constant speed of 35 knots(nautical miles per hour) and the other is headed directly south at a constand speed of 22 knots. Express the distance d between the boats as a funtion of the time t.

(35,0)(0,22)
f(t)=sqrt(35^2 + 22^2) There isn't any t at the RHS
=sqrt(1709)
=41.34

Is that right?
Nearly! You've calculated the distance after 1 hour traveling time.

From $speed=\dfrac{distance}{elapsed\ time}$ you'll get:

$distance = speed \cdot (elapsed\ time)$

Let t denote the elapsed time in hours then the legs of the right triangle are:

$d_E=35 \cdot t$ ....... $d_S=22 \cdot t$

and the distance between the two ships is:

$f(t)=\sqrt{(35t)^2+(22t)^2}=t\cdot \sqrt{1709}$

3. so is t*sqrt(1709) what you get after solving the function or is the function supposed to be two parts?

4. Originally Posted by ninobrn99
so is t*sqrt(1709) what you get after solving the function or is the function supposed to be two parts?

I only simplified the first rough calculations:

\begin{aligned}
f(t)&=\sqrt{(35t)^2+(22t)^2}\\&=t\cdot \sqrt{1709}\end{aligned}

5. so then this isnt the base function to find all other values of t?
Sorry, this stuff is really confusing me!

6. Originally Posted by ninobrn99
so then this isnt the base function to find all other values of t?
Sorry, this stuff is really confusing me!
No, this function gives you the distance between the two ships with respect to the elapsed time. So after

1 hour ---> d = 41.34 nmi

2 hours ---> d = 82.68 nmi

3 hours ---> d = 124.02 nmi

and so on.

If you want to calculate t you have to solve the equation of the function for t.

7. Originally Posted by earboth
No, this function gives you the distance between the two ships with respect to the elapsed time. So after

1 hour ---> d = 41.34 nmi

2 hours ---> d = 82.68 nmi

3 hours ---> d = 124.02 nmi

and so on.

If you want to calculate t you have to solve the equation of the function for t.
okay, I understand having to find f(t) by plugging in numbers, but I thought the question just wanted the base function.
Originally Posted by ninobr99
Express the distance d between the boats as a funtion of the time t.

8. Originally Posted by ninobrn99
so then this isnt the base function to find all other values of t?
Sorry, this stuff is really confusing me!
No, this is the base function (by the way: What is a base function and what is the difference between a function and a base function?) to find values of f(t) by plugging in values for t

9. by base function, I mean the function of the problem. What you translate from word to math.

10. Originally Posted by ninobrn99
by base function, I mean the function of the problem. What you translate from word to math.
The base function is:

$f(t)=\sqrt{1709} \cdot t$

Since $\sqrt{1709}$ is a constant the graph of f is a straight line passing through the origin.

11. so then, that is the answer to the question?"Express the distance d between the boats as a funtion of the time t."?

Thank you again for being patient!

12. Originally Posted by ninobrn99
so then, that is the answer to the question?"Express the distance d between the boats as a funtion of the time t."?

Thank you again for being patient!
It's the expression you've been talking about over and over. You think of the two boats as moving along the legs of a right triangle and the distance between them is the hypotenuse of that triangle. earboth calculated the general length of this hypotenuse, based of time which is what you want. If you don't get the answer then you don't fully understand the question I think, which is fine. Just ask us to clarify something.

13. Originally Posted by ninobrn99
so then, that is the answer to the question?"Express the distance d between the boats as a funtion of the time t."?
You've got it!

14. Originally Posted by Jameson
It's the expression you've been talking about over and over. You think of the two boats as moving along the legs of a right triangle and the distance between them is the hypotenuse of that triangle. earboth calculated the general length of this hypotenuse, based of time which is what you want. If you don't get the answer then you don't fully understand the question I think, which is fine. Just ask us to clarify something.
I understand that part. What was confusing me was the pythagorean theorem and the distance formula. That and yes, I do not fully understand it. It's the only question that has been irking me for the past 2 days! I could just as easily miss the question, but I'm not that type of person. I'd like to understand it since I still have about 6 weeks left of class.

15. Originally Posted by ninobrn99
I understand that part. What was confusing me was the pythagorean theorem and the distance formula. That and yes, I do not fully understand it. It's the only question that has been irking me for the past 2 days! I could just as easily miss the question, but I'm not that type of person. I'd like to understand it since I still have about 6 weeks left of class.
Well I'm glad you're the type to make sure you get them all! You'll fit in here well. You understood the basic idea of how to use the right triangle formed by the boats to solve for the distance, you just forgot to show how the triangle will change with time, which is what earboth added. I think you see from earboth's post the reasoning. The distance formula is really just the pythagorean theorem in disguise, which makes it one less thing to memorize if you think of it that way.