1. ## Boat problem

A straight river is 20m wide. The velocity of the river at (x,y) is v=[-(3x(20-x))/100]j m/min, 0<=x<=20. A boat leaves the shore at (0,0) and travels through the water with a constant velocity. It arrives at the opposite shore at (20,0). The speed of the boat is always radical 20 m/min.

Find the velocity of the boat.
Find the location of the boat at time t.

2. ## Velocity vectors

Hello antman
Originally Posted by antman
A straight river is 20m wide. The velocity of the river at (x,y) is v=[-(3x(20-x))/100]j m/min, 0<=x<=20. A boat leaves the shore at (0,0) and travels through the water with a constant velocity. It arrives at the opposite shore at (20,0). The speed of the boat is always radical 20 m/min.

Find the velocity of the boat.
Find the location of the boat at time t.

Suppose that the velocity of the boat relative to the water is $p\vec{i} + q\vec{j}$, where $p$ and $q$ are constants. Then $p^2 + q^2 = 20$, since the speed of the boat (through the water, I assume that's what the question means) is a constant $\sqrt{20}$.

Then the velocity of the boat relative to the bank = velocity of boat relative to water + velocity of water relative to bank

$= p\vec{i} + \left(q - \frac{3x(20-x)}{100}\right) \vec{j}$

$= \frac{d\vec{r}}{dt}$, where $\vec{r}$ is the position vector of the boat at time $t$

$= \frac{dx}{dt}\vec{i}+ \frac{dy}{dt}\vec{j}$

$\Rightarrow \frac{dx}{dt}= p$ and $\frac{dy}{dt}= q - \frac{3x(20-x)}{100}$

$\Rightarrow x = pt$, since $x = 0$ when $t = 0$

$\Rightarrow$ time taken to cross $= \frac{20}{p}$

and $\frac{dy}{dt}= q - \frac{3pt(20-pt)}{100}$

Solve for y by integrating. Then use $y = 0$ when $t = 0$ and $t = \frac{20}{p}$.

This gives $q = 2$, and (from $p^2 + q^2 = 20$) $p = 4$.

So the velocity of the boat is $4\vec{i}+2\vec{j}$, and $\vec{r} = x\vec{i} +y\vec{j} = ...$

Can you fill in the gaps?

3. I multiplied dy/dt out to = q-[(60pt-3pt^2)/100]. Then I tried to integrate and got [(q^2)/2]-[((30pt^2)-(pt^3))/100] but wasn't able to get q=2 so I know I'm doing something wrong.

Would the position of the boat just be the position vector 4ti+2tj?

4. ## Velocity vectors

Hello antman
Originally Posted by antman
I multiplied dy/dt out to = q-[(60pt-3pt^2)/100]. Then I tried to integrate and got [(q^2)/2]-[((30pt^2)-(pt^3))/100]...
You're integrating with respect to $t$, so (when simplified) it's

$y = qt - \frac{3}{10}pt^2 + \frac{1}{100}p^2t^3 + c$

(and again, $y = 0$ when $t = 0$, so $c=0$)

Now plug in $y = 0$ when $t = \frac{20}{p}$ to find $q$.

Originally Posted by antman
Would the position of the boat just be the position vector 4ti+2tj?
No. $4\vec{i}+ 2\vec{j}$ is the velocity of the boat relative to the water. Use the expressions for $x$ and $y$ in the equation

$\vec{r} = x\vec{i} + y\vec{j}$

5. Im sorry, im still not sure how you were able to get q=2 with both q and p being unknown variables. As you can tell, I am very confused. Thank you for all of your help.

6. Nevermind! I got that part. So the position would be 4txi+2tyj? or just 4tx+2ty?

7. ## Velocity vectors

Hello antman
Originally Posted by antman
Nevermind! I got that part. So the position would be 4txi+2tyj? or just 4tx+2ty?
The velocity $p\vec{i}+q\vec{j}$, which we now know is $4\vec{i} + 2\vec{j}$, is the velocity of the boat relative to the water. (See the start of my original post.) In other words, if we are in a boat ourselves that is not moving through the water - so to us, the water appears to be still - the first boat's velocity appears to be $4\vec{i} + 2\vec{j}$. Since the units here are m/min, that means that the first boat is moving at $4$ m/min faster across the river than we are, and $2$ m/min faster upstream (since $\vec{i}$ and $\vec{j}$ are unit vectors across and upstream, respectively).

But the water is, of course, moving relative to the bank of the river, and will carry the boat (and us, in our imaginary boat) with it. This is why we need the vector velocity equation

• velocity of boat relative to bank = velocity of boat relative to water + velocity of water relative to bank

to find the boat's movement relative to the bank. Now $x$ and $y$ are measured relative to the bank, and we have found that

$x = pt$ and $y = qt -\frac{3}{10}pt^2+\frac{1}{100}p^2t^3$

So these equations, with $p=4$ and $q =2$, give the location of the boat at time $t$. If you like, you can express this as a single vector equation using $\vec{r} = x\vec{i} + y\vec{j}$, or

$\vec{r} = 4t\vec{i} + \left(2t -1.2t^2+0.16t^3\right)\vec{j}$