MHB What is the Best Landing Point for the Messenger on a 5km Beach Journey?

[leprofece]

A Ship anchored to 6 Km from a beach rectilinue, (infront of a town, located on the beach. A Messenger, in the? in?
in the manor as long as possible, from the ship until a position B, located on the beach. The Messenger leaves the ship in a Rowing boat and can develop a speed 3 km/h, and is known that on the beach can walk at a rate of 5 kph. find out where should get land if you know that ithe distance AB 5 Km?.
answer 1,5 km before get after doing similar problem I must do a plot ok and i get t = t1 + t2
then frac{x}{3}+\frac{y}{5}
Bc = 5 -\sqrt{x^2-6 }
I got x/3-\frac{x}{\5sqrt{x^2-6}}
derivating \frac{1}{3}-5sqrt{x^2-6}
But I don't get the answer

 

[Prove It]

A Ship anchored to 6 Km from a beach rectilinue, (infront of a town, located on the beach. A Messenger, in the? in?
in the manor as long as possible, from the ship until a position B, located on the beach. The Messenger leaves the ship in a Rowing boat and can develop a speed 3 km/h, and is known that on the beach can walk at a rate of 5 kph. find out where should get land if you know that ithe distance AB 5 Km?.
answer 1,5 km before get after doing similar problem I must do a plot ok and i get t = t1 + t2
then frac{x}{3}+\frac{y}{5}
Bc = 5 -\sqrt{x^2-6 }
I got x/3-\frac{x}{\5sqrt{x^2-6}}
derivating \frac{1}{3}-5sqrt{x^2-6}
But I don't get the answer

Your setup has a lot of mistakes. In your drawing you should have a vertical line that is 6 units long and a horizontal line that is 5 units long. Somewhere on the horizontal line you choose a point to draw another line to the top of the vertical line. This distance away from B we will call x, so the remaining part of the horizontal line is ( 5 - x ).

Notice that you now have a right angle triangle, so the hypotenuse can be found using Pythagoras as $\displaystyle \begin{align*} \sqrt{ 6^2 + (5 - x)^2 } = \sqrt{36 + 25 - 10x + x^2} = \sqrt{x^2 - 10x + 61} \end{align*}$.

The distance traveled on land is x, if done at 5 km/h, it takes x/5 hours.

The distance traveled on water is $\displaystyle \begin{align*} \sqrt{x^2 - 10x + 61} \end{align*}$, if done at 3 km/h, it takes $\displaystyle \begin{align*} \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$ hours.

So the total time is

$\displaystyle \begin{align*} t = \frac{x}{3} + \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$

This is what you need to differentiate, set equal to 0, etc to find the x value to get the minimum time.

 

[MarkFL]

You may find this thread to be of interest:

http://mathhelpboards.com/math-notes-49/using-snells-law-determine-fastest-escape-route-4178.html

 

[leprofece]

Your setup has a lot of mistakes. In your drawing you should have a vertical line that is 6 units long and a horizontal line that is 5 units long. Somewhere on the horizontal line you choose a point to draw another line to the top of the vertical line. This distance away from B we will call x, so the remaining part of the horizontal line is ( 5 - x ).

Notice that you now have a right angle triangle, so the hypotenuse can be found using Pythagoras as $\displaystyle \begin{align*} \sqrt{ 6^2 + (5 - x)^2 } = \sqrt{36 + 25 - 10x + x^2} = \sqrt{x^2 - 10x + 61} \end{align*}$.

The distance traveled on land is x, if done at 5 km/h, it takes x/5 hours.

The distance traveled on water is $\displaystyle \begin{align*} \sqrt{x^2 - 10x + 61} \end{align*}$, if done at 3 km/h, it takes $\displaystyle \begin{align*} \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$ hours.

So the total time is

$\displaystyle \begin{align*} t = \frac{x}{3} + \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$

This is what you need to differentiate, set equal to 0, etc to find the x value to get the minimum time.

I think something is wrong
why do you write the same denominator?
when I derive this I got \frac{1}{3}+\frac{x-5}{sqrt{x^2-10x+61}
}
When I equated to 0 this I got (x-5)^2 = x^2-10x+61
Ichecked it in wolpfrang alpha and we got a complex root

And my secon doubt is why don't you write 3 and 5 in denominator?? I mean

$\displaystyle \begin{align*} t = \frac{x}{3} + \frac{\sqrt{x^2 - 10x + 61}}{5} \end{align*}$

Deriving this that I got x = 16,6 after equating to 0

Deriving if $\displaystyle \begin{align*} t = \frac{x}{5} + \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$
I got 18,6,3
if I introduce this in $\displaystyle \begin{align*} t = \frac{x}{5} + \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$
I got x= 7,05
And I don't get 400/3 yhat is the right answer
By the method that sugested markflo it isnot allowed in my country in this level

 

[MarkFL]

Let's see what we get if we apply the first method to which I linked, which I derived using only calculus and trigonometry, which must be allowed here. :D

The distance (in km) we want is:

$$x=5-6\cot\left(\cos^{-1}\left(\frac{3}{5}\right)\right)=5-6\cdot\frac{3}{4}=\frac{1}{2}$$

You cited an answer of:

$$\frac{3}{2}$$

But, if you take the time function Prove It gave (after correcting for the minor typo), which is:

$$t=\frac{x}{5}+\frac{\sqrt{x^2 - 10x + 61}}{3}$$

differentiate with respect to $x$, equate to zero, and discard the solution outside the desired range, you find:

$$x=\frac{1}{2}$$

You say you got 7.05 (which is not in the desired range) but you don't show what you did. I suggest differentiating again, and if you do not get the correct answer, post your work so we can find where you are going wrong. :D

 

[Prove It]

I think something is wrong
why do you write the same denominator?
when I derive this I got \frac{1}{3}+\frac{x-5}{sqrt{x^2-10x+61}
}
When I equated to 0 this I got (x-5)^2 = x^2-10x+61
Ichecked it in wolpfrang alpha and we got a complex root

And my secon doubt is why don't you write 3 and 5 in denominator?? I mean

$\displaystyle \begin{align*} t = \frac{x}{3} + \frac{\sqrt{x^2 - 10x + 61}}{5} \end{align*}$

Deriving this that I got x = 16,6 after equating to 0

Deriving if $\displaystyle \begin{align*} t = \frac{x}{5} + \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$
I got 18,6,3
if I introduce this in $\displaystyle \begin{align*} t = \frac{x}{5} + \frac{\sqrt{x^2 - 10x + 61}}{3} \end{align*}$
I got x= 7,05
And I don't get 400/3 yhat is the right answer
By the method that sugested markflo it isnot allowed in my country in this level

Yes that was just a typo, the first term is x/5.

 

[leprofece]

Let's see what we get if we apply the first method to which I linked, which I derived using only calculus and trigonometry, which must be allowed here. :D

The distance (in km) we want is:

$$x=5-6\cot\left(\cos^{-1}\left(\frac{3}{5}\right)\right)=5-6\cdot\frac{3}{4}=\frac{1}{2}$$

You cited an answer of:

$$\frac{3}{2}$$

But, if you take the time function Prove It gave (after correcting for the minor typo), which is:

$$t=\frac{x}{5}+\frac{\sqrt{x^2 - 10x + 61}}{3}$$

differentiate with respect to $x$, equate to zero, and discard the solution outside the desired range, you find:

$$x=\frac{1}{2}$$

You say you got 7.05 (which is not in the desired range) but you don't show what you did. I suggest differentiating again, and if you do not get the correct answer, post your work so we can find where you are going wrong. :D

greetings markflo how are you ?'
Right I got 1/2 and then may I substitute in $$t=\frac{x}{5}+\frac{\sqrt{x^2 - 10x + 61}}{3}$$ to get the right answer ??
Ok I tried and I didnt got 400/3 my total time was 38/5
so where must I substitute or introduce 1/2?

 

[MarkFL]

Hello! I am fine, how about you?

Doesn't that supposedly correct time of $$\frac{400}{3}\text{ hr}$$ seem outrageously high to you? Even if the entire journey was across water it would only take slightly longer than the minimum time possible, which is 2.6 hours. And it would take 3 hours if the person in the boat headed straight for the shore. How could it possible take 133 hours 20 minutes?

I get:

$$t_{\min}=t\left(\frac{1}{2}\right)=\frac{13}{5}\text{ hr}$$

By the way, in your working you should demonstrate that the critical value is at a local minimum. :D

 

[leprofece]

Hello! I am fine, how about you?

Doesn't that supposedly correct time of $$\frac{400}{3}\text{ hr}$$ seem outrageously high to you? Even if the entire journey was across water it would only take slightly longer than the minimum time possible, which is 2.6 hours. And it would take 3 hours if the person in the boat headed straight for the shore. How could it possible take 133 hours 20 minutes?

I get:

$$t_{\min}=t\left(\frac{1}{2}\right)=\frac{13}{5}\text{ hr}$$

By the way, in your working you should demonstrate that the critical value is at a local minimum. :D

I am not asked the tme but find out where should get land if you know that ithe distance AB 5 Km?.
answer 1,5 km before getting or reaching B
I get confused and it is not 400/3
in fact the answer is in kilometers no
in hours

 

[MarkFL]

Haven't we already established that the point at which we should land is 1/2 km from B?

 

[leprofece]

Haven't we already established that the point at which we should land is 1/2 km from B?

0.5 km y no 1,5 km

If I introduce 0.5 in {\frac{ \sqrt(x^2-10x+61) }{5}}
iGet the 1,5 Km
and also 6- 4,5 = 1,5
what do you think??