sailing sailing in the deep blue sea

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maeveoneill

Junior Member
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Sep 24, 2005
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a sailboat travelling northeast at 8 km/h tacks east 5 km, then north 5 km. when it is northeast of its starting point, how fast is its distance from that point increasing?

i tried solving this on my own and got:


z² = x² + y²
z² = 5² + 5²
z² = 50
z = square root of 50

z² = x² + y²
2z dz/dt = 2x dx/dt + 2y dy/dt *2s cancel out*
square root of 500 dz/dt = 10(20) + (20)(20)
dz/dt = 600/ square root of 500
= 6/ square root of 5 km/h

this is wrong. can you tell me where i went wrong. the correct answer is 4/square root of 2 km/h[/i]
 
I don't understand the question. Is it 8 kmh NE or 5 E + 5 N? Quite a bit different.

When it is NE of the starting point, which way is it going? Are we finishing a N leg or startign an E leg?
 
it goes 5 km east and then 5 km north. as a result it is travelling northeast. its speed while travelling 5km east and 5km north is 8km/h.


this should create a triagonal. we are trying to find the rate of change of the distance between teh starting point and end point which is after it has travelled both east and north for a total of 10 km.

help?
 
Now it makes even less sense. \(\displaystyle \sqrt{5^{2}+5^{2}}\,=\,7.07\). It's just not that close to 8.
 
Hello, maeveoneill!

Are you sure of the wording of this peoblem.
. . It's rather silly . . . I'll reword it to make my point.

A sailboat travelling northeast at 8 km/hr.
When it is northeast of its starting point, . . (Isn't it always northeast of its starting point?)
how fast is its distance from that point increasing? . . (um, a wild guess ... maybe 8 km/h?)
Click to expand...
Okay, enough sarcasm . . .

I must assume that the sailboat is moving at 8 km/hr in any direction.

You already have the equation: .\(\displaystyle z^2\:=\:x^2\,+\,y^2\)

Then: .\(\displaystyle \L2z\left(\frac{dz}{dt}\right)\:=\:2x\left(\frac{dx}{dt}\right)\,+\,2y\left(\frac{dy}{dt}\right)\)

And: .\(\displaystyle \L\frac{dz}{dt}\:=\:\frac{1}{z}\left[x\left(\frac{dx}{dt}\right)\,+\,y\left(\frac{dy}{dt}\right)\right]\)


At that instant: .\(\displaystyle x=5,\,y=5,\;\frac{dx}{dt}= 8,\,\frac{dy}{dt}=8\) . . . and: \(\displaystyle z\:=\:\sqrt{5^2\,+\,5^2}\:=\:5\sqrt{2}\)


Therefore: .\(\displaystyle \L\frac{dz}{dt}\:=\:\frac{1}{5\sqrt{2}}\left[5(8)\,+\,5(8)\\right]\:=\:\frac{80}{5\sqrt{2}}\:=\:\frac{16}{\sqrt{2}}\:=\:8\sqrt{2}\) km/hr.
 
I have problems reading the new stuff but I think you are looking at 1/z not 1/2
-----------------
Gene
 
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