Calculus - optimization

Fliffi

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Given the point P(10,2) and the line y=2x-3, find the point Q on the line such that PQ is the shortest possible difference
 
This is a calculus problem? Maybe.

How might you approach it if you never had heard of the calculus?

With the calculus, can you write an expression for the distance from P to ANY point on the line?
 
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Pick an arbitrary point (x,y). What is the distance from (x,y) and point P? Is there a constraint on the point (x,y)? If yes, include that constraint into the distance from (x,y) to point P.

Please show us your work in finding the distance from (x,y) to point P.
 
well i couldnt figure out a way to do it with calculus but if u kniw that a perpenidcular line is 2
Pick an arbitrary point (x,y). What is the distance from (x,y) and point P? Is there a constraint on the point (x,y)? If yes, include that constraint into the distance from (x,y) to point P.

Please show us your work in finding the distance from (x,y) to point P.

Well i couldnt do it with calculus but

a perpindicular line is the gradient * the perpindicular gradient = -1
we want to find a perpindicular line from 10,2 to a point of y = 2x-3
the gradient is 2 so 2*x=-1 x= -0.5

form a line y-2 = -0.5(x-10)
y=-0.5x+7

2x+3=-0.5x+7
x=4

y(4)=2*4-3
=5

so the point that intersects at the shortest possible point is (4,5)
 
Okay, now how do you pass your calculus class?

Do you remember the Distance Formula? It should help.
 
In year 12 not really calculus xD

yuh so Speed * Time = Distance

so square root of (4-10)^2+(5-2)^2

comes out to be square root of 45
 
In year 12 not really calculus xD

yuh so Speed * Time = Distance

so square root of (4-10)^2+(5-2)^2

comes out to be square root of 45
That's nice, but are you sure this isn't in a textbook section called "Applications of the 1st Derivative"?
 
well i couldnt figure out a way to do it with calculus but if u kniw that a perpenidcular line is 2


Well i couldnt do it with calculus but

a perpindicular line is the gradient * the perpindicular gradient = -1
we want to find a perpindicular line from 10,2 to a point of y = 2x-3
the gradient is 2 so 2*x=-1 x= -0.5

form a line y-2 = -0.5(x-10)
y=-0.5x+7

2x+3=-0.5x+7
x=4

y(4)=2*4-3
=5

so the point that intersects at the shortest possible point is (4,5)
You certainly can find the distance between (x,y) and (10,2). It is \(\displaystyle d=\sqrt{(x-10)^2-(y-2)^2)}\). Now the constraint is y=2x-3 so just update that distance formula and get \(\displaystyle d=\sqrt{(x-10)^2-(2x-3-2)^2)}\).
Can you now finish from here?
 
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