Minimum distance between two curves (1 Viewer)

[king.rafa]

Hey guys

What is the minimum distance between y=3x^2 and y=2x-1. I have used the distance formula, but the problem is that I have no idea what to put for the x values. The y value in the distance formula is 3x^2-2x+1 but what do i put in the for the x values.

Cheers

 

[Timothy.Siu]

f(x)=3x^2-2x+1
f'(x)=6x-2
=0 when x=1/3

1/3 is a turning point, and since 3x^2-2x+1 has no zeros, it is either totally positive or negative, and hence when x=1/3 is the minimum distance.
y=1/3-2/3+1=2/3

distance=2/3

 

[cutemouse]

I'm sure that you use the perp. dist formula somehow... Will have a go later.

 

[Timothy.Siu]

I'm sure that you use the perp. dist formula somehow... Will have a go later.

lol

 

[Drongoski]

Hey guys

What is the minimum distance between y=3x^2 and y=2x-1. I have used the distance formula, but the problem is that I have no idea what to put for the x values. The y value in the distance formula is 3x^2-2x+1 but what do i put in the for the x values.

Cheers

Is the answer: 2/(3sqr(5)) ?? from point (2/3, 1) on the parabola to the line.


Edit

I misread quadratic as y = 3x^2 -2x + 1 so got above answer.

 

[king.rafa]

f(x)=3x^2-2x+1
f'(x)=6x-2
=0 when x=1/3

1/3 is a turning point, and since 3x^2-2x+1 has no zeros, it is either totally positive or negative, and hence when x=1/3 is the minimum distance.
y=1/3-2/3+1=2/3

distance=2/3

thats what i did, but how do i use the distance formula to solve

 

[Timothy.Siu]

thats what i did, but how do i use the distance formula to solve

that is the distance (if its correct), because that function finds the distance between the curves, and i found the minimum point was x=1/3 y=2/3
the y value being the minimum distnace

 

[king.rafa]

that is the distance (if its correct), because that function finds the distance between the curves, and i found the minimum point was x=1/3 y=2/3
the y value being the minimum distnace

but when you have (x2-x1)^2 in that part of the formula, what do you do with that

 

[GUSSSSSSSSSSSSS]

but when you have (x2-x1)^2 in that part of the formula, what do you do with that

NO NO he HAS NOT used distance formula at all

he has created one function, by solving the 2 similtaneously
and by there being NO ROOTS, this proves that the two curves never touch
and he has then found the MINIMUM value for this curve, and the 'y' value at this point will correspond to the minimum distance....

 

[king.rafa]

im still confused. how can you solve a problem like that.

isnt that like x+1=x+2?

NO NO he HAS NOT used distance formula at all

he has created one function, by solving the 2 similtaneously
and by there being NO ROOTS, this proves that the two curves never touch
and he has then found the MINIMUM value for this curve, and the 'y' value at this point will correspond to the minimum distance....

 

[azureus88]

Tim, the "minimum" distance you found was only the minimum difference in y values between y=3x^2 and y=2x-1. But the real minimum distance is the perpendicular distance.

I think Drongoski is correct.

 

[king.rafa]

lol anyone got the full solution and working?

im getting really confused here

 

[undalay]

f(x)=3x^2-2x+1
f'(x)=6x-2
=0 when x=1/3

1/3 is a turning point, and since 3x^2-2x+1 has no zeros, it is either totally positive or negative, and hence when x=1/3 is the minimum distance.
y=1/3-2/3+1=2/3

distance=2/3

Not exactly saying you are wrong, but this method only compares both curves vertically.
Where in some cases the vertical minimum may not be the absolute minimum.

 

[azureus88]

To find perpendicular distance, we need the gradient of y=2x-1 to be the same as the gradient of the tangent to y=3x^2.

dy/dx = 6x = 2
so x=1/3

subbing it back into y=3x^2, we get coordinates (1/3, 1/3)

using perpendicular distance with 2x-y-1=0, we get 2/(3sqrt5) as suggested by Drongoski.

 

[GUSSSSSSSSSSSSS]

Not exactly saying you are wrong, but this method only compares both curves vertically.
Where in some cases the vertical minimum may not be the absolute minimum.

yes good point

i shall redo a solution then..

 

[king.rafa]

Not exactly saying you are wrong, but this method only compares both curves vertically.
Where in some cases the vertical minimum may not be the absolute minimum.

how do you answer it? lol in this case it actually does give the absolute minimum. the answers correspond to the book. how would you answer it to make sure your answer was the absolute minimum.

the question is from the year 11 cambridge 3 unit page 298 question 22

my bad: the complete question:

the point x,y lies on the cruve y=3x^2. find the coordinates of p so that the distance from p to the line y=2x-1 is a minimum

 

[azureus88]

2/3 is not the absolute minimum. The answer in your book is probably wrong as you'll notice 2/(3sqrt5) is smaller than 2/3.

 

[undalay]

perp distance = |Am+Bn+C| / sqrt(A^2 + B^2)

Let our line be 2x-y-1
Let our point be (x,3x^2)

distance =|2x -3x^2 -1| / rt5

distance = (3x^2 - 2x +1) / rt 5

Differentiating the numerator: 6x - 2
Thus minimum distance at x = 1/3, or when numerator = 2/3

min distance = 2/3rt5

 

[oly1991]

I'm sure that you use the perp. dist formula somehow... Will have a go later.

lol

wasn't so funny after all

 

[cutemouse]

I'm sure that you use the perp. dist formula somehow... Will have a go later.

lol

It's funny how you made an absolute fool out of yourself =D

One word for you mate: Novice