Find min/max distance between (0,4) and y=x^2/4

  • Thread starter Poppynz
  • Start date
In summary, the problem is to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3. The formula for the distance is y(distance)=√((x-0)^2+((x^2)/4)^2), which simplifies to y(distance)=(x^2+(X^4)/16+16)^0.5. The suggested approach is to calculate the derivative of y^2 instead of y, which eliminates the square root and makes the problem easier to solve.
  • #1
Poppynz
6
0
I am trying to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3.

This is what i have so far

y(distance)=√((x-0)^2+((x^2)/4)^2)
=(x^2+(X^4)/16+16)^0.5
then calculate y'
then make y'=0
then determine if max or min using 2nd derivative test

The problem i am having is that the answers i get for x when y' = 0 are imaginary. i have computed this problem using MATLAB and get the same answers except that they are real and have isolated the problem to my simplification of y.

matlab gives it as y=0.25(-16x^2+x^4+256)^0.5

Could some one please explain how to simplify y=(x^2+(X^4)/16+16)^0.5 to y=0.25(-16x^2+x^4+256)^0.5 or suggest a better way to approach this problem

Thanks in advance
 
Physics news on Phys.org
  • #2
Poppynz said:
y(distance)=√((x-0)^2+((x^2)/4)^2)
=(x^2+(X^4)/16+16)^0.5

I think you meant [(x^2)/4-4]^2, not [(x^2)/4]^2. Also, you might find it easier to calculate the derivative of y^2 instead of y. Squaring the distance eliminates the square root, which is a major pain in the neck.
 

Related to Find min/max distance between (0,4) and y=x^2/4

1. How do you find the minimum and maximum distance between (0,4) and y=x^2/4?

To find the minimum and maximum distance between (0,4) and y=x^2/4, you can use calculus techniques. First, find the derivative of the distance function and set it equal to 0 to find the critical points. Then, plug the critical points into the distance function to find the minimum and maximum distances.

2. What is the formula for finding the distance between a point and a parabola?

The formula for finding the distance between a point (x1,y1) and a parabola y=x^2 is given by D = |y1 - x1^2|. This formula can be derived by using the Pythagorean theorem and the distance formula.

3. Can you use the Pythagorean theorem to find the distance between a point and a parabola?

Yes, the Pythagorean theorem can be used to find the distance between a point and a parabola. By treating the parabola as the hypotenuse of a right triangle with one leg being the distance from the point to the x-axis and the other leg being the distance from the point to the y-axis, we can apply the Pythagorean theorem to find the distance.

4. What is the significance of finding the minimum and maximum distance between a point and a parabola?

Finding the minimum and maximum distance between a point and a parabola can provide valuable information about the behavior of the parabola. It can also be useful in optimization problems, where finding the shortest or longest distance between a point and a parabola is necessary.

5. Can the minimum and maximum distance between a point and a parabola be the same?

Yes, it is possible for the minimum and maximum distance between a point and a parabola to be the same. This can happen when the point lies on the parabola, resulting in a distance of 0. It can also occur when the point is located at the vertex of the parabola, where the minimum and maximum distances are equal.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
642
  • Calculus and Beyond Homework Help
Replies
1
Views
579
  • Calculus and Beyond Homework Help
Replies
1
Views
482
  • Calculus and Beyond Homework Help
Replies
2
Views
696
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
950
  • Calculus and Beyond Homework Help
Replies
2
Views
785
  • Calculus and Beyond Homework Help
Replies
3
Views
393
Back
Top