Solution of the unfinished problem in tutorial session The idea of this solution is to use calc...

1. Find the coordinates of the point (x,y) on the line 2x + 6y = 1 which is closest to the point (1, 3).

1.1. From the previews exercise we have that: 1) δ² = (x-1)²+(y-3)² and 2) 2x + 6y = 1. If we find the derivatives respect to x we find: from 1) 2δ(dδ/dx) = 2(x-1) + 2(y-3)(dy/dx) and from 2) 2+ 6(dy/dx) = 0

1.2. Therefore we have that

1.3. 3) dy/dx = −2/6 = −1/3.

1.4. Since we want to minimize δ, we want to find the points in the line for which dδ/dx =0.

1.5. We can then make dδ/dx =0 in from 1) and put equation 3) to get:

1.6. 5) 0 = 2(x-1) + 2(y-3)(-1/3) = 2x + (-2/3)y

1.7. Multiplying by 3/2 the equality: 0= 3x - y and so 3x = y

1.8. That is the closes point (x,y) in the line 2x+6y = 1 is the point such that 3x=y. Then to know the values we do 3x=y in the equation of the line: 2x + 6(3x) =1 then 20x= 1 and x = 1/20 and then y=3/20

1.9. The point we are looking for is (1/20, 3/20)

2. Part 1. The point (x,y) lies on the line 2x + 6y = 1. Express the distance from the point (1, 3) to the point (x,y) as a function of x.

2.1. Graph of the line and point

2.2. Let δ be the distance form the point (1,3) to the point (x,y) in the line. Then we have that: δ² = (x-1)²+(y-3)² but form the equation of the line we get: y= (1-2x)/6 and so δ² = (x-1)² + ((1-2x)/6 − 3)² Taking the square root on both sides we get δ as a function of x

3. Now try to find the distance by yourself and tell me why this is in fact a minimal distance! Note that you can also use the solution of the problem on part 1 to find the minimum of the function δ in therms of x. Why don't you try?