Solution of the unfinished problem in tutorial session The idea of this solution is to use calculus with implicit differentiationby Rogelio Yoyontzin

1. 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.

1.1. Graph of the line and point

1.2. 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

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

2.1. 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

2.2. Therefore we have that

2.3. 3) dy/dx = −2/6 = −1/3.

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

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

2.6. 5) 0 = 2(x-1) + 2(y-3)(-1/3) = 2x + (-2/3)y

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

2.8. 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

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

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?