# Name: Math 110-003 Student number: Quiz 5 (Mar 31) ```Name:
Student number:
Math 110-003
Quiz 5 (Mar 31)
Write full solutions to the questions below. It is to your own benefit to write as clearly and
legibly as possible.
1. In this problem, you will find the shortest distance between the line y = x − 1 and the
parabola y = x2 .
(a) In this part, let a be fixed number (a parameter) and let (a, a2 ) be a fixed point
on the parabola y = x2 . Let (x, x − 1) be a point on the line y = x − 1. Then the
squared distance between (x, x − 1) and (a, a2 ) is
d2 (x) = (x − a)2 + (x − 1 − a2 )2 .
(1)
Differentiate d2 with respect to x and set the resulting expression to 0. Then solve
the resulting equation to find x in terms of a.
Hint: do not expand d2 before differentiating it.
(b) Substitute the expression for x found in (a) back into the formula for d2 above
(again, it is advised that you do not expand d2 ). Now d2 should be a function of
a alone. Differentiate d2 with respect to a this time, set the resulting expression
for 0 and solve a. There should be no x in your solution.
Hint: instead of expanding, try factoring.
(c) Plug the value of a found in (b) back into the expression for d2 found in (b). Take
the square root of the resulting value of d2 in order to find the shortest distance
between the line and the parabola.
Solution.
(a) We have
d 2
d = 2(x − a) + 2(x − 1 − a2 ) = 4x − 2(a2 + a + 1).
dx
Setting this equal to 0 and solving for x gives
1
1
1
1
x = (a2 + a + 1) = a2 + a + .
2
2
2
2
(2)
(3)
(b) With x as above, we have
d2 = (a2 /2 − a/2 + 1/2)2 + (−a2 /2 + a/2 − 1/2)
= (a2 /2 − a/2 + 1/2)2 + (−(a2 /2 − a/2 + 1/2))2
= (a2 /2 − a/2 + 1/2)2 + (a2 /2 − a/2 + 1/2)2
= 2(a2 /2 − a/2 + 1/2)2
1
= (a2 − a + 1)2 .
2
(4)
(5)
(6)
(7)
(8)
Differentiating with respect to a, we get
d 2
d = (a2 − a + 1)(2a − 1).
da
(9)
Name:
Student number:
Math 110-003
Quiz 5 (Mar 31)
In order for this to be 0, either the factor a2 − a + 1 must equal 0 or 2a − 1 must
equal 0. Since (−1)2 − 4(1)(1) &lt; 0, the equation a2 − a + 1 = 0 has no solutions,
so it must be the case that 2a − 1 = 0, i.e. that
a = 1/2.
(10)
(c) Plugging a = 1/2 into the expression for d2 found above, we get
1
1
1
d2 = ((1/2)2 − (1/2) + 1)2 = (1/4 − 1/2 + 1)2 = (3/4)2 .
2
2
2
(11)
3
d= √
4 2
(12)
Thus,
is the minimum distance.
Demo.
For a visualization of the idea behind this problem, see
https://www.desmos.com/calculator/fxifb9vwh5.
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