Expert: Paul Klarreich - 9/15/2007

Question
1. There are two tangent lines to the curve y = 4x - x^2 that go through 2,5. Find the equations of both.

2. A fly is crawling from left to right along the top of the curve y = 7 - x^2. A spider waits at the point  4,0. Find the distance between the two insects when they first see each other  

Answer
Questioner:   Casey
Category:  Calculus
Private:  No
 
Subject:  Calculus problems
Question:  1. There are two tangent lines to the curve y = 4x - x^2 that go through (2,5). Find the equations of both.

2. A fly is crawling from left to right along the top of the curve y = 7 - x^2. A spider waits at the point  4,0. Find the distance between the two insects when they first see each other

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Hi, Casey,

Strange as it may seem, these two examples are really the same problem.

1. There are two tangent lines to the curve y = 4x - x^2 that go through (2,5). Find the equations of both.

If a line is a solution then:

1. It is tangent to  y = 4x - x^2 at  some  x, so its slope is  m = 4 - 2x.

2. It passes through the point (x,4x - x^2)

3. It also passes through the point (2,5).

The slope of a line through (a, 4x - x^2) and (2,5) is:

   5 - 4x + x^2
m = -------------
      2 - x

But we also have:  m = 4 - 2x, so set those equal:

   5 - 4x + x^2
m = ------------- = 4 - 2x
      2 - x

(4 - 2x)(2 - x) = 5 - 4x + x^2

8 - 8x + 2x^2 = 5 - 4x + x^2

x^2 - 4x + 3 = 0

(x - 3)(x - 1) = 0

x = 3  or  x = 1

I think you can handle the rest.  You just want the slopes at these values of x, from  m = 4 - 2x, then use (2,5) for the point in each case.

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2. A fly is crawling from left to right along the top of the curve y = 7 - x^2. A spider waits at the point  4,0. Find the distance between the two insects when they first see each other.

This is the same problem, basically -- the line between them is tangent to the curve and passing through  (4,0).

The slope of this line, given by the derivative, is  m = -2x.

The line passes through (x, 7 - x^2) and (4,0) so its slope is:
   0 - (7 - x^2)   x^2 - 7
m = ------------- = ---------
      4 - x         4 - x

Set those equal:

x^2 - 7
--------- = -2x
 4 - x

x^2 - 7 = -2x(4 - x)

x^2 - 7 = -8x + 2x^2

0 = 7 - 8x + x^2

0 = x^2 - 8x + 7

(x - 7)(x - 1) = 0

x = 7,  x = 1.

Obviously  x = 1, y = 6 is the solution (fly) point, so now all you need is the distance between (1, 6)  and (4,0) = sqrt(9 + 36) = sqrt(45) = 3 sqrt(5)