Find Tangent Line to Two Parabolas

[ApostleO]

SOLVED: Find Tangent Line to Two Parabolas

With help from daon2 I was able to solve this problem which had given me quite a bit of trouble. Thanks, daon2! Please let me know if you notice any mistakes.

Two small arches have the shape of parabolas. The first is given by f(x)=1-x² for -1≤x≤1 and the second by g(x)=4-(4-4)² for 2≤x≤6. A board is placed on top of these arches so it rests on both (Figure 24). What is the slope of the board? Hint: Find the tangent line to y=f(x) that intersects y=g(x) in exactly one point.Fig. 24

Excuse my paint illustration. Not to scale, obviously.​

Step 1: Assume two points on the tangent line T, A and B, at (A, f(A)) and (B,g(B)), are the points at which the tangent line touch the graphs for f(x) and g(x). Find the derivatives of each of the curves in terms of these points.

f'(A)=-2A
g'(B)=-2B+8

Step 2: Recognize that these two slopes must be equal to describe the same tangent line.

-2A=-2B+8
A+4=B

Step 3: Express both points A and B in relation to A.

A: (A, 1-A[SUP]2[/SUP])
B: (B, 4-(B-4)[SUP]2[/SUP])
B: (A+4, 4-((A+4)-4)[SUP]2
[/SUP]B: (A+4, 4-A[SUP]2[/SUP])

Step 4: Express the slope in terms of Δy/Δx, considering that A and B are on the same line.

m=((4-A2)-(1-A2))/((A+4)-A)
m=3/4

Step 5: Use the slope, and the fact that points A and B both fall on line T, and each fall as the tangent points of contact for f(x) and g(x) respectively, find points A and B. (Actually, I realized I only needed one of these two points, so I will only show the work for one.)

m=3/4=-2A
-3/8=A

f(-3/8)=1-(-3/8)[SUP]2[/SUP]=55/64

A: (-3/8, 55/64)

Step 6: Knowing one of the points on line T and its slope, finish its equation.

y=mx+c
55/64=3/4*(-3/8)+c
55/64=-9/32+c
73/64=c

y=3/4x+73/64

Answer: y=3/4x+73/64

This was completed thanks to daon2. Thanks, daon2!

I punched the equations for the two curves and this line into my graphing calculator, and it seems correct. Would anyone be so kind as to point out any errors I may have made?

 

[daon2]

You will need to find colinear points (a, f(a)) and (b, g(b)), such that the points belong to a line with slope m=f'(a)=g'(b).

Using your derivatives:

f'(a) = -2a
g'(b) = -2b+8

And we need -2a= -2b + 8, i.e. b=a+4.

Plug these into f and g to get f(a)=1-a^2, g(b) = g(a+4) = 4-a^2.

So (a, 1-a^2) and (a+4, 4-a^2) lie on the same line. What is the slope of that line?

Once you have that, you can find a,b and then the equation.

 

[ApostleO]

Awesome, thanks daon2! That was exactly the hint I needed. I think I got it, unless you see any errors above.

EDIT: I edited my original post, and apparently it somehow deleted itself... and after I had written up everything so neatly to help anyone else who ran into this problem. Anyway, I got it thanks to your help.