Continuity and Differentiability

The neighborhood young people are building a skating ramp. At the moment, it rises 8 feet vertically as it moves 10 feet horizontally (See figure 1).

When they tried it out they realized that the ramp did not work too well because of the sharp corner created at the intersection of the street and the ramp. One of the young people was a very good calculus student. She realized that what they needed was a smooth transition between the ramp and the street. She proposed smoothing the sharp corner using a parabolic connection going in the same direction as the ramp (tangent to the ramp) when it touched it and going in the same direction of the street (tangent to the street) when it met it. She determines that 3 feet would be enough for the connection to work.

Now we need to help her find the appropriate parabolic arc. It may be helpful for us to set up a coordinate system with some reference point as the origin. From it, write equations for both lines and the parabola. Don’t forget to use derivatives and write equations using the derivatives as well. Find all the unknowns and graph the solutions to check your results. Remember that there are more than one way to solve this problem.

Consider the following figure with reference points:

Figure 2

The general equation of a parabola is f(x) = a x2 + b x + c. The general equation of a line is y = m x + b. In the following setting we have two lines; let’s call them L1(x) = m1 x + b1 and L2(x) = m2 x + b2, and let L1 be the ramp and L2 be the street. From the information given we know the slope of L1, namely 8/10 = 0.8. From our reference coordinate system we know the y-intercept for L1, namely 0. Therefore, we know the complete equation of L1: L1(x) = 0.8 x.

We also know that the slope of L2 is 0, since it is a horizontal line. Therefore, the equation of L2 is L2(x)=b2.

We may take the derivatives with respect to x of all three functions and get: L1’(x) = 0.8, L2’(x) = 0, and f’(x) = 2a x + b.

So we have the following equations:

f(0) = 0 (continuity condition on parabola and ramp)

f’(0) = L1’(0) = 0.8 (smoothness condition for parabola and ramp)

f’(-3) = L2’(-3) = 0 (smoothness condition for parabola and street)

These three equations give us a system of unknowns to be solved:

a(0)2 + b(0) + c = 0 (first condition) implies c = 0

2a(0) + b = 0.8 (second condition) implies b = 0.8

2a(-3) + b = 2a(-3) + 0.8 = 0 (third condition) implies a = 2/15

We can obtain b2 from continuity with the parabola, that is, we have f(-3) = L2(-3) and f(-3) = 2(-3)2 / 15 + 8(-3) / 10 = -1.2. Therefore, the complete equations are:

f(x) = 2x2 / 15 + 0.8x

L1(x) = 0.8x

L2(x) = -1.2

We can check this solution by graphing it on the calculator using the feature which allows you to graph a function on a specific range. Note: Square brackets are located above the + and - keys. To access them press SHIFT + and SHIFT -.

Enter the following into your graph list:

Y1 = 2 X2 ¸ 15 + 0.8 X, [-3,0]

Y2 = 0.8 X, [0,10]

Y3 = -1.2, [-10,-3]

Make Y1, the parabola, orange by using the cursor arrows to highlight Y1 and pressing F4 (COLR) F2 (Orng). Also, turn on the numerical derivative and take off the axes by pressing SHIFT SETUP and using your cursor arrows to move down until you see derivative and press F1 (On) then continue down to axes and press F2 (0ff). Press EXIT.

The initial viewing window works very well for this problem so press SHIFT F3 (V-Window) F1 (INIT) EXIT F6 (DRAW).

Press SHIFT F1 (Trace). We can check continuity and differentiability.