Lesson 13.7: Help for Exercises 2123 on page 817
Let's look at a problem that is similar to Exercises 2123. Consider an object that is at (3, 12) at time t = 0 and at (28, 20) at time t = 6.
Start by establishing a pair of ordered pairs of the form (t,x) and use this information to determine the equation of a line of the form x = mt + b. At t = 0, (t, x) = (0, 3) and at t = 6 (t, x) = (6, 28). The slope of the line between these two points is .
Next find the value of b:
x = 4.17t + b 
Substitute 4.17 for m 
3 = 4.17•0 + b 
Substitute 3 for x and 0 for t 
3 = b 
Simplify. b is by itself 
The first parametric equation is x = 4.17t + 3. The next step is to establish a pair of ordered pairs of the form (t,y) and use this information to determine the equation of a line of the form (y = mt + b). Using a similar procedure, you can find that the second parametric equation is y = 1.33t + 12.
These parametric equations can also be expressed in the form and . If you draw a sketch of the original two points, you can determine the value of using the tangent function:
Next set vcos17.7° equal to 4.17 (the slope of the first parametric equation) and solve for v = 4.4. This results in the pair of parametric equations x = (4.4cos17.7°)t + 3 and y = (4.4cos17.7°)t + 12.
Use either of these approaches to solve Exercises 2123.
