Algebra 2

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Chapter 13 : Trigonometric Ratios and Functions
13.7 Problem Solving Help

Lesson 13.7: Help for Exercises 21-23 on page 817

Let's look at a problem that is similar to Exercises 21-23. 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 21-23.

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