10.3. Parameters and Parameter Elimination http://www.ck12.org
- Use your calculator to model the race in #2.
Answers: - Use the fact that a point plus a vector yields another point. A vector between these points is< 4 − 1 , 8 − 3 >=<
3 , 5 >
Thus the point (1, 3) plusttimes the vector< 3 , 5 >will produce the point (4, 8) whent=1 and the point (1, 3)
whent=0.
(x,y) = ( 1 , 3 )+t·< 3 , 5 >, for 0≤t≤ 1
You than then break up this vector equation into parametric form.
x= 1 + 3 t
y= 3 + 5 t
0 ≤t≤ 1- First draw a picture and then represent each character with a set of parametric equations.
The tortoise’s position is (-101, 0) att=0 and (-97.8, 0) att=1. You can deduce that the equation modeling the
tortoise’s position is:
x 1 =− 101 + 3. 2 ·t
y 1 = 0The hare’s position is (101, 0) att=21 and (91.2, 0) att=22. Note that it does not make sense to make equations
modeling the hare’s position before 21 seconds have elapsed because the Hare is napping and not moving. You can
set up an equation to solve for the hare’s theoretical starting position had he been running the whole time.
x 2 =b− 9. 8 t
101 =b− 9. 8 · 21
305. 8 =bThe hare’s position equation aftert=21 can be modeled by:
x 2 = 305. 8 − 9. 8 ·t
y 2 = 0The tortoise crossesx=0 whent≈ 31 .5. The hare crossesx=0 whent≈ 31 .2. The hare wins by about 1.15 feet.
- There are many settings you should know for parametric equations that bring questions like this to life. The
TI-84 has features that allow you to see the race happen.
First, set the mode to simultaneous graphing. This will show both the tortoise and hare’s position at the same time.