1. 10.1 Parametric Equations

2. In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function. Remember: To graph, if there are trig functions in the parametric equations, plug in t values between 0 and 2Ө. If not, try different t- values between –10 and 10.

3. Graph the parametric curve represented by x = t 2 – 1 andy = 1 – t 2 -6 -4 -2 0 2 4 6 -6-4-2246 txy –23–3 –100 0 1 100 23–3

4. Write the Cartesian equation of the parametric curve represented by x = t 2 – 1 andy = 1 – t 2 Solve for tx = t 2 – 1 ±(x + 1) 1/2 = t Substitute t into the other expression y = 1 – (±(x + 1) 1/2 ) 2 y = 1 – (x + 1) y = –x

5. The formula for finding the slope of a parametrized curve is: This makes sense if we think about canceling dt. If f and g have derivatives at t, then the parametrized curve also has a derivative at t.

6. The formula for finding the slope of a parametrized curve is: We assume that the denominator is not zero.

7. To find the second derivative of a parametrized curve, we find the derivative of the first derivative: 1.Find the first derivative ( dy/dx ). 2. Find the derivative of dy/dx with respect to t. 3. Divide by dx/dt.

8. Example:

9. 1.Find the first derivative ( dy/dx ).

10. 2. Find the derivative of dy/dx with respect to t. Quotient Rule

11. 3. Divide by dx/dt.

12. The equation for the length of a parametrized curve is similar to our previous “length of curve” equation: (Notice the use of the Pythagorean Theorem.) (proof on pg. 533)

13. Remember: –A particle is AT REST when both dy/dt and dx/dt are 0.

14. Likewise, the equations for the surface area of a parametrized curve are similar to our previous “surface area” equations: