First of all, we need to know what it means for a parameterize curve to be differentiable. FACT:

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1 CALCULUS WITH PARAMETERIZED CURVES In calculus I we learned how to differentiate and integrate functions. In the chapter covering the applications of the integral, we learned how to find the length of a curve and the surface area of the surface that is generated when the curve is revolved about an axis. Now we will learn how to apply these concepts to parameterized curves. We will start with the concept of the derivative of parameterized curves. DERIVATIVE OF PARAMETERIZED CURVES First of all, we need to know what it means for a parameterize curve to be differentiable. FACT: A parameterized curve x = f (t), y = g (t) is differentiable at t = t 0 if f and g are differentiable at t = t 0. The curve is differentiable if it is differentiable at every parameter value. The curve is smooth if f ' and g' are continuous and not simultaneously zero. At a point on a differentiable parameterized curve where y is also a differentiable function of x, then the derivatives dx/ dt, dy/ dt, and dy/ dx are related by the chain rule. FACT: EXAMPLE 1: Find the equation for the line tangent to the given curve at the given point. First we will find dy/ dt and dx/ dt. Now evaluate each derivative at the given value of t. To do this, I will draw the reference triangle for this angle.

2 Now substitute everything into the formula for the derivative. Next, we need to find the point of the curve at the given value of t. Now to write the equation of the tangent line to the curve at the given point. EXAMPLE 2: Find the equation for the line tangent to the given curve at the given point. First of all, we will find dy/ dt and dx/ dt.

3 Now evaluate each derivative at t = 3. Now sub them into the formula for dy/ dx. Next find the point on the curve when t = 3. Now to write the equation of the tangent line to the curve when t = 3. y - 3 = - 2(x - 2) y - 3 = -2x + 4 y = -2x + 7 Sometimes we will need to find the second derivative of a parameterized curve, so here is the formula that we will use to find this derivative. EXAMPLE 3: Find d 2 y/ dx 2 of x = 2t 2 + 3, y = t 4. First, I will find dy/ dx.

4 Now to find dy'/ dt where y' = dy/ dx. Now put everything together. EXAMPLE 4: Find d 2 y/ dx 2 of x = cos t, y = 1 + sin t. First, find dy/ dx. Now to find dy'/ dt where y' = dy/ dx. Now put everything together. EXAMPLE 5: Assuming that the equations

5 define x and y implicitly as differentiable functions x = f (t) and y = g (t). Find the slope of the curve x = f (t), y = g (t) at t = 0. In this problem, both x and y are implicitly defined in terms of t, so we will have to do implicit differentiation to find dy/ dt and dx/ dt. First find dx/ dt. Every time we take a derivative of x tack on dx/ dt. Then solve for dx/ dt. Next find dy/ dt. Now to find what x and y equals when t = 0.

6 The square root of x cannot equal -1/ 2, since it is the principle root, so x = 0 when t = 0. Now plug these values into dy/ dt and dx/ dt. LENGTH OF CURVES How do we find the length of a parameterized curve? Using the formula that we learned in calculus I, we can modify it to get the following formula. FACT: LENGTH OF A PARAMETERIZED CURVE If a smooth curve x = f (t), y = g (t), a t b, is traversed exactly once as t increases from a to b, then the curve's length is

7 EXAMPLE 6: Find dy/ dt and dx/ dt. EXAMPLE 7: Find dy/ dt and dx/ dt.

8 SURFACE AREA If we modify the formulas for the surface area that we learned in calculus I, we will have the formulas for the surface area generated by revolving a parameterized curve about an axis. FACT: SURFACE AREA If a smooth curve x = f (t), y = g (t), a t b, is traversed exactly once as t increases from a to b, then the areas of the surfaces generated by revolving the curve about the coordinate axes are as follows. 1. Revolution about the x-axis (y 0) 2. Revolution about the y-axis (x 0) EXAMPLE 8: Find the area of the surface generated by about the y-axis. First of all, I will find dy/ dt and dx/ dt.

9 In this set of supplemental notes, I have covered how to take the derivative of a set of parametric equation, find the length of a parametric curve, and find the surface area of a parametric curve that is revolved about one of the coordinate axis. The last two topics are topics that are covered in calculus I. The required formulas have been modified to work with parametric curves.