**10.2** Calculus with Parametric

### Curves

### Tangents

### Tangents

*Suppose f and g are differentiable functions and we want to *
find the tangent line at a point on the parametric curve
*x = f(t), y = g(t), where y is also a differentiable function of x.*

Then the Chain Rule gives

### Tangents

*If dx/dt ≠ 0, we can solve for dy/dx:*

Equation 1 (which you can remember by thinking of

*canceling the dt’s) enables us to find the slope dy/dx of the *
tangent to a parametric curve without having to eliminate

### Tangents

We see from (1) that the curve has a horizontal tangent

*when dy/dt = 0 (provided that dx/dt ≠ 0) and it has a vertical *
*tangent when dx/dt = 0 (provided that dy/dt ≠ 0).*

This information is useful for sketching parametric curves.

*It is also useful to consider d*^{2}*y/dx*^{2}. This can be found by
*replacing y by dy/dx in Equation 1:*

### Example 1

*A curve C is defined by the parametric equations x = t*^{2},
*y = t*^{3} *– 3t.*

*(a) Show that C has two tangents at the point (3, 0) and *
find their equations.

*(b) Find the points on C where the tangent is horizontal or *
vertical.

(c) Determine where the curve is concave upward or downward.

*Example 1 – Solution*

*(a) Notice that y = t*^{3} *– 3t = t(t*^{2} *– 3) = 0 when t = 0 or t = *
*Therefore the point (3, 0) on C arises from two values of *
*the parameter, t = and t = . *

*This indicates that C crosses itself at (3, 0).*

*Example 1 – Solution*

Since

*the slope of the tangent when t = is *

*dy/dx* *, so the equations of the *

tangents at (3, 0) are

cont’d

*Example 1 – Solution*

*(b) C has a horizontal tangent when dy/dx = 0, that is, when *
*dy/dt = 0 and dx/dt ≠ 0. Since dy/dt = 3t*^{2 }– 3, this

*happens when t*^{2} *= 1, that is, t = ±1.*

*The corresponding points on C are (1, –2) and (1, 2).*

*C has a vertical tangent when dx/dt = 2t = 0, that is, *
*t = 0. (Note that dy/dt *≠ 0 there.)

*The corresponding point on C is (0, 0).*

cont’d

*Example 1 – Solution*

(c) To determine concavity we calculate the second derivative:

*Thus the curve is concave upward when t > 0 and concave *
*downward when t < 0.*

cont’d

*Example 1 – Solution*

(d) Using the information from parts (b) and (c), we sketch
*C in Figure 1.*

**Figure 1**

cont’d

### Areas

### Areas

*We know that the area under a curve y = F(x) from a to b is*
*A = F(x) dx, where F(x) ≥ 0.*

If the curve is traced out once by the parametric equations
*x = f(t) and y = g(t), *α *≤ t ≤* β, then we can calculate an area
formula by using the Substitution Rule for Definite Integrals
as follows:

### Example 3

Find the area under one arch of the cycloid
*x = r(*θ – sin θ*) y = r(1 – cos *θ)
(See Figure 3.)

*Example 3 – Solution*

One arch of the cycloid is given by 0 ≤ θ ≤ 2π.

*Using the Substitution Rule with y = r(1 – cos *θ) and
*dx = r(1 – cos *θ*)d*θ, we have

*Example 3 – Solution*

_{cont’d}

### Arc Length

### Arc Length

*We already know how to find the length L of a curve C *
*given in the form y = F(x), a* *≤ x ≤ b.*

*If F*′ is continuous, then

*Suppose that C can also be described by the parametric *
*equations x = f(t) and y = g(t), *α *≤ t ≤* β, where

### Arc Length

*This means that C is traversed once, from left to right, as t*
increases from α to β *and f(*α*) = a, f(*β*) = b.*

Putting Formula 1 into Formula 2 and using the Substitution Rule, we obtain

*Since dx/dt > 0, we have*

### Arc Length

*Even if C can’t be expressed in the form y = F(x), Formula *
3 is still valid but we obtain it by polygonal approximations.

We divide the parameter interval [α, β*] into n subintervals *
of equal width ∆t.

*If t*_{0}*, t*_{1}*, t*_{2}*, . . . , t** _{n}* are the endpoints

*of these subintervals, then x*

_{i}*= f(t*

*)*

_{i}*and y*

_{i}*= g(t*

*) are the coordinates of*

_{i}*points P*

_{i}*(x*

_{i}*, y*

_{i}*) that lie on C and the*

*polygon with vertices P*

*, P*

*, . . . , P*

### Arc Length

*We define the length L of C to be the limit of the lengths of *
*these approximating polygons as n* → :

*The Mean Value Theorem, when applied to f on the interval *
*[t*_{i –1}*, t*_{i}*], gives a number in (t*_{i –1}*, t** _{i}*) such that

*f(t*_{i}*) – f(t*_{i –1}*) = f′( ) (t*_{i }*– t** _{i –1}*)

### Arc Length

If we let ∆x_{i}*= x*_{i}*– x** _{i –1}* and ∆y

_{i}*= y*

_{i }*– y*

*, this equation becomes*

_{i –1}*∆x*_{i}*= f′( ) ∆t*

*Similarly, when applied to g, the Mean Value Theorem *
*gives a number in (t*_{i –1}*, t** _{i}*) such that

*∆y*_{i}*= g′( ) ∆t*

### Arc Length

Therefore

and so

### Arc Length

The sum in (4) resembles a Riemann sum for the function but it is not exactly a Riemann sum

because ≠ in general.

*Nevertheless, if f′ and g′ are continuous, it can be shown *
that the limit in (4) is the same as if and were equal,
namely,

### Arc Length

Thus, using Leibniz notation, we have the following result, which has the same form as Formula 3.

Notice that the formula in Theorem 5 is consistent with the
*general formulas L = ∫ ds and (ds)*^{2 }*= (dx)*^{2} *+ (dy)*^{2}.

### Arc Length

Notice that the integral gives twice the arc length of the
*circle because as t increases from 0 to 2*π, the point

*(sin 2t, cos 2t) traverses the circle twice. *

*In general, when finding the length of a curve C from a *

parametric representation, we have to be careful to ensure
*that C is traversed only once as t increases from *α to β.

### Example 5

*Find the length of one arch of the cycloid x = r(*θ – sin θ),
*y = r(1 – cos *θ).

Solution:

From Example 3 we see that one arch is described by the parameter interval 0 ≤ θ ≤ 2π.

Since

and

*Example 5 – Solution*

We have

cont’d

*Example 5 – Solution*

To evaluate this integral we use the identity
sin^{2}*x = (1 – cos 2x) with *θ *= 2x, which gives *
1 – cos θ = 2 sin^{2}(θ/2).

Since 0 ≤ θ ≤ 2π, we have 0 ≤ θ/2 ≤ π and so sin(θ/2) ≥ 0.

Therefore

cont’d

*Example 5 – Solution*

and so

cont’d

### Surface Area

### Surface Area

*Suppose the curve c given by the parametric equations *
*x = f(t),y = g(t), *α *≤ t ≤* β*, where f′, g′ are continuous, *
*g(t) ≥ 0, is rotated about the x-axis. If C is traversed exactly*
*once as t increases from *α to β*, then the area of the *

resulting surface is given by

*The general symbolic formulas S = ∫ 2*π*y ds and*

*S = ∫ 2*π*x ds are still valid, but for parametric curves we use*

### Example 6

*Show that the surface area of a sphere of radius r is 4*π*r*^{2}.
Solution:

The sphere is obtained by rotating the semicircle
*x = r cos t* *y = r sin t* 0 ≤ t ≤ π
*about the x-axis. *

Therefore, from Formula 6, we get

*Example 6 – Solution*

_{cont’d}