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 d2y/dx2. 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 = t2, y = t3 – 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 = t3 – 3t = t(t2 – 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 = 3t2 – 3, this
happens when t2 = 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)dθ, we have
Example 3 – Solution
cont’dArc 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 t0, t1, t2, . . . , tn are the endpoints of these subintervals, then xi = f(ti) and yi = g(ti) are the coordinates of points Pi(xi, yi) 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 [ti –1, ti], gives a number in (ti –1, ti) such that
f(ti) – f(ti –1) = f′( ) (ti – ti –1)
Arc Length
If we let ∆xi = xi – xi –1 and ∆yi = yi – yi –1, this equation becomes
∆xi = f′( ) ∆t
Similarly, when applied to g, the Mean Value Theorem gives a number in (ti –1, ti) such that
∆yi = 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 2π, 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 sin2x = (1 – cos 2x) with θ = 2x, which gives 1 – cos θ = 2 sin2(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 = ∫ 2πy ds and
S = ∫ 2πx ds are still valid, but for parametric curves we use
Example 6
Show that the surface area of a sphere of radius r is 4πr2. 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
