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# 10.2

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## 10.2 Calculus with Parametric

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### 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

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### 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

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### 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:

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### 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.

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### 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).

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### 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

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### 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

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### 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

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### Example 1 – Solution

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

Figure 1

cont’d

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### 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:

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### Example 3

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

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### 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

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cont’d

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### 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

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### 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

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### 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

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### 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)

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### 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

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Therefore

and so

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### 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,

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### 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.

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### 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.

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### 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

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We have

cont’d

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### 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

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and so

cont’d

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### 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

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### 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

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### Example 6 – Solution

cont’d

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