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Chapter 5 · Continuity and Differentiability

Exercise 5.6 · Q3: dydx\frac{dy}{dx} when x=sin⁡tx=\sin t⁠, y=cos⁡2ty=\cos 2t

Exercise 5.6Parametric differentiation · Section 5.6
The question

If xx and yy are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dydx\dfrac{dy}{dx}⁠.

3. x=sin⁡tx = \sin t⁠, y=cos⁡2ty = \cos 2t

Exam answer dxdt=cos⁡t,dydt=−2sin⁡2t\begin{aligned} \frac{dx}{dt} &= \cos t, \\ \frac{dy}{dt} &= -2\sin 2t \end{aligned} dydx=dy/dtdx/dt=−2sin⁡2tcos⁡t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin 2t}{\cos t}

Using sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t⁠,

dydx=−4sin⁡tcos⁡tcos⁡t\frac{dy}{dx} = \frac{-4\sin t\cos t}{\cos t}

∴ dydx=−4sin⁡t\therefore\ \dfrac{dy}{dx} = -4\sin t (cos⁡t≠0)(\cos t \ne 0)

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Divide the two tt-rates:

dydx=dy/dtdx/dt=−2sin⁡2tcos⁡t,\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin 2t}{\cos t},

wherever cos⁡t≠0\cos t \neq 0⁠. Then the double-angle formula sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t produces a factor cos⁡t\cos t on top that cancels the one underneath, leaving −4sin⁡t-4\sin t⁠.

Given, to find, and the plan
Given
x=sin⁡tx = \sin t⁠, y=cos⁡2ty = \cos 2t⁠, with parameter tt⁠.
To find
dydx\dfrac{dy}{dx}⁠, without eliminating tt⁠.
Plan
Differentiate xx⁠; differentiate yy with the chain rule; find where dxdt=0\frac{dx}{dt} = 0⁠; divide; simplify with sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t⁠; look at the excluded points on a picture.

Basics you need

1Derivatives of sine and cosine

ddtsin⁡t=cos⁡t\frac{d}{dt}\sin t = \cos t and ddtcos⁡t=−sin⁡t\frac{d}{dt}\cos t = -\sin t (Table 5.3). Basics: derivative rules

2Chain rule

If y=f(u)y = f(u) and u=g(t)u = g(t)⁠, then dydt=dydu⋅dudt\frac{dy}{dt} = \frac{dy}{du}\cdot\frac{du}{dt} (Theorem 4, Section 5.3.1). For cos⁡2t\cos 2t⁠: outer function cos⁡u\cos u⁠, inner function u=2tu = 2t⁠. Lesson: chain rule

3The parametric formula (Section 5.6)

dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}⁠, provided dxdt≠0\frac{dx}{dt} \neq 0⁠. Lesson: parametric differentiation

4Double-angle formula and zeros of cosine

sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t for every tt⁠. And cos⁡t=0\cos t = 0 exactly when t=(2n+1)π2t = (2n+1)\frac{\pi}{2}⁠, n∈Zn \in \Z⁠. Basics: trigonometric identities

Solution

  1. Differentiate xx with respect to tt

    dxdt=ddtsin⁡t=cos⁡t.\frac{dx}{dt} = \frac{d}{dt}\sin t = \cos t.
    The parametric formula needs the tt-rate of xx⁠; this one comes straight from the table.
    ddtsin⁡t=cos⁡t\frac{d}{dt}\sin t = \cos t (Table 5.3).

  2. Differentiate yy with respect to tt⁠: chain rule

    y=cos⁡2ty = \cos 2t is a function of a function: outer function cos⁡u\cos u⁠, inner function u=2tu = 2t⁠. So

    dydt=ddu(cos⁡u)⋅ddt(2t)=−sin⁡u⋅2=−2sin⁡2t.\begin{aligned} \frac{dy}{dt} &= \frac{d}{du}(\cos u)\cdot\frac{d}{dt}(2t) \\ &= -\sin u\cdot 2 = -2\sin 2t. \end{aligned}
    The angle is 2t2t⁠, not tt⁠, so the derivative of cosine alone is not enough: we must multiply by the rate at which the angle 2t2t changes, which is 22⁠. Forgetting this factor is the most common slip here.
    Chain rule, Theorem 4, Section 5.3.1. Lesson: chain rule

  3. Find where the formula may be used

    We need dxdt=cos⁡t≠0\frac{dx}{dt} = \cos t \neq 0⁠, that is

    t≠(2n+1)π2,n∈Z.t \neq (2n+1)\frac{\pi}{2}, \quad n \in \Z.
    We are going to divide by cos⁡t\cos t⁠, so the values where it is zero are set aside before dividing.
    Basic 4: the zeros of cosine are the odd multiples of π2\frac{\pi}{2}⁠.

  4. Divide

    For cos⁡t≠0\cos t \neq 0⁠,

    dydx=dy/dtdx/dt=−2sin⁡2tcos⁡t.\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin 2t}{\cos t}.
    The yy-rate goes on top and the xx-rate underneath. This is already a correct answer; the next step makes it simpler.
    Parametric formula, Section 5.6. Lesson: parametric differentiation

  5. Simplify with the double-angle formula

    −2sin⁡2tcos⁡t=−2⋅2sin⁡tcos⁡tcos⁡t=−4sin⁡t.\begin{aligned} \frac{-2\sin 2t}{\cos t} &= \frac{-2\cdot 2\sin t\cos t}{\cos t} \\ &= -4\sin t. \end{aligned}

    Cancelling cos⁡t\cos t is allowed because cos⁡t≠0\cos t \neq 0 (Step 3). Hence

    dydx=−4sin⁡t(cos⁡t≠0).\frac{dy}{dx} = -4\sin t \qquad \big(\cos t \neq 0\big).
    The top has the angle 2t2t and the bottom the angle tt⁠. Writing sin⁡2t\sin 2t in terms of the single angle tt reveals the common factor cos⁡t\cos t⁠. Note that −4sin⁡t-4\sin t itself is defined for every tt⁠, but our derivation used cos⁡t≠0\cos t \neq 0⁠, so the condition stays attached to the answer.
    sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t (Class XI). Basics: trigonometric identities

  6. Look at the curve and the excluded points

    Since x=sin⁡tx = \sin t⁠, the point never leaves the strip −1≤x≤1-1 \leq x \leq 1⁠. At t=π2t = \frac{\pi}{2} the point is (sin⁡π2,cos⁡π)=(1,−1)(\sin\frac{\pi}{2}, \cos\pi) = (1, -1)⁠, and at t=−π2t = -\frac{\pi}{2} it is (−1,−1)(-1, -1)⁠. At these values dxdt=cos⁡t=0\frac{dx}{dt} = \cos t = 0 and also dydt=−2sin⁡2t=−2sin⁡(±π)=0\frac{dy}{dt} = -2\sin 2t = -2\sin(\pm\pi) = 0⁠: the moving point stops at an end of the arc and turns back, so the formula gives no slope there.

    x y O −1 1 −1 1 t = π/6 t = 0 t = π/2 t = −π/2 t = π/6: slope −4 sin(π/6) = −2 t = 0: slope −4 sin 0 = 0 t = ±π/2: cos t = 0, the point turns back
    The curve x=sin⁡tx = \sin t⁠, y=cos⁡2ty = \cos 2t (equal scales). At t=π6t = \frac{\pi}{6}⁠, the point (12,12)\big(\frac12, \frac12\big)⁠, the slope is −4sin⁡π6=−2-4\sin\frac{\pi}{6} = -2⁠; at t=0t = 0 the slope is 00⁠. The ends (±1,−1)(\pm1, -1) are reached at t=±π2t = \pm\frac{\pi}{2}⁠.
    The picture explains the excluded values: they are not a flaw in the method, they are the turning-back points at the two ends of the arc.
    The slope of a curve is the slope of its tangent. Basics: slope at a point
Result dydx=−2sin⁡2tcos⁡t=−4sin⁡t,\frac{dy}{dx} = \frac{-2\sin 2t}{\cos t} = -4\sin t,

for every tt with cos⁡t≠0\cos t \neq 0⁠, that is t≠(2n+1)π2t \neq (2n+1)\frac{\pi}{2}⁠, n∈Zn \in \Z⁠. (Textbook answer: −4sin⁡t-4\sin t⁠.)

Check it with numbers

A short chord. At t=0.5t = 0.5 the point is (0.4794255, 0.5403023)(0.4794255,\ 0.5403023) and the predicted slope is −4sin⁡0.5≈−1.9177022-4\sin 0.5 \approx -1.9177022⁠. The points at t=0.499t = 0.499 and t=0.501t = 0.501 are (0.478547716, 0.541984166)(0.478547716,\ 0.541984166) and (0.480302881, 0.538618284)(0.480302881,\ 0.538618284)⁠, so

ΔyΔx=−0.0033658820.001755165≈−1.91770.\frac{\Delta y}{\Delta x} = \frac{-0.003365882}{0.001755165} \approx -1.91770.

Eliminating tt⁠, only as a check. cos⁡2t=1−2sin⁡2t\cos 2t = 1 - 2\sin^2 t⁠, so y=1−2x2y = 1 - 2x^2⁠. Then dydx=−4x=−4sin⁡t\frac{dy}{dx} = -4x = -4\sin t⁠. The same answer.

Common mistakes
  1. Forgetting the chain-rule factor. ddtcos⁡2t=−2sin⁡2t\frac{d}{dt}\cos 2t = -2\sin 2t⁠, not −sin⁡2t-\sin 2t⁠. Without the 2 the answer comes out as −2sin⁡t-2\sin t⁠.
  2. Sign slip. The derivative of cosine is −-sine; dropping the minus gives +4sin⁡t+4\sin t⁠.
  3. Cancelling cos⁡t\cos t without a condition. It is allowed only because cos⁡t≠0\cos t \neq 0 where the formula is used.
  4. Eliminating tt as the solution. y=1−2x2y = 1 - 2x^2 is a check, not what the question asks for.
What to take away
  1. Parametric slope: differentiate both coordinates with respect to the parameter, then divide.
  2. When the angles on top and bottom differ (2t2t and tt⁠), rewrite in one angle to find common factors.
  3. Excluded parameter values often mark special points of the curve, here the two turning-back ends.

Every fact used, and where it comes from

FactUsed inSource
ddtsin⁡t=cos⁡t\frac{d}{dt}\sin t = \cos t⁠, ddtcos⁡t=−sin⁡t\frac{d}{dt}\cos t = -\sin tSteps 1, 2Table 5.3 (Basics: derivative rules)
Chain ruleStep 2Theorem 4, Section 5.3.1 (Lesson: chain rule)
cos⁡t=0  ⟺  t=(2n+1)π2\cos t = 0 \iff t = (2n+1)\frac{\pi}{2}Steps 3, 6Class XI, Trigonometric Functions
dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}⁠, dxdt≠0\frac{dx}{dt} \neq 0Steps 3, 4Section 5.6 (Lesson: parametric differentiation)
sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t⁠; cos⁡2t=1−2sin⁡2t\cos 2t = 1 - 2\sin^2 t (check only)Step 5, checkClass XI, double-angle formulas (Basics: trigonometric identities)
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