Chapter 5 · Continuity and Differentiability

5.4 Parametric Forms and Second-Order Derivatives

When $x$ and $y$ both depend on a parameter $t$, the slope of the curve is $\tfrac{dy}{dx}=\tfrac{dy/dt}{dx/dt}$. Differentiating a derivative once more gives the second-order derivative $\tfrac{d^2y}{dx^2}$, which measures how fast the slope itself is changing.

Mission control does not describe a rocket's flight by an equation $y=f(x)$. It records where the rocket is at each instant: its horizontal distance $x(t)$ and its height $y(t)$, both as functions of time $t$. The same is true of a cricket ball tracked by Hawk-Eye, a satellite in orbit, or a valve on a bicycle wheel, which traces the arches of a cycloid as the wheel rolls. Describing a curve through a third variable — a parameter — is often far simpler than eliminating it, and it lets a curve loop back, cross itself or have vertical tangents, which no single function $y=f(x)$ can do. The chain rule tells us how to read off the slope $\tfrac{dy}{dx}$ directly from the two rates $\tfrac{dx}{dt}$ and $\tfrac{dy}{dt}$.

The second idea of this section answers a question every driver understands: a speedometer gives the rate of change of distance, but how quickly is the speed changing? Differentiating the derivative gives the second-order derivative — acceleration in mechanics, the bending of a curve in geometry, and the heart of the second-derivative test for maxima and minima in Chapter 6. Many Board questions ask you to prove that a function satisfies a relation between $y$, $\tfrac{dy}{dx}$ and $\tfrac{d^2y}{dx^2}$; these relations are the differential equations of Chapter 9.

Curves in Parametric Form

📘 Definition 5.6 — Parametric form of a curve

If the coordinates of a point on a curve are given as functions of a third variable $t$, $$x=f(t),\qquad y=g(t),\qquad t\in I$$ (an interval), the curve is said to be given in parametric form, and $t$ is the parameter. Each value of $t$ gives one point $(f(t),g(t))$.

curve	parametric form	Cartesian equation
circle, radius $a$	$x=a\cos t,\ y=a\sin t$	$x^2+y^2=a^2$
ellipse	$x=a\cos t,\ y=b\sin t$	$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
parabola	$x=at^2,\ y=2at$	$y^2=4ax$
cycloid	$x=a(t-\sin t),\ y=a(1-\cos t)$	(no simple one)
projectile	$x=(u\cos\alpha)t,\ y=(u\sin\alpha)t-\tfrac12gt^2$	a parabola

📐 Theorem 5.9 — Derivative of a function in parametric form

Let $x=f(t)$ and $y=g(t)$, where $f$ and $g$ are differentiable, and suppose $y$ is a differentiable function of $x$ along the curve. At every $t$ with $f'(t)\ne0$, $$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{g'(t)}{f'(t)}.$$

Why. By the chain rule, $\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$; divide by $\dfrac{dx}{dt}\ne0$. $\blacksquare$

Where $\dfrac{dx}{dt}=0$ but $\dfrac{dy}{dt}\ne0$, the point is moving straight up or down and the tangent is vertical; where both are $0$ the point momentarily stops, and the curve may have a sharp point (a cusp).

✏️ Example 1 — Slopes from parameters

Find $\dfrac{dy}{dx}$: (a) $x=a\cos\theta$, $y=a\sin\theta$ (b) $x=at^2$, $y=2at$ (c) $x=4t$, $y=\dfrac4t$ (d) the cycloid $x=a(\theta-\sin\theta)$, $y=a(1-\cos\theta)$

💡 Show solution▶

(a) $\dfrac{dx}{d\theta}=-a\sin\theta$, $\dfrac{dy}{d\theta}=a\cos\theta$, so $\dfrac{dy}{dx}=\dfrac{a\cos\theta}{-a\sin\theta}=-\cot\theta$ ($\sin\theta\ne0$). Check with Section 5.2: $-\dfrac xy=-\dfrac{a\cos\theta}{a\sin\theta}$ ✓.

(b) $\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac1t$ ($t\ne0$). At the vertex ($t=0$) the tangent is vertical.

(d) $\dfrac{dx}{d\theta}=a(1-\cos\theta)$, $\dfrac{dy}{d\theta}=a\sin\theta$. Using $1-\cos\theta=2\sin^2\tfrac\theta2$ and $\sin\theta=2\sin\tfrac\theta2\cos\tfrac\theta2$: $$\dfrac{dy}{dx}=\dfrac{\sin\theta}{1-\cos\theta}=\cot\dfrac\theta2\quad(\theta\ne0,2\pi,\dots).$$ At $\theta=2\pi$ both rates vanish: the cycloid has a cusp where the valve touches the road.

✏️ Example 2 — Parametric and implicit agree

The astroid is $x=a\cos^3t$, $y=a\sin^3t$. Find $\dfrac{dy}{dx}$ from the parametric form, and check it using its Cartesian equation $x^{2/3}+y^{2/3}=a^{2/3}$.

💡 Show solution▶

Parametric $\dfrac{dx}{dt}=-3a\cos^2t\sin t$, $\dfrac{dy}{dt}=3a\sin^2t\cos t$, so $\dfrac{dy}{dx}=\dfrac{3a\sin^2t\cos t}{-3a\cos^2t\sin t}=-\tan t$ (where $\sin t\cos t\ne0$).

Implicit $\tfrac23x^{-1/3}+\tfrac23y^{-1/3}\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{y^{1/3}}{x^{1/3}}=-\dfrac{a^{1/3}\sin t}{a^{1/3}\cos t}=-\tan t$ ✓.

Note At $t=0,\tfrac\pi2,\pi,\tfrac{3\pi}2$ both $\tfrac{dx}{dt}$ and $\tfrac{dy}{dt}$ vanish: the four cusps of the astroid (see Interactive 1).

✏️ Example 3 — The flight of a cricket ball

A ball is hit so that, $t$ seconds later, it is $x=20t$ m along the ground and $y=15t-4.9t^2$ m above the bat (air resistance ignored). (a) Find $\dfrac{dy}{dx}$ in terms of $t$. (b) At what angle does the ball leave the bat? (c) When is it moving horizontally, and how high is it then?

💡 Show solution▶

(a) $\dfrac{dx}{dt}=20$, $\dfrac{dy}{dt}=15-9.8t$, so $\dfrac{dy}{dx}=\dfrac{15-9.8t}{20}$.

(b) At $t=0$ the slope is $\tfrac{15}{20}=0.75$, so the angle with the ground is $\tan^{-1}0.75\approx36.9^\circ$.

(c) Horizontal when $\dfrac{dy}{dx}=0$: $t=\dfrac{15}{9.8}\approx1.53$ s. Then $y=15(1.531)-4.9(1.531)^2\approx11.5$ m — the top of the flight.

The point $(x(t),y(t))$ moves as $t$ increases. Its velocity has a horizontal part $\tfrac{dx}{dt}$ (red arrow) and a vertical part $\tfrac{dy}{dt}$ (blue arrow); the green tangent has slope $\tfrac{dy}{dx}=\tfrac{dy/dt}{dx/dt}$. Watch the cusps of the cycloid and the astroid, where both arrows shrink to nothing, and the top of the ball's flight, where the blue arrow vanishes.

Curve:$t$:

$t$—

$\tfrac{dx}{dt}$—

$\tfrac{dy}{dt}$—

$\tfrac{dy}{dx}$—

Second-Order Derivatives

📘 Definition 5.7 — Second-order derivative

Let $y=f(x)$ be differentiable. If its derivative $f'$ is itself differentiable, the derivative of $f'$ is the second-order derivative of $f$: $$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right),$$ also written $f''(x)$, $y''$ or $D^2y$. We also use the short forms $y_1=\dfrac{dy}{dx}$ and $y_2=\dfrac{d^2y}{dx^2}$.

If $s(t)$ is the position of a car, $s'(t)$ is its velocity and $s''(t)$ its acceleration. On a graph, $f''(x)$ is the rate at which the slope changes as you move to the right: where $f''\gt0$ the slope increases and the tangent turns anticlockwise — the curve bends upward (concave up, like a cup); where $f''\lt0$ the tangent turns clockwise and the curve bends downward (concave down, like a cap).

✏️ Example 4 — Differentiating twice

Find $\dfrac{d^2y}{dx^2}$: (a) $y=x^3+\tan x$ (b) $y=e^x\sin x$ (c) $y=\log x$ ($x\gt0$) (d) $y=\sin^{-1}x$ ($|x|\lt1$)

💡 Show solution▶

(a) $y_1=3x^2+\sec^2x$. Then $y_2=6x+2\sec x\cdot\sec x\tan x=6x+2\sec^2x\tan x$.

(b) $y_1=e^x\sin x+e^x\cos x=e^x(\sin x+\cos x)$. Then $y_2=e^x(\sin x+\cos x)+e^x(\cos x-\sin x)=2e^x\cos x$.

(c) $y_1=\dfrac1x=x^{-1}$, so $y_2=-x^{-2}=-\dfrac{1}{x^2}$: negative for every $x\gt0$, so $\log x$ always bends downward.

(d) $y_1=(1-x^2)^{-1/2}$, so $y_2=-\tfrac12(1-x^2)^{-3/2}\cdot(-2x)=\dfrac{x}{(1-x^2)^{3/2}}$.

✏️ Example 5 — Proving a relation between $y$, $y_1$ and $y_2$

(a) If $y=2\sin3x-5\cos3x$, show that $\dfrac{d^2y}{dx^2}+9y=0$. (b) If $y=(\sin^{-1}x)^2$, $|x|\lt1$, show that $(1-x^2)y_2-xy_1=2$. (c) If $y=e^{2x}(a+bx)$, show that $y_2-4y_1+4y=0$.

💡 Show solution▶

(a) $y_1=6\cos3x+15\sin3x$ and $y_2=-18\sin3x+45\cos3x=-9(2\sin3x-5\cos3x)=-9y$. So $y_2+9y=0$.

(b) step 1 $y_1=2\sin^{-1}x\cdot\dfrac{1}{\sqrt{1-x^2}}$. Clear the root first: $\sqrt{1-x^2}\,y_1=2\sin^{-1}x$.

(b) step 2 Differentiate again (product rule on the left): $\sqrt{1-x^2}\,y_2+\dfrac{-x}{\sqrt{1-x^2}}\,y_1=\dfrac{2}{\sqrt{1-x^2}}$. Multiply by $\sqrt{1-x^2}$: $(1-x^2)y_2-xy_1=2$. ✓ (Squaring or clearing denominators before the second differentiation keeps the algebra short.)

(c) $y_1=2e^{2x}(a+bx)+be^{2x}=2y+be^{2x}$. Differentiating, $y_2=2y_1+2be^{2x}=2y_1+2(y_1-2y)=4y_1-4y$. So $y_2-4y_1+4y=0$ — the constants $a,b$ have disappeared, as they must for a relation true for every $a$ and $b$.

✏️ Example 6 — Second derivatives of parametric and implicit curves

(a) For $x=at^2$, $y=2at$, find $\dfrac{d^2y}{dx^2}$. (b) For $x^2+y^2=25$, show that $\dfrac{d^2y}{dx^2}=-\dfrac{25}{y^3}$.

💡 Show solution▶

(a) first $\dfrac{dy}{dx}=\dfrac1t$ (Example 1(b)). This is a function of $t$, not of $x$.

(a) second By the chain rule, $\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac1t\right)=\dfrac{d}{dt}\left(\dfrac1t\right)\cdot\dfrac{dt}{dx}=-\dfrac{1}{t^2}\cdot\dfrac{1}{2at}=-\dfrac{1}{2at^3}$.

(b) $y_1=-\dfrac xy$. Quotient rule: $y_2=-\dfrac{y\cdot1-x\,y_1}{y^2}=-\dfrac{y+\frac{x^2}{y}}{y^2}=-\dfrac{y^2+x^2}{y^3}=-\dfrac{25}{y^3}$, using the equation of the circle in the last step.

Check (b) on the top half $y=\sqrt{25-x^2}$ at $x=3$: $y_2=-\tfrac{25}{64}\approx-0.3906$; a second difference $\dfrac{y(3.001)-2y(3)+y(2.999)}{0.001^2}$ gives $-0.3906$ ✓.

🧭 Problem-solving strategy — second derivatives

Explicit $y=f(x)$: differentiate, simplify, differentiate again.
Parametric: find $\dfrac{dy}{dx}$ as a function of $t$, then $\dfrac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$. It is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$.
Implicit: differentiate $y_1$ again, treating $y$ as a function of $x$; then replace $y_1$ by its formula and simplify with the original equation.
“Show that … = 0”: clear roots and denominators from the $y_1$ equation before differentiating again, and try to write $y_2$ in terms of $y$ and $y_1$.

The curve is coloured by the sign of $f''$: green where $f''\gt0$ (bending upward), red where $f''\lt0$ (bending downward). The lower strip plots the slope $f'(x)$. As the point moves, see the tangent turn anticlockwise on green stretches (the slope graph climbs) and clockwise on red ones (it falls). Where the colour changes, $f''=0$ and the slope graph turns round.

Function:$x$:

$x$—

$f'(x)$ (slope)—

$f''(x)$—

the tangent is turning—

⚠️ Common mistakes when studying alone

1. Dividing the second derivatives. For $x=f(t)$, $y=g(t)$: $\dfrac{d^2y}{dx^2}\ne\dfrac{g''(t)}{f''(t)}$. Differentiate $\dfrac{dy}{dx}$ with respect to $t$ and divide by $\dfrac{dx}{dt}$ once more.

2. Forgetting that $\dfrac{dy}{dx}$ is a function of $t$. $\dfrac{d}{dx}\left(\dfrac1t\right)$ is not $0$, and it is not $-\dfrac{1}{t^2}$; it is $-\dfrac{1}{t^2}\cdot\dfrac{dt}{dx}$.

3. Squaring the first derivative. $\dfrac{d^2y}{dx^2}$ is not $\left(\dfrac{dy}{dx}\right)^2$. For $y=x^3$, $y_2=6x$ but $(y_1)^2=9x^4$.

4. Using a parametric slope where $\dfrac{dx}{dt}=0$. There the tangent is vertical (or the curve has a cusp); the formula does not give a number.

5. Leaving $y_1$ inside an implicit $y_2$. Substitute the formula for $y_1$ and simplify with the curve's equation (Example 6(b)).

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026; not in the CBSE Class 12 syllabus or JEE Main. JEE Advanced uses two classical theorems that need exactly the hypotheses of this chapter.

Rolle's theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f'(c)=0$ for at least one $c$ in $(a,b)$. Picture: a smooth curve that starts and ends at the same height has a horizontal tangent somewhere in between.

Mean value theorem (Lagrange). If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then for at least one $c$ in $(a,b)$ $$f'(c)=\dfrac{f(b)-f(a)}{b-a}.$$ Picture: somewhere the tangent is parallel to the chord joining the end points. (If you drive $100$ km in $2$ hours, at some instant your speedometer read exactly $50$ km/h.)

Worked example. For $f(x)=x^2-4x$ on $[1,4]$: $f$ is a polynomial, so the hypotheses hold. The chord slope is $\dfrac{f(4)-f(1)}{4-1}=\dfrac{0-(-3)}{3}=1$, and $f'(c)=2c-4=1$ gives $c=2.5\in(1,4)$. The hypotheses matter: $|x|$ on $[-1,1]$ has $f(-1)=f(1)$ but no horizontal tangent — it is not differentiable at $0$.

Exercises

Parametric forms. Find $\dfrac{dy}{dx}$ (and the requested values).

1 $x=2t+1$, $y=t^2-3$
2 $x=3\cos\theta$, $y=2\sin\theta$
3 $x=t+\dfrac1t$, $y=t-\dfrac1t$
4 $x=e^t\cos t$, $y=e^t\sin t$
5 $x=\log t$, $y=t^2+1$ ($t\gt0$)
6 $x=\sin2t$, $y=\cos2t$
7 $x=\sec t$, $y=\tan t$; the value at $t=\tfrac\pi4$
8 $x=2(\theta-\sin\theta)$, $y=2(1-\cos\theta)$; the value at $\theta=\tfrac\pi2$
9 A thrown ball. $x=30t$, $y=40t-5t^2$ (metres, seconds). Find $\dfrac{dy}{dx}$ at $t=2$, and the time when the ball moves horizontally.

💡 Show solutions▶

1 $\dfrac{2t}{2}=t$

2 $\dfrac{2\cos\theta}{-3\sin\theta}=-\dfrac23\cot\theta$

3 $\dfrac{1+1/t^2}{1-1/t^2}=\dfrac{t^2+1}{t^2-1}$ ($t\ne\pm1$)

4 $\dfrac{dx}{dt}=e^t(\cos t-\sin t)$, $\dfrac{dy}{dt}=e^t(\sin t+\cos t)$, so $\dfrac{dy}{dx}=\dfrac{\sin t+\cos t}{\cos t-\sin t}$.

5 $\dfrac{2t}{1/t}=2t^2$

6 $\dfrac{-2\sin2t}{2\cos2t}=-\tan2t$

7 $\dfrac{\sec^2t}{\sec t\tan t}=\dfrac{\sec t}{\tan t}=\operatorname{cosec}t$; at $\tfrac\pi4$: $\sqrt2$.

8 As in Example 1(d), $\dfrac{dy}{dx}=\cot\dfrac\theta2$; at $\theta=\tfrac\pi2$: $\cot\tfrac\pi4=1$.

9 $\dfrac{dy}{dx}=\dfrac{40-10t}{30}$; at $t=2$: $\dfrac{20}{30}=\dfrac23$. Horizontal when $40-10t=0$: $t=4$ s.

Find the second-order derivative $\dfrac{d^2y}{dx^2}$.

1 $y=x^4-3x^2+2$
2 $y=x\cos x$
3 $y=\log(1+x^2)$
4 $y=e^{3x}\sin2x$
5 $y=\tan^{-1}x$
6 $y=x^2\log x$ ($x\gt0$)
7 $y=\sin(x^2)$
8 $y=\dfrac{1}{1+x}$

💡 Show solutions▶

1 $y_1=4x^3-6x$, $y_2=12x^2-6$

2 $y_1=\cos x-x\sin x$, $y_2=-\sin x-\sin x-x\cos x=-2\sin x-x\cos x$

3 $y_1=\dfrac{2x}{1+x^2}$, $y_2=\dfrac{2(1+x^2)-2x\cdot2x}{(1+x^2)^2}=\dfrac{2(1-x^2)}{(1+x^2)^2}$

4 $y_1=e^{3x}(3\sin2x+2\cos2x)$; $y_2=3e^{3x}(3\sin2x+2\cos2x)+e^{3x}(6\cos2x-4\sin2x)=e^{3x}(5\sin2x+12\cos2x)$

5 $y_1=(1+x^2)^{-1}$, $y_2=-\dfrac{2x}{(1+x^2)^2}$

6 $y_1=2x\log x+x$, $y_2=2\log x+2+1=2\log x+3$

7 $y_1=2x\cos(x^2)$, $y_2=2\cos(x^2)-4x^2\sin(x^2)$

8 $y_1=-(1+x)^{-2}$, $y_2=\dfrac{2}{(1+x)^3}$

III

Relations, parametric and implicit second derivatives, and motion.

1 If $y=A\cos2x+B\sin2x$, show that $\dfrac{d^2y}{dx^2}+4y=0$.
2 If $y=e^{-x}(a\cos x+b\sin x)$, show that $y_2+2y_1+2y=0$.
3 If $y=\log\!\left(x+\sqrt{1+x^2}\right)$, show that $(1+x^2)y_2+xy_1=0$.
4 If $y=5e^{2x}-3e^{-x}$, show that $y_2-y_1-2y=0$.
5 For $x=a\cos t$, $y=b\sin t$, find $\dfrac{d^2y}{dx^2}$.
6 For $x=t^2$, $y=t^3$, find $\dfrac{d^2y}{dx^2}$.
7 For $x^2-y^2=9$, show that $\dfrac{d^2y}{dx^2}=-\dfrac{9}{y^3}$.
8 Find the values of $k$ for which $y=e^{kx}$ satisfies $y_2-5y_1+6y=0$.
9 Motion. A particle moves along a line with position $s(t)=t^3-6t^2+9t$ metres at time $t$ seconds. Find its velocity and acceleration, the time at which the acceleration is zero, and the velocity at that time.

💡 Show solutions▶

1 $y_1=-2A\sin2x+2B\cos2x$, $y_2=-4A\cos2x-4B\sin2x=-4y$.

2 $y_1=-y+e^{-x}(-a\sin x+b\cos x)$. Differentiate: $y_2=-y_1-e^{-x}(-a\sin x+b\cos x)+e^{-x}(-a\cos x-b\sin x)=-y_1-(y_1+y)-y=-2y_1-2y$, using $e^{-x}(-a\sin x+b\cos x)=y_1+y$. So $y_2+2y_1+2y=0$.

3 $y_1=\dfrac{1}{\sqrt{1+x^2}}$ (Section 5.3, Exercise I.7), so $\sqrt{1+x^2}\,y_1=1$. Differentiate: $\sqrt{1+x^2}\,y_2+\dfrac{x}{\sqrt{1+x^2}}y_1=0$; multiply by $\sqrt{1+x^2}$: $(1+x^2)y_2+xy_1=0$.

4 $y_1=10e^{2x}+3e^{-x}$, $y_2=20e^{2x}-3e^{-x}$. $y_2-y_1-2y=(20-10-10)e^{2x}+(-3-3+6)e^{-x}=0$.

5 $\dfrac{dy}{dx}=-\dfrac ba\cot t$. Then $\dfrac{d^2y}{dx^2}=\dfrac{\frac ba\operatorname{cosec}^2t}{-a\sin t}=-\dfrac{b}{a^2\sin^3t}$.

6 $\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$, so $\dfrac{d^2y}{dx^2}=\dfrac{3/2}{2t}=\dfrac{3}{4t}$ ($t\ne0$).

7 $y_1=\dfrac xy$; $y_2=\dfrac{y-xy_1}{y^2}=\dfrac{y-\frac{x^2}{y}}{y^2}=\dfrac{y^2-x^2}{y^3}=-\dfrac{9}{y^3}$.

8 $y_1=ky$, $y_2=k^2y$, so $(k^2-5k+6)y=0$. Since $y=e^{kx}\ne0$: $k^2-5k+6=0$, $k=2$ or $k=3$.

9 $v=s'(t)=3t^2-12t+9$ m/s, $a=s''(t)=6t-12$ m/s². $a=0$ at $t=2$ s, when $v=12-24+9=-3$ m/s (moving backwards at $3$ m/s).

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: the Board sets a 2-mark parametric derivative (often evaluated at a given $\theta$), a second-derivative “show that” relation for 3 marks, and occasionally a projectile case study. JEE asks for parametric second derivatives and relations such as $(1-x^2)y_2-xy_1=\dots$.

E1. Board · 1 mark If $x=3t^2$ and $y=6t$, then $\dfrac{d^2y}{dx^2}$ equals (A) $-\dfrac{1}{t^2}$ (B) $-\dfrac{1}{6t^3}$ (C) $\dfrac{1}{6t^3}$ (D) $0$

💡 Show solution▶

Idea $\dfrac{dy}{dx}=\dfrac{6}{6t}=\dfrac1t$; then $\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}\left(\dfrac1t\right)\div\dfrac{dx}{dt}=-\dfrac{1}{t^2}\cdot\dfrac{1}{6t}$.

Answer (B). Trap (A) differentiates $\tfrac1t$ with respect to $t$ and forgets to divide by $\tfrac{dx}{dt}$; trap (D) divides $\tfrac{d^2y}{dt^2}=0$ by $\tfrac{d^2x}{dt^2}$.

E2. Board · 2 marks Find $\dfrac{dy}{dx}$ at $\theta=\tfrac\pi2$ for $x=a(\theta+\sin\theta)$, $y=a(1-\cos\theta)$.

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Rates $\dfrac{dx}{d\theta}=a(1+\cos\theta)$, $\dfrac{dy}{d\theta}=a\sin\theta$. (1 mark)

Answer $\dfrac{dy}{dx}=\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{2\sin\frac\theta2\cos\frac\theta2}{2\cos^2\frac\theta2}=\tan\dfrac\theta2$; at $\theta=\tfrac\pi2$ this is $\tan\tfrac\pi4=1$. (1 mark)

E3. Board · 3 marks If $y=3e^{2x}+2e^{3x}$, prove that $\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}+6y=0$.

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Derivatives $y_1=6e^{2x}+6e^{3x}$, $y_2=12e^{2x}+18e^{3x}$. (1½ marks)

Substitute $y_2-5y_1+6y=(12-30+18)e^{2x}+(18-30+12)e^{3x}=0+0=0$. (1½ marks)

E4. Board · Case study A model rocket at a school science fair is launched so that, $t$ seconds later, it is $x=50t$ m downrange and $y=50\sqrt3\,t-5t^2$ m high. (i) Find $\dfrac{dy}{dx}$ in terms of $t$. [2 marks] (ii) At what angle to the ground is it launched? [1 mark] (iii) When does it reach its greatest height? [1 mark]

💡 Show solution▶

(i) $\dfrac{dx}{dt}=50$, $\dfrac{dy}{dt}=50\sqrt3-10t$, so $\dfrac{dy}{dx}=\sqrt3-\dfrac t5$.

(ii) At $t=0$ the slope is $\sqrt3=\tan60^\circ$: launched at $60^\circ$.

(iii) The path is horizontal when $\sqrt3-\tfrac t5=0$: $t=5\sqrt3\approx8.66$ s. (The height is then $375$ m.)

E5. JEE Main If $y=e^{\sin^{-1}x}$, $|x|\lt1$, then $(1-x^2)y_2-xy_1$ equals (A) $0$ (B) $y$ (C) $-y$ (D) $2y$

💡 Show solution▶

Method $y_1=\dfrac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}=\dfrac{y}{\sqrt{1-x^2}}$, so $\sqrt{1-x^2}\,y_1=y$.

Again $\sqrt{1-x^2}\,y_2-\dfrac{x}{\sqrt{1-x^2}}y_1=y_1$. Multiply by $\sqrt{1-x^2}$: $(1-x^2)y_2-xy_1=\sqrt{1-x^2}\,y_1=y$.

Answer (B). Trap (A) is the answer for $y=\sin^{-1}x$ itself. ⏱ Fast route: test $x=0$: $y=1$, $y_1=1$, and differentiating $y_1=y(1-x^2)^{-1/2}$ gives $y_2=y_1(1-x^2)^{-1/2}+xy(1-x^2)^{-3/2}=1$ there; the expression is $1=y$, which rules out (A), (C) and (D).

E6. JEE Advanced (One or more options correct.) For the curve $x=2\cos t-\cos2t$, $y=2\sin t-\sin2t$, (A) $\dfrac{dy}{dx}=\tan\dfrac{3t}{2}$ (B) the tangent at $t=\tfrac\pi3$ is vertical (C) the tangent at $t=\tfrac{2\pi}{3}$ is horizontal (D) $\dfrac{d^2y}{dx^2}=0$ at $t=\tfrac{2\pi}3$

💡 Show solution▶

(A) True $\dfrac{dy}{dt}=2(\cos t-\cos2t)=4\sin\tfrac{3t}2\sin\tfrac t2$ and $\dfrac{dx}{dt}=2(\sin2t-\sin t)=4\cos\tfrac{3t}2\sin\tfrac t2$, so $\dfrac{dy}{dx}=\tan\dfrac{3t}{2}$ (where $\sin\tfrac t2\ne0$).

(B) True, (C) True At $t=\tfrac\pi3$: $\cos\tfrac{3t}2=\cos\tfrac\pi2=0$ while $\tfrac{dy}{dt}=4\cdot1\cdot\tfrac12\ne0$ — vertical. At $t=\tfrac{2\pi}3$: $\tan\pi=0$ — horizontal.

(D) False $\dfrac{d^2y}{dx^2}=\dfrac{\frac32\sec^2\frac{3t}2}{dx/dt}$. At $t=\tfrac{2\pi}3$: $\sec^2\pi=1$ and $\tfrac{dx}{dt}=2\left(-\tfrac{\sqrt3}2-\tfrac{\sqrt3}2\right)=-2\sqrt3$, giving $-\dfrac{\sqrt3}{4}\ne0$.

Answer (A), (B), (C).

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. From a phone video of a thrown ball, can you measure $g$, the acceleration due to gravity, using parametric derivatives?

Why it matters. Sports analysts (Hawk-Eye in cricket and tennis), traffic engineers and space agencies all reconstruct motion from position data sampled at fixed times. Velocities and accelerations are then estimated by numerical first and second derivatives — exactly $\tfrac{dx}{dt}$, $\tfrac{dy}{dt}$ and $\tfrac{d^2y}{dt^2}$ from this section.

What to do.

Film a ball thrown across a wall with a metre stick in view, holding the phone still and side-on (most phones record at $30$ frames per second, so frames are $\tfrac1{30}$ s apart).
Step through the video frame by frame and record the ball's position $(x,y)$ in metres for $15$–$25$ frames. (The free program Tracker does this automatically.)
In a spreadsheet, estimate $\tfrac{dx}{dt}$ and $\tfrac{dy}{dt}$ by central differences $\tfrac{x_{k+1}-x_{k-1}}{2\Delta t}$, and the slope of the path $\tfrac{dy}{dx}=\tfrac{dy/dt}{dx/dt}$. Where is the slope zero?
Estimate $\tfrac{d^2y}{dt^2}$ by $\tfrac{y_{k+1}-2y_k+y_{k-1}}{(\Delta t)^2}$. Average it, compare with $-9.8\ \text{m/s}^2$, and check that $\tfrac{d^2x}{dt^2}$ is close to $0$.

Tools: Phone camera, metre stick, spreadsheet; optionally the free Tracker video-analysis program.

Hand in: Your data table, plots of $x(t)$, $y(t)$ and the path, your estimate of $g$ with a percentage error, and a paragraph on why second differences are so much noisier than first differences.

Going further. Fit $y=at^2+bt+c$ to your data (spreadsheet trendline) and compare $2a$ with your second-difference estimate. Which is more reliable, and why?

Instructor note: careful work gives $g$ within $5$–$10\%$. Position errors of about $1$ cm become errors of order $\tfrac{0.01}{(1/30)^2}=9\ \text{m/s}^2$ in a single second difference, so individual values scatter wildly and averaging (or a quadratic fit, $g\approx-2a$) is essential — a vivid lesson on why second derivatives amplify noise.

🧪 Self-check quiz

0/5 correct

Q1. If $x=t^2$ and $y=2t$, then $\dfrac{dy}{dx}=$

Q2. $\dfrac{d^2}{dx^2}(\sin x)=$

Q3. If $y=e^{3x}$, then $y_2=$

Q4. For $x=f(t)$, $y=g(t)$, the second derivative $\dfrac{d^2y}{dx^2}$ equals

Q5. $y=A\cos3x+B\sin3x$ satisfies

← 5.3 Exponential and Logarithmic Functions; Logarithmic Differentiation Chapter 6 — Application of Derivatives →