Sometimes a curve is too tangled to write as y = f ( x ) y = f(x) y = f ( x ) third variable — a parameter t t t x x x y y y x = x ( t ) , y = y ( t ) x = x(t), \qquad y = y(t) x = x ( t ) , y = y ( t ) t t t ( x ( t ) , y ( t ) ) (x(t), y(t)) ( x ( t ) , y ( t )) slope of that curve in the x y xy x y x x x y y y t t t t t t divide .
Intuition Speed ratio = slopeImagine a bug walking along the curve. In a tiny instant d t dt d t
it moves d x = x ˙ d t dx = \dot x\, dt d x = x ˙ d t
it moves d y = y ˙ d t dy = \dot y\, dt d y = y ˙ d t
The slope of its path is "rise over run" = d y d x = y ˙ d t x ˙ d t \dfrac{dy}{dx} = \dfrac{\dot y\, dt}{\dot x\, dt} d x d y = x ˙ d t y ˙ d t d t dt d t cancels — slope doesn't care how fast you walk, only the ratio of vertical to horizontal speed.
We start from the chain rule , which is the only machinery we need.
y y y x x x x x x t t t d y d t = d y d x ⋅ d x d t \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} d t d y = d x d y ⋅ d t d x
This is just the chain rule applied to y y y t t t through x x x d x d t \dfrac{dx}{dt} d t d x d x d t ≠ 0 \dfrac{dx}{dt}\neq 0 d t d x = 0
Common mistake Steel-man the seductive wrong answerIt feels right to write
d 2 y d x 2 = ? y ¨ x ¨ \frac{d^2y}{dx^2} \stackrel{?}{=} \frac{\ddot y}{\ddot x} d x 2 d 2 y = ? x ¨ y ¨ y ˙ / x ˙ \dot y/\dot x y ˙ / x ˙ This is wrong.
Why it's tempting: the pattern looks self-similar.
The fix: d 2 y d x 2 \dfrac{d^2y}{dx^2} d x 2 d 2 y differentiate d y d x \dfrac{dy}{dx} d x d y x x x — not with respect to t t t t t t
Derivation. Let y ′ ≡ d y d x y' \equiv \dfrac{dy}{dx} y ′ ≡ d x d y t t t d 2 y d x 2 = d d x ( d y d x ) = d d x ( y ′ ) \frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dx}(y') d x 2 d 2 y = d x d ( d x d y ) = d x d ( y ′ )
But we only know how to differentiate things w.r.t. t t t same parametric trick on y ′ y' y ′ d d x ( y ′ ) = d ( y ′ ) / d t d x / d t \frac{d}{dx}(y') = \frac{\,d(y')/dt\,}{\,dx/dt\,} d x d ( y ′ ) = d x / d t d ( y ′ ) / d t
Now y ′ = y ˙ / x ˙ y' = \dot y/\dot x y ′ = y ˙ / x ˙ t t t d d t ( y ˙ x ˙ ) = y ¨ x ˙ − y ˙ x ¨ x ˙ 2 \frac{d}{dt}\!\left(\frac{\dot y}{\dot x}\right) = \frac{\ddot y\,\dot x - \dot y\,\ddot x}{\dot x^{2}} d t d ( x ˙ y ˙ ) = x ˙ 2 y ¨ x ˙ − y ˙ x ¨
Divide by x ˙ \dot x x ˙
Worked example 1 — A point on a circlex = cos t , y = sin t x = \cos t,\; y = \sin t x = cos t , y = sin t
x ˙ = − sin t \dot x = -\sin t x ˙ = − sin t y ˙ = cos t \dot y = \cos t y ˙ = cos t Why? differentiate each w.r.t. t t t d y d x = cos t − sin t = − cot t \dfrac{dy}{dx} = \dfrac{\cos t}{-\sin t} = -\cot t d x d y = − sin t cos t = − cot t Why this step? slope = speed ratio.Check: at t = π / 2 t = \pi/2 t = π /2 ( 0 , 1 ) (0,1) ( 0 , 1 ) = − cot ( π / 2 ) = 0 =-\cot(\pi/2)=0 = − cot ( π /2 ) = 0 x ¨ = − cos t , y ¨ = − sin t \ddot x = -\cos t,\ \ddot y = -\sin t x ¨ = − cos t , y ¨ = − sin t d 2 y d x 2 = ( − sin t ) ( − sin t ) − ( cos t ) ( − cos t ) ( − sin t ) 3 = sin 2 t + cos 2 t − sin 3 t = − 1 sin 3 t \dfrac{d^2y}{dx^2} = \dfrac{(-\sin t)(-\sin t) - (\cos t)(-\cos t)}{(-\sin t)^3} = \dfrac{\sin^2 t + \cos^2 t}{-\sin^3 t} = \dfrac{-1}{\sin^3 t} d x 2 d 2 y = ( − sin t ) 3 ( − sin t ) ( − sin t ) − ( cos t ) ( − cos t ) = − sin 3 t sin 2 t + cos 2 t = sin 3 t − 1 Why this step? plug into the boxed formula; numerator simplifies by sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1
Worked example 2 — A parabolay = x 2 y=x^2 y = x 2 x = t , y = t 2 x = t,\; y = t^2 x = t , y = t 2
x ˙ = 1 , y ˙ = 2 t ⇒ d y d x = 2 t = 2 x \dot x = 1,\ \dot y = 2t \Rightarrow \dfrac{dy}{dx} = 2t = 2x x ˙ = 1 , y ˙ = 2 t ⇒ d x d y = 2 t = 2 x x ¨ = 0 , y ¨ = 2 ⇒ d 2 y d x 2 = ( 2 ) ( 1 ) − ( 2 t ) ( 0 ) 1 3 = 2 \ddot x = 0,\ \ddot y = 2 \Rightarrow \dfrac{d^2y}{dx^2} = \dfrac{(2)(1)-(2t)(0)}{1^3} = 2 x ¨ = 0 , y ¨ = 2 ⇒ d x 2 d 2 y = 1 3 ( 2 ) ( 1 ) − ( 2 t ) ( 0 ) = 2 Why care? it confirms the formula reproduces the ordinary answer when x = t x=t x = t
Worked example 3 — Cycloid-type, wherex ¨ ≠ 0 \ddot x\neq 0 x ¨ = 0 x = t − sin t , y = 1 − cos t x = t - \sin t,\; y = 1 - \cos t x = t − sin t , y = 1 − cos t
x ˙ = 1 − cos t , y ˙ = sin t ⇒ d y d x = sin t 1 − cos t \dot x = 1 - \cos t,\ \dot y = \sin t \Rightarrow \dfrac{dy}{dx} = \dfrac{\sin t}{1-\cos t} x ˙ = 1 − cos t , y ˙ = sin t ⇒ d x d y = 1 − cos t sin t x ¨ = sin t , y ¨ = cos t \ddot x = \sin t,\ \ddot y = \cos t x ¨ = sin t , y ¨ = cos t d 2 y d x 2 = cos t ( 1 − cos t ) − sin t ⋅ sin t ( 1 − cos t ) 3 = cos t − cos 2 t − sin 2 t ( 1 − cos t ) 3 = cos t − 1 ( 1 − cos t ) 3 = − 1 ( 1 − cos t ) 2 \dfrac{d^2y}{dx^2} = \dfrac{\cos t(1-\cos t) - \sin t\cdot\sin t}{(1-\cos t)^3} = \dfrac{\cos t - \cos^2 t - \sin^2 t}{(1-\cos t)^3} = \dfrac{\cos t - 1}{(1-\cos t)^3} = \dfrac{-1}{(1-\cos t)^2} d x 2 d 2 y = ( 1 − cos t ) 3 cos t ( 1 − cos t ) − sin t ⋅ sin t = ( 1 − cos t ) 3 cos t − cos 2 t − sin 2 t = ( 1 − cos t ) 3 cos t − 1 = ( 1 − cos t ) 2 − 1 Why this step? used cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 cos t − 1 = − ( 1 − cos t ) \cos t - 1 = -(1-\cos t) cos t − 1 = − ( 1 − cos t ) x ¨ y ˙ \ddot x\, \dot y x ¨ y ˙ essential — drop it and you get a wrong answer.
Common mistake Other classic traps
Forgetting x ˙ \dot x x ˙ Vertical tangents (e.g. circle at ( ± 1 , 0 ) (\pm1,0) ( ± 1 , 0 ) d y / d x dy/dx d y / d x Mixing up which derivative goes on top. It's always y ˙ / x ˙ \dot y/\dot x y ˙ / x ˙ for , over the variable you differentiate against ).Using x ˙ 2 \dot x^2 x ˙ 2 x ˙ 3 \dot x^3 x ˙ 3 in the 2nd derivative. Steel-man: you'd be using d d t ( y ′ ) \frac{d}{dt}(y') d t d ( y ′ ) forgetting the last ÷ x ˙ \div\dot x ÷ x ˙ .
Recall Feynman: explain to a 12-year-oldImagine driving a car. A GPS tells you your east–west speed and your north–south speed separately. How steep is your path on the map? Just divide north-speed by east-speed — that's the slope. You never needed an equation for the road! For "how the slope itself is bending" (the second derivative), you can't just divide the accelerations ; you have to track how the slope changes as you move east, which means dividing by your east-speed one extra time — that's the mysterious "cube" x ˙ 3 \dot x^3 x ˙ 3
Mnemonic Remember the formulasFirst: "dot-y over dot-x ."
Second: "Top differentiated by bottom, minus bottom by top, all over x ˙ \dot x x ˙ " — same as quotient rule but with an extra x ˙ \dot x x ˙ "cube on the bottom or you've been fooled."
What is d y d x \dfrac{dy}{dx} d x d y x ( t ) , y ( t ) x(t),y(t) x ( t ) , y ( t ) d y / d t d x / d t = y ˙ x ˙ \dfrac{dy/dt}{dx/dt} = \dfrac{\dot y}{\dot x} d x / d t d y / d t = x ˙ y ˙ , valid when
x ˙ ≠ 0 \dot x\neq 0 x ˙ = 0 .
Why does the parameter d t dt d t d y / d x dy/dx d y / d x Slope is a ratio of vertical to horizontal change; the common factor
d t dt d t divides out, so walking speed is irrelevant.
Write the parametric second derivative formula. d 2 y d x 2 = y ¨ x ˙ − y ˙ x ¨ x ˙ 3 \dfrac{d^2y}{dx^2} = \dfrac{\ddot y\,\dot x - \dot y\,\ddot x}{\dot x^{3}} d x 2 d 2 y = x ˙ 3 y ¨ x ˙ − y ˙ x ¨ .
Why is d 2 y d x 2 ≠ y ¨ x ¨ \dfrac{d^2y}{dx^2}\neq \dfrac{\ddot y}{\ddot x} d x 2 d 2 y = x ¨ y ¨ Because
d 2 y / d x 2 d^2y/dx^2 d 2 y / d x 2 means differentiating
d y / d x dy/dx d y / d x w.r.t.
x x x (through
t t t ), giving a quotient-rule result divided again by
x ˙ \dot x x ˙ .
Where does the power x ˙ 3 \dot x^3 x ˙ 3 x ˙ 2 \dot x^2 x ˙ 2 from the quotient rule on
y ˙ / x ˙ \dot y/\dot x y ˙ / x ˙ , times one more
x ˙ \dot x x ˙ from the final
÷ d x / d t \div\, dx/dt ÷ d x / d t .
What happens geometrically when x ˙ = 0 \dot x = 0 x ˙ = 0 Vertical tangent;
d y / d x dy/dx d y / d x is undefined (slope
→ ∞ \to\infty → ∞ ).
For x = cos t , y = sin t x=\cos t,\,y=\sin t x = cos t , y = sin t d y / d x dy/dx d y / d x General method to get d 2 y / d x 2 d^2y/dx^2 d 2 y / d x 2 d y / d x dy/dx d y / d x Differentiate
d y / d x dy/dx d y / d x w.r.t.
t t t , then divide by
d x / d t dx/dt d x / d t .
Chain rule — the engine behind both formulas.Quotient rule — used to differentiate y ˙ / x ˙ \dot y/\dot x y ˙ / x ˙ Implicit differentiation — alternative when no parameter is available.Tangents and normals — slope d y / d x dy/dx d y / d x Concavity and second derivative — sign of d 2 y / d x 2 d^2y/dx^2 d 2 y / d x 2 Cycloid , Parametric curves — natural homes for this technique.
differentiate w.r.t. x not t
via chain rule and quotient rule
Parameter t drives x and y
Chain rule dy/dt = dy/dx times dx/dt
First derivative dy/dx = ydot/xdot
xdot not zero else vertical tangent
Tempting wrong ydotdot/xdotdot
ydotdot xdot minus ydot xdotdot over xdot cubed
Intuition Hinglish mein samjhoDekho, kabhi-kabhi curve itna complicated hota hai ki use seedha y = f ( x ) y=f(x) y = f ( x ) t t t x = x ( t ) x=x(t) x = x ( t ) y = y ( t ) y=y(t) y = y ( t ) t t t ( x , y ) (x,y) ( x , y ) d t dt d t x x x y y y d y d x = y ˙ x ˙ \frac{dy}{dx}=\frac{\dot y}{\dot x} d x d y = x ˙ y ˙ d t dt d t
Ab sabse bada trap: second derivative. Bahut log galti se d 2 y d x 2 = y ¨ x ¨ \frac{d^2y}{dx^2}=\frac{\ddot y}{\ddot x} d x 2 d 2 y = x ¨ y ¨ galat hai! Reason: d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y d y d x \frac{dy}{dx} d x d y fir se x x x differentiate karna, t t t y ˙ x ˙ \frac{\dot y}{\dot x} x ˙ y ˙ t t t x ˙ \dot x x ˙ y ¨ x ˙ − y ˙ x ¨ x ˙ 3 \frac{\ddot y\,\dot x-\dot y\,\ddot x}{\dot x^{3}} x ˙ 3 y ¨ x ˙ − y ˙ x ¨ cube (x ˙ 3 \dot x^3 x ˙ 3
Yeh kyun important hai? Circle, cycloid, projectile path — yeh sab natural roop se parametric form me aate hain. Tangent line, concavity (up/down bending), aur physics me velocity-acceleration sab yahi se nikalta hai. Practice ke liye circle (x = cos t , y = sin t x=\cos t,y=\sin t x = cos t , y = sin t 0 0 0
