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JEE Mains Maths · Differential Equations

Curves and Growth from Rates

Setting up the equation from words: a curve described by its tangent, normal or the area under it, and a quantity whose rate of change is proportional to itself or to its distance from a fixed value.

All Differential Equations notesJEE Mains Maths strategy

Why this matters

Thirteen PYQs, nine of them multiple choice. Once the equation is written it is separable or linear; the marks are in translating the condition. Nine describe a curve through its tangent, normal or area, and four are growth, cooling or population models. Two ideas cover the page.

Concept 1 of 2: Curves from tangent, normal and area conditions

Write the tangent at (x,y)(x,y) as Y−y=y′(X−x)Y-y=y'(X-x) and read off what the condition needs: its xx-intercept x−yy′x-\frac{y}{y'}, its yy-intercept y−xy′y-xy', or the normal Y−y=−1y′(X−x)Y-y=-\frac{1}{y'}(X-x). An area condition ∫axy dt=F(x,y)\int_a^xy\,dt=F(x,y) becomes an equation after differentiating in xx. The result is separable, homogeneous or linear.

Definition

  • Tangent intercepts: xx-axis at x−yy′x-\frac{y}{y'}, yy-axis at y−xy′y-xy'.
  • Slope of the normal: −1y′-\frac{1}{y'}.
  • Normals through a fixed point (a,b)(a,b): (x−a)+(y−b)y′=0(x-a)+(y-b)y'=0, a circle centred there.
  • Area ∫axy dt=F(x,y)⇒y=ddxF(x,y)\int_a^xy\,dt=F(x,y)\Rightarrow y=\frac{d}{dx}F(x,y).

Tangent at (x, y)

Y−y=dydx(X−x):X-int=x−yy′,  Y-int=y−xy′Y-y=\frac{dy}{dx}(X-x):\quad X\text{-int}=x-\frac{y}{y'},\ \ Y\text{-int}=y-xy'

Worked example

The tangent at every point of a curve meets the yy-axis at height 2y2y. The curve passes through (1,1)(1,1). Find it.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q158Moderate

Example 1 · Differential Equations · Curves and Growth from Rates

A particle is moving in the xy-plane along a curve C passing through the point (3,3)(3,3). The tangent to the curve CC at the point PP meets the XX-axis at QQ. If the yy-axis bisects the segment PQPQ, then CC is a parabola with

Midpoint and intercept are not the same

'The yy-axis bisects PQPQ' means the midpoint has x=0x=0, i.e. x+(x−yy′)=0x+\left(x-\frac{y}{y'}\right)=0. Setting the intercept itself to zero is a different condition.
Drill 8 more on curves from tangent, normal and area conditions

Concept 2 of 2: Growth, decay and cooling

'Rate proportional to the amount' is dNdt=kN\frac{dN}{dt}=kN, so N=N0ektN=N_0e^{kt}. 'Rate proportional to the difference from a fixed value AA' — Newton's cooling, or a population with a constant loss — is dTdt=−k(T−A)\frac{dT}{dt}=-k(T-A), so T−A=(T0−A)e−ktT-A=(T_0-A)e^{-kt}. Equal time steps multiply the difference by the same factor, which is usually quicker than finding kk.

Definition

  • N′=kN⇒N=N0ektN'=kN\Rightarrow N=N_0e^{kt}; doubling time ln⁡2k\frac{\ln2}{k}.
  • T′=−k(T−A)⇒T−A=(T0−A)e−ktT'=-k(T-A)\Rightarrow T-A=(T_0-A)e^{-kt}.
  • P′=aP−b⇒P−ba=(P0−ba)eatP'=aP-b\Rightarrow P-\frac ba=\left(P_0-\frac ba\right)e^{at}.
  • Equal intervals: the ratio of successive differences is constant.

Newton's law of cooling

dTdt=−k(T−A) ⇒ T−A=(T0−A) e−kt\frac{dT}{dt}=-k(T-A)\ \Rightarrow\ T-A=(T_0-A)\,e^{-kt}

Worked example

A body at 100∘100^\circ cools to 60∘60^\circ in 10 minutes in a room at 20∘20^\circ. Find its temperature after 20 minutes.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q158Moderate

Example 2 · Differential Equations · Curves and Growth from Rates

The temperature T(t)T(t) of a body at time t=0t = 0 is 160∘160^{\circ} FF and it decreases continuously as per the differential equation dTdt=−K(T−80)\frac{dT}{dt}= - K(T - 80), where KK is positive constant. If T(15)=120∘FT(15) =120^{\circ}F, then T(45)T(45) is equal to

The difference decays, not the temperature

In cooling, T−AT-A is multiplied by e−kte^{-kt}, not TT itself. Halving the temperature instead of its excess over the room gives a wrong answer.
Drill 3 more on growth, decay and cooling

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Curves from tangent, normal and area conditions

    Tangent at (x, y)

    Y−y=dydx(X−x):X-int=x−yy′,  Y-int=y−xy′Y-y=\frac{dy}{dx}(X-x):\quad X\text{-int}=x-\frac{y}{y'},\ \ Y\text{-int}=y-xy'
  • Growth, decay and cooling

    Newton's law of cooling

    dTdt=−k(T−A) ⇒ T−A=(T0−A) e−kt\frac{dT}{dt}=-k(T-A)\ \Rightarrow\ T-A=(T_0-A)\,e^{-kt}

Watch out for (2)

Test yourself on Differential Equations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Differential Equations chapter testDrill all 13 questions