Porisma

Worked solution

Apply initial data and an inverse-square gradient law

(a) Solve cos⁡2x dy/dx=1\cos^2x\,dy/dx=1 for 0<x<π/20<x<\pi/2, given y=0y=0 when x=π/4x=\pi/4, expressing y as a function of x.

(b) Find the equation of a curve through (1,2)(1,2) and (1/4,−10)(1/4,-10) whose gradient is inversely proportional to x2x^2.

CJC · Lecture · Self-Practice 1

(a) Integrate and use the initial condition

On the given interval dy/dx=sec⁡2xdy/dx=\sec^2x. Thus y=tan⁡x+Cy=\tan x+C. At x=π/4x=\pi/4, 0=1+C0=1+C, so y=tan⁡x−1y=\tan x-1 on 0<x<π/20<x<\pi/2.

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(b) Determine the proportionality and integration constants

Write dy/dx=k/x2dy/dx=k/x^2 for x≠0x\ne0. Integration gives y=−k/x+Cy=-k/x+C. The points give 2=−k+C2=-k+C and −10=−4k+C-10=-4k+C. Subtracting gives k=4k=4, then C=6C=6. Therefore y=−4/x+6y=-4/x+6. The connected branch through both given positive x-values has x>0x>0.

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Common mistakes

Self-Practice 1 contains two asks across two pages; the second curve must also be determined.

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