Porisma

Worked solution

Use two point conditions and a slowing-distance model

(a) Find y=f(x)y=f(x) through (1,0)(1,0) and (−1,2)(-1,2) with dy/dx=3x2−x−2+Cdy/dx=3x^2-x^{-2}+C. marks

(b) Perry jogs a distance xx kilometres in tt hours. The difference between his initial speed and his speed at time tt is proportional to distance covered. He starts at 55 km/h and travels at 44 km/h after 22 km. Find the exact time for a 55 km jog. marks

[2013/YJC/I/11 (modified)]

CJC · Review problems · Question 5

Answer

(a) y=x3+x−1−3x+1y=x^3+x^{-1}-3x+1; (b) 2ln⁡22\ln2 hours.

(a) Integrate the given family and solve the constants

y=x3+x−1+Cx+Dy=x^3+x^{-1}+Cx+D. The two points give C+D=−2C+D=-2 and −C+D=4-C+D=4, hence C=−3C=-3, D=1D=1 and y=x3+x−1−3x+1y=x^3+x^{-1}-3x+1, x≠0x\ne0. This retains the source convention of a common algebraic formula and constants across its two disconnected x-domains.

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(b) Express the speed reduction in terms of distance

5−dx/dt=kx5-dx/dt=kx. At x=2x=2, dx/dt=4dx/dt=4 gives 1=2k1=2k, so k=1/2k=1/2 and dx/dt=5−x/2=(10−x)/2dx/dt=5-x/2=(10-x)/2.

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(b) Separate, apply the start condition and evaluate

dt/dx=2/(10−x)dt/dx=2/(10-x), hence t=−2ln⁡∣10−x∣+Ct=-2\ln|10-x|+C. At t=0t=0, x=0x=0, C=2ln⁡10C=2\ln10. During the jog 0≤x<100\le x<10, so t=2ln⁡[10/(10−x)]t=2\ln[10/(10-x)]. At x=5x=5, t=2ln⁡2t=2\ln2 hours. The model approaches the limiting distance 1010 km only as time tends to infinity.

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Common mistakes

The rate is speed in kilometres per hour and the proportionality is to distance, not to elapsed time.

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