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Worked solution

Differentiate a rational and logarithmic parametrisation

A curve has x=t/(1+t)x=t/(1+t), y=ln⁡(cos⁡t)y=\ln(\cos t), where the printed exclusions are t≠−1,π/2t\ne-1,\pi/2. Find dy/dxdy/dx in terms of tt. marks

[2009/CJC/Promo/10(a)]

CJC · Review problems · Question 4

Answer

dy/dx=−(1+t)2tan⁡tdy/dx=-(1+t)^2\tan t on t≠−1t\ne-1, cos⁡t>0\cos t>0.

Differentiate each coordinate

dx/dt=[(1+t)−t]/(1+t)2=1/(1+t)2dx/dt=[(1+t)-t]/(1+t)^2=1/(1+t)^2. Also dy/dt=−sin⁡t/cos⁡t=−tan⁡tdy/dt=-\sin t/\cos t=-\tan t.

How did this part go?

Divide in the correct order and state the real domain

(dy/dx)=(dy/dt)/(dx/dt)=−(1+t)2tan⁡t(dy/dx)=(dy/dt)/(dx/dt)=-(1+t)^2\tan t. For a real curve the logarithm also requires cos⁡t>0\cos t>0, in addition to t≠−1t\ne-1. The source exclusions alone are not a complete real-domain description; the derivative holds wherever the original real parametrisation is defined.

How did this part go?

Common mistakes

The logarithm needs a positive argument; excluding only pi/2 is not enough to describe its real domain.

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