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Worked solution

Solve the exact parameter values for a prescribed gradient

A curve has x=3t−1/tx=3t-1/t, y=t+4/(3t)y=t+4/(3t). Find the exact value(s) of tt at which the gradient is 1/61/6. marks

[2008/YJC/Promo/12(a)]

CJC · Review problems · Question 5

Answer

t=±3t=\pm\sqrt3.

Form and simplify the derivative ratio

For t≠0t\ne0, dx/dt=3+1/t2=(3t2+1)/t2dx/dt=3+1/t^2=(3t^2+1)/t^2 and dy/dt=1−4/(3t2)=(3t2−4)/(3t2)dy/dt=1-4/(3t^2)=(3t^2-4)/(3t^2). Thus dy/dx=(3t2−4)/[3(3t2+1)]dy/dx=(3t^2-4)/[3(3t^2+1)].

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Set the ratio equal to the given slope

(3t2−4)/[3(3t2+1)]=1/6(3t^2-4)/[3(3t^2+1)]=1/6 gives 6(3t2−4)=3(3t2+1)6(3t^2-4)=3(3t^2+1), then 6t2−8=3t2+16t^2-8=3t^2+1, so t2=3t^2=3. Both t=3t=\sqrt3 and t=−3t=-\sqrt3 are allowed, with dx/dt>0dx/dt>0 at each.

How did this part go?

Common mistakes

A squared parameter equation generally gives both positive and negative values unless the original domain excludes one.

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