Question

a point moves on the x - axis in such a way that its velocity at time t (t > 0) is given by v = \frac{\ln t}{t}. at what value of t does v attain its maximum?
a 1
b e^{1/2}
c e
d e^{3/2}
e there is no maximum value for v.

Explanation:

Step1: Find the derivative of \(v\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = \ln t\), \(u^\prime=\frac{1}{t}\), and \(v=t\), \(v^\prime = 1\). Then \(v^\prime=\frac{\frac{1}{t}\times t-\ln t\times1}{t^{2}}=\frac{1 - \ln t}{t^{2}}\)

Step2: Set the derivative equal to zero

Set \(v^\prime=\frac{1-\ln t}{t^{2}} = 0\). Since \(t^{2}>0\) for \(t>0\), we solve \(1-\ln t = 0\).

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Using the property \(y = \ln x\Leftrightarrow x = e^{y}\), we get \(t = e\)

Step3: Check the second - derivative (optional for multiple - choice)

Find the second - derivative \(v^{\prime\prime}\). Using the quotient rule again on \(v^\prime=\frac{1-\ln t}{t^{2}}\), let \(u = 1-\ln t\), \(u^\prime=-\frac{1}{t}\), and \(v=t^{2}\), \(v^\prime = 2t\). Then \(v^{\prime\prime}=\frac{-\frac{1}{t}\times t^{2}-(1 - \ln t)\times2t}{t^{4}}=\frac{-t-2t + 2t\ln t}{t^{4}}=\frac{2\ln t-3}{t^{3}}\)
When \(t = e\), \(v^{\prime\prime}=\frac{2\times1 - 3}{e^{3}}=-\frac{1}{e^{3}}<0\), so \(v\) has a maximum at \(t = e\)

Answer:

C. \(e\)