- Given that \(y = \ln [t + \sqrt{(1 + t^2)}]\), show that \(\frac{dy}{dt} = \frac{1}{\sqrt{(1+t^2)}}\). [3]
The curve \(C\) has parametric equations
$$x = \frac{1}{\sqrt{(1+t^2)}}, \quad y = \ln [t + \sqrt{(1 + t^2)}], \quad t \in \mathbb{R}.$$
A student was asked to prove that, for \(t > 0\), the gradient of the tangent to \(C\) is negative.
The attempted proof was as follows:
$$y = \ln \left(t + \frac{1}{x}\right)$$
$$= \ln \left(\frac{tx + 1}{x}\right)$$
$$= \ln (tx + 1) - \ln x$$
$$\therefore \frac{dy}{dx} = \frac{t}{tx + 1} - \frac{1}{x}$$
$$= \frac{\frac{t}{x}}{t + \frac{1}{x}} - \frac{1}{x}$$
$$= \frac{t\sqrt{(1+t^2)}}{t + \sqrt{(1+t^2)}} - \sqrt{(1 + t^2)}$$
$$= -\frac{(1+t^2)}{t + \sqrt{(1+t^2)}}$$
As \((1 + t^2) > 0\), and \(t + \sqrt{(1 + t^2)} > 0\) for \(t > 0\), \(\frac{dy}{dx} < 0\) for \(t > 0\).
- Identify the error in this attempt.
- Give a correct version of the proof. [6]
- Prove that \(\ln [-t + \sqrt{(1 + t^2)}] = -\ln [t + \sqrt{(1 + t^2)}]\). [3]
- Deduce that \(C\) is symmetric about the \(x\)-axis and sketch the graph of \(C\). [3]