Problems

1. The continuous function \(\f\) is defined by \[ \tan \f(x) = x \ \ \ \ \ (-\infty < x <\infty) \] and \(\f(0)=\pi\). Sketch the curve \(y=\f(x)\).
2. The continuous function \(\g\) is defined by \[ \tan \g(x) = \frac x {1+x^2} \ \ \ \ \ \ (-\infty < x <\infty) \] and \(\g(0)=\pi\). Sketch the curves \(y= \dfrac x {1+x^2} \ \) and \(y=\g(x)\).
3. The continuous function \(\h \) is defined by \(\h (0)=\pi\) and \[ \tan \h (x)= \frac x {1-x^2}\, \ \ \ \ \ (x \ne \pm 1) \,. \] (The values of \(\h (x)\) at \(x=\pm1\) are such that \(\h (x)\) is continuous at these points.) Sketch the curves \(y= \dfrac x {1-x^2} \ \) and \(y=\h (x)\).
4. [Not on original exam] The continuous functions \(\h_1\) and \(\h_2\) are defined by: \(\h_1(0)=\h_2(0)=\pi \), \[ \tan \h_1(x) = \frac {x+x^4} {1+x^2+x^4} \ \ \ \ \ \text{and} \ \ \ \ \ \ \tan \h_2(x) = \frac {4x-x^3} {1-x^4} \,. \] for values of \(x\) at which the right hand sides are defined. Find \(\lim\limits_{x\to\infty}\h_1(x)\) and \(\lim\limits_{x\to\infty}\h_2(x)\,\).

Show Solution

---

Solution:

1. \(\,\)
   

   + - ↺
2. \(\,\)
   

   + - ↺
3. \(\,\)
   

   + - ↺
4. Note that \(\frac{x+x^4}{1+x^2+x^4}\) is continuous, and nicely behaved on \((-\infty, \infty)\) so we can see that \(\lim_{x \to \infty} h_1(x) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\). \(\frac{4x-x^3}{1-x^4}\) on the other hand has asymptotes at \(\pm 1\). So as as \(x \to 1\), \(h_1(x) \to \pi + \frac{\pi}{2} = \frac{3\pi}{2}\). Then as \(x \to \infty\) we increase by another \(\frac{\pi}{2}\), so \(\lim_{x \to \infty} h_2(x) = 2\pi\)

An alternative way to think about the last two parts is to consider \(h\) as giving the (continuous) argument (shifted by \(\pi\)) of \((1-t^2)+it\) (blue), \((1+t^2+t^4)+i(t+t^4)\) (orange) or \((1-t^4)+i(4t-t^3)\) (green). We can see the orange line never wraps around the origin, so the argument is always easy to find. The blue does one full circuit, from \(-\pi\) to \(\pi\) (or \(0\) to \(2\pi\) in our world. And the green line also does a full \(2\pi\) loop.

+ - ↺

Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]

Find the area of the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x \le a\). What happens to this area as \(a\) tends to infinity? Find the volume of the solid obtained when the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x\le a\), is rotated through \(2 \pi\) radians about the \(x\)-axis. What happens to this volume as \(a\) tends to infinity?