Few hard questions

**biloon** · September 24th 2010, 08:43 AM

1) Use the definition of derivative to show that function y = g(x) implicitly defined by equation: $y^{2/3} + x^{2/3} = 1$ is not differentiable at point (0,1)

2) Find all points at which the tangent lines to the curve are horizontal for equation $2(x^2 + y^2)^2 = x^2 - y^2$

3) Let $f(x) = sin x / (1+cos x)$ , Show that $f'(x) = 1/(1+cos x)$

These questions are from my university's past exams. It won't be much problem for me, IF they give me more time to solve it... The first two problems are given like only about 10 minute to solve, and the last one is not even 5 minute (Given 180 min for 17 problems, but last one is like part a of the problem)

So, it either means 1) I am doing it the long way through, or 2) the university examiners don't expect anyone to pass the exam... Therefore, I want to ask for opinions on how to solve these types of complex questions in a more time-efficient way.

PS: I used 1-2 hours just to pull that lim h -> 0 of $((1-h^{2/3})^{3/2} - 1 )/h$ from the first problem

**emakarov** · September 24th 2010, 03:08 PM

1) Use the definition of derivative to show that function y = g(x) implicitly defined by equation: $y^{2/3} + x^{2/3} = 1$ is not differentiable at point (0,1)

As you wrote, $y(x)=(1-x^{2/3})^{3/2}$ . Using Maclaurin series for $(1+x)^\alpha$ , we get $y(x)=1-3x^{2/3}/2+o(x^{2/3})$ . Therefore, $(y(x)-y(0))/x=-3/(2x^{1/3})+o(1/x^{1/3})$ . This expression tends to $-\infty$ when $x\to0^+$ and to $\infty$ when $x\to0^+$ , so there is no limit.

It also helps to have a sketch of the graph of y(x). This function is symmetric w.r.t. the vertical x = 0 axis because x occurs squared. Also, the behavior of $x^{2/3}$ near 0 is similar to that of $\sqrt{x}$ , in particular, $(x^{2/3})'\to\infty$ when $x\to0^+$ . Therefore, $1-x^{2/3}$ has a sharp angle at (0,1), which is its maximum point. I believe that raising everything to power 3/2 does not fundamentally change this behavior.

3) Let $f(x) = \sin x / (1+\cos x)$ , Show that $f'(x) = 1/(1+\cos x)$

This is pretty straightforward using the formula for the derivative of a fraction and the fact that $\sin^2x+\cos^2x=1$ .

**biloon** · September 24th 2010, 07:38 PM

I have never used Maclaurin series before. Therefore, I don't really understand the term $o(x^{2/3})$ , so if you can explain me, then that would be great. Also I did use L'Hopital Rule for solving 1st question and I check limit left-hand = right-hand to prove it's not differentiable. However, when I apply the L'Hopital Rule, I get a constant as the limit, not infinity. Therefore, I'm not sure if I'm applying it in the right situation or not. For the 3rd problem, i think they want a proof? I'm not sure, but for me.. using the quotient derivative is kind of "too easy for an exam question".

**mr fantastic** · September 25th 2010, 02:36 AM

Originally Posted by biloon

[snip] For the 3rd problem, i think they want a proof? I'm not sure, but for me.. using the quotient derivative is kind of "too easy for an exam question".

Nonsense. You're asked to find a derivative. The method is not prescribed. Therefore using the quotient rule is the obvious approach to take. A competent student will get it done easily in under 5 minutes.

Originally Posted by biloon

[snip]

2) Find all points at which the tangent lines to the curve are horizontal for equation $2(x^2 + y^2)^2 = x^2 - y^2$

[snip]

The clock starts ...... now:

Using implicit differentiation:

$\displaystyle 4(x^2 + y^2) \left( 2x + 2y \frac{dy}{dx} \right) = 2x + 2y \frac{dy}{dx}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{x(1 - 4x^2 - 4y^2)}{y(2x^2 + 2y^2 - 1)}$ .

Solve $\displaystyle \frac{dy}{dx} = 0$ subject to $2(x^2 + y^2)^2 = x^2 - y^2$ :

$x(1 - 4x^2 - 4y^2) = 0$ .... (A)

$2(x^2 + y^2)^2 = x^2 - y^2$ .... (B)

From (A):

Case 1: x = 0. Substitute into (B): y = 0. However, dy/dx is indeterminant at (0, 0) so that point is probably a crunode. (Certainly a crunode if you recognise the given relation as defining a lemniscate).

Case 2: $4x^2 + 4y^2 = 1$ . Substitute into (B) etc. You have 5 minutes left, plenty of time. Test that dy/dx is not indeterminant for the solutions. (Here's a short cut: solve 4x^2 + 4y^2 = 1 and 2(x^2 + y^2)^2 = x^2 - y^2 - Wolfram|Alpha)

**HallsofIvy** · September 25th 2010, 03:41 AM

A "crunode" is a point at which a graph intersects itself. (I had to look that up!)

Also called a "double point". (Which I do recognize.)

**emakarov** · September 25th 2010, 08:39 AM

I have never used Maclaurin series before. Therefore, I don't really understand the term $o(x^{2/3})$ , so if you can explain me, then that would be great.

I suppose it would not help telling that Maclaurin series is Taylor series at x = 0. Little-o notation $f(x)=o(g(x))$ when $x\to\alpha$ means that $f(x)$ becomes negligible compared to $g(x)$ when $x\to\alpha$ .

Also I did use L'Hopital Rule for solving 1st question and I check limit left-hand = right-hand to prove it's not differentiable. However, when I apply the L'Hopital Rule, I get a constant as the limit, not infinity. Therefore, I'm not sure if I'm applying it in the right situation or not.

Using l'Hôpital's rule to find $\displaystyle\lim_{x\to0}\frac{(1-x^{2/3})^{3/2}-1}{x}$ seems justified because the limits of the nominator and denominator are both 0. Further, $\displaystyle[(1-x^{2/3})^{3/2}-1]'=3/2(1-x^{2/3})^{1/2}(-(2/3)x^{-1/3})=-\frac{(1-x^{2/3})^{1/2}}{x^{1/3}}$ . Now, the nominator tends to 1, so the limit is $-\infty$ when $x\to0^+$ , and it is $+\infty$ when $x\to0^-$ .

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