1. ## BC calc problems

1973 -
Given the curve $\displaystyle x + xy +2y^2 = 6$, find the coordinates of all other points on this curve with slop equal to the slope at (2,1)

1970 -
the function $\displaystyle f$ is defined for all x not equal to 0, in the interval $\displaystyle -1<x<1$ by $\displaystyle f(x)=\frac{x-\sin 2x}{\sin x}$

a) how should $\displaystyle f(0)$ be defined in order that $\displaystyle f$ be continuous for all x in the interval $\displaystyle -1<x<1$?
b) With $\displaystyle f(0)$ defined as in part a), use the definition of the derivative to determine whether $\displaystyle f'(0)$ exists. Show your work.

For 1973, ive been able to find the derivative as $\displaystyle \frac{dy}{dx} = \frac{-y-1}{x+4y}$. However, ive tried solving it with the original equation but cannot get an answer

For 1970, I've got $\displaystyle f(0) = -1$, but using the definition of the derivitive gets really really messy and i cant get the h to cancel out.

all help appreciated!!

2. Originally Posted by dizzystudent
1973 -
Given the curve $\displaystyle x + xy +2y^2 = 6$, find the coordinates of all other points on this curve with slop equal to the slope at (2,1)
That's the correct formula for the derivative. Set x = 2, y = 1 to find the slope. Then set the derivative equal to this slope to get an equation for the unknown point(s). It will be linear, so you can solve it for x or y (you choose). The other equation is the original equation $\displaystyle x + xy +2y^2 = 6$. When you eliminate one of the two unknowns, you will be left with a quadratic equation for the other one. One solution corresponds to the point (2,1) which you know already, the other one is new.
1970 -
the function $\displaystyle f$ is defined for all x not equal to 0, in the interval $\displaystyle -1<x<1$ by $\displaystyle f(x)=\frac{x-\sin 2x}{\sin x}$
$\displaystyle \lim_{x \to 0} f(x) = -1$ is correct. Mathcad tells me the limit of the difference quotient is 0. Are you allowed to use L'Hopital's rule? Otherwise the only thing you can do is use $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$.

3. thanks for helping me with those... but i still need help with part B on my second question.

4. Originally Posted by hpe
$\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 0$.
It's 1. I see you fixed it though.

5. Originally Posted by dizzystudent
thanks for helping me with those... but i still need help with part B on my second question.
Agreed. This is very messy, but post what you've done and I'll see if I can help work it out.

6. Originally Posted by Jameson
Agreed. This is very messy, but post what you've done and I'll see if I can help work it out.
Ok, here goes. Define $\displaystyle f(x) = \frac{x-\sin 2x}{\sin x}$ for $\displaystyle x \ne 0$, $\displaystyle f(0) = -1$. The goal is to show that $\displaystyle f'(0) = 0$, using the definition.

So set up the difference quotient:
$\displaystyle \frac{\frac{x-\sin 2x}{\sin x} + 1}{x}=\frac{x - \sin 2x + \sin x}{x \sin x}= \frac{x - 2 \sin x \cos x + \sin x}{x \sin x}$.

Recall that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$; thus it suffices to determine the limit of $\displaystyle \frac{x - 2 \sin x \cos x + \sin x}{x^2} = \frac{x - \sin x}{x^2} + \frac{2\sin x}{x} \cdot \frac{1 - \cos x}{x}$.

Since $\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{(1 - \cos x)(1+ \cos x)}{x(1 + \cos x)}$ which becomes $\displaystyle \lim_{x \to 0} \frac{-\sin^2 x}{x(1 + \cos x)} = 0$, we only need to discuss $\displaystyle \lim_{x \to 0}\frac{x - \sin x}{x^2}$.

Now look at the proof of $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$. In the course of that proof, the estimate $\displaystyle \sin x \cos x \le x \le \tan x$ is shown.

See the picture below: the big right triangle has an angle x and cathetes 1 and $\displaystyle \tan x$. The red right triangle has area $\displaystyle \frac{1}{2}\sin x \cos x$, the red plus blue sector has area $\displaystyle \frac{x}{2}$, and red + blue + yellow has an area equal to $\displaystyle \frac{1}{2} \tan x$.

It follows that $\displaystyle x\cos x \le \sin x \le x\sec x$.

This implies $\displaystyle \frac{x - x \sec x}{x^2} \le \frac{x - \sin x}{x^2} \le \frac{x - x\cos x}{x^2}$ or $\displaystyle \frac{\cos x - 1}{x \cos x} = \frac{1 - \sec x}{x} \le \frac{x - \sin x}{x^2} \le \frac{1 - \cos x}{x}$ .

As shown above, $\displaystyle \lim_{x \to 0}\frac{1-\cos x}{x} = 0$. Therefore also $\displaystyle \lim_{x \to 0}\frac{x - \sin x}{x^2} = 0$.

QED