BC calc problems

**dizzystudent** · November 11th 2005, 08:28 AM

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

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

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

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

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

all help appreciated!!

**hpe** · November 11th 2005, 05:59 PM

Originally Posted by dizzystudent

1973 -
Given the curve $x + xy +2y^2 = 6$ , 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 $x + xy +2y^2 = 6$ . 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 $f$ is defined for all x not equal to 0, in the interval $-1<x<1$ by $f(x)=\frac{x-\sin 2x}{\sin x}$

$\lim_{x \to 0} f(x) = -1$ 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 $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ .

**dizzystudent** · November 12th 2005, 11:40 AM

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

**Jameson** · November 12th 2005, 11:48 AM

Originally Posted by hpe

$\lim_{x \to 0} \frac{\sin(x)}{x} = 0$ .

It's 1. I see you fixed it though.

**Jameson** · November 12th 2005, 11:52 AM

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.

**hpe** · November 19th 2005, 05:41 AM

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 $f(x) = \frac{x-\sin 2x}{\sin x}$ for $x \ne 0$ , $f(0) = -1$ . The goal is to show that $f'(0) = 0$ , using the definition.

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

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

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

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

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

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

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

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

QED

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