# Limit and tangent problem

• Oct 31st 2009, 09:04 AM
Cursed
Limit and tangent problem
I'm unsure how to go about either problem. Can I get some direction? Thanks. (Happy)

1. Find the values of the constants $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}$

My attempt:
$\displaystyle f'(0) = \lim_{x\to0} f(x) = \lim_{x\to0} \frac{f(x) - f(a)}{x-a}$

$\displaystyle f(x) = \sqrt[3]{ax +b}$
$\displaystyle f'(x) = \frac{1}{3}a(ax+b)^{-2/3}$
$\displaystyle f'(0) = \frac{1}{3}ab^{-2/3}$

$\displaystyle \frac{1}{3}ab^{-2/3} = \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}$

$\displaystyle \frac{1}{3}ab^{-2/3} = \frac{5}{12}$

$\displaystyle ab^{-2/3} = \frac{5}{4}$

$\displaystyle a = \frac{5b^{2/3}}{4}$

2. Show that the length of the portion of a-ny tangent line to the asteroid $\displaystyle x^{2/3} + y^{2/3} = a^{2/3}$ cut off by the coordinate axes is constant.

My attempt:

$\displaystyle x^{2/3} + y^{2/3} = a^{2/3}$

$\displaystyle \frac{2}{3}x^{-1/3} +\frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0$

$\displaystyle x^{-1/3} +y^{-1/3}\frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = \frac{-x^{-1/3}}{y^{-1/3}}$

$\displaystyle \frac{dy}{dx} = \frac{-x^{-1/3}}{[(a^{2/3}-x^{2/3})^{3/2}]^{1/3}} = \frac{-x^{-1/3}}{(a^{2/3}-x^{2/3})^2}$
• Oct 31st 2009, 04:50 PM
redsoxfan325
Quote:

Originally Posted by Cursed
I'm unsure how to go about either problem. Can I get some direction? Thanks. (Happy)

1. Find the values of the constants $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}$

My attempt:
$\displaystyle f'(0) = \lim_{x\to0} f(x) = \lim_{x\to0} \frac{f(x) - f(a)}{x-a}$

$\displaystyle f(x) = \sqrt[3]{ax +b}$
$\displaystyle f'(x) = \frac{1}{3}a(ax+b)^{-2/3}$
$\displaystyle f'(0) = \frac{1}{3}ab^{-2/3}$

$\displaystyle \frac{1}{3}ab^{-2/3} = \lim_{x\to0} \frac{\sqrt[3]{ax +b} -2}{x} =\frac{5}{12}$

$\displaystyle \frac{1}{3}ab^{-2/3} = \frac{5}{12}$

$\displaystyle ab^{-2/3} = \frac{5}{4}$

$\displaystyle a = \frac{5b^{2/3}}{4}$

$\displaystyle b$ must equal $\displaystyle 8$, so that the numerator approaches zero as the denominator does. Otherwise, the function will blow up as $\displaystyle x\to0$.

Now use L'Hopital's Rule.

Spoiler:
$\displaystyle \lim_{x\to0}\frac{\sqrt[3]{ax+8}-2}{x}=\lim_{x\to0}\frac{a}{3}(ax+8)^{-2/3}$.

Plugging in $\displaystyle x=0$ gives:

$\displaystyle \frac{a}{3}\cdot8^{-2/3}=\frac{a}{12}$

So $\displaystyle a=5$, $\displaystyle b=8$.
• Oct 31st 2009, 05:01 PM
redsoxfan325
Quote:

Originally Posted by Cursed
I'm unsure how to go about either problem. Can I get some direction? Thanks. (Happy)

2. Show that the length of the portion of a-ny tangent line to the asteroid $\displaystyle x^{2/3} + y^{2/3} = a^{2/3}$ cut off by the coordinate axes is constant.

My attempt:

$\displaystyle x^{2/3} + y^{2/3} = a^{2/3}$

$\displaystyle \frac{2}{3}x^{-1/3} +\frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0$

$\displaystyle x^{-1/3} +y^{-1/3}\frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = \frac{-x^{-1/3}}{y^{-1/3}}$

$\displaystyle \frac{dy}{dx} = \frac{-x^{-1/3}}{[(a^{2/3}-x^{2/3})^{3/2}]^{1/3}} {\color{red}~\neq~} \frac{-x^{-1/3}}{(a^{2/3}-x^{2/3})^2}$

Not sure I understand what the problem is asking, but there is a small error in your last calculation.

$\displaystyle \frac{dy}{dx} = \frac{-x^{-1/3}}{y^{-1/3}}=-\frac{y^{1/3}}{x^{1/3}}=-\frac{[(a^{2/3}-x^{2/3})^{3/2}]^{1/3}}{x^{1/3}}=-\frac{\sqrt{a^{2/3}-x^{2/3}}}{x^{1/3}}$ $\displaystyle =-\frac{\sqrt{a^{2/3}-x^{2/3}}}{\sqrt{x^{2/3}}}=-\sqrt{\frac{a^{2/3}}{x^{2/3}}-1}=-\sqrt{(a/x)^{2/3}-1}$

Whatever calculation you need to do, the derivative is probably easiest to work with in that form.