Limits word problems

[wind]

1. the tangent to the curve xy=4 at a point P in the first quadrent meets the x-axis at A and the y-axsis at B. Prove that the area of triangle AOB, where O is the origin, is independant of the position P.

I drew a picture

The notation confuses me is xy=4 the same as y=4/x?
All I could think of doing nest was finding the limit of the tangent, but how would that help if it is essentually, the slope of the tangent?

y=4/x

m= lim f( a+h) - f(a)/h
x-0

I don't know what to do.....

now what do I do?

 

[daon]

I think you are confused. Your drawing couldn't possibly be more wrong.

Given that y is a function of x, and xy=4, there is no value for y when x is zero. The graph you have drawn closely resembles y=x^2+something.

 

[wind]

^ya, I Know that is not the curve, I was just tying to get an idea for the triangle, does that matter for this question?......What do I do to solve this, now i'm even more confused, it said "meets the x-axis at A and the y-axsis at B" so that would mean there is and x and y intercept right? then, It cant be y=4/x right? so what does xy=4 mean?

And what about the second question....

 

[galactus]

\(\displaystyle \L\\\lim_{x\to\0}\frac{\sqrt[3]{ax+1}-1}{x}\)


You can use L'Hopital's Rule because you have an indeterminate form.

Differentiate numerator and denominator, getting:

\(\displaystyle \L\\\lim_{x\to\0}\frac{a}{3(ax+1)^{\frac{2}{3}}}\)

Factor out constants a/3:

\(\displaystyle \L\\\frac{a}{3}\lim_{x\to\0}(ax+1)^{\frac{-2}{3}}\)

You can see what \(\displaystyle (ax+1)^{\frac{-2}{3}}\) approaches as x approaches 0, can't you?.

See what the limit is now?.

 

[wind]

umm.... I don't understand...what do I do with the a/3 in front of the limit? and we don't know how to find the derivitive (sp?) yet , is there another way this problem could be solved...I was working on a solution but I guess this is wrong I multiplied by the reciprical 3 times to get rid of the 3rd root... can you do stuff like that?

(ax + 1) ^ -2/3
=1??
the limit is 1??

 

[galactus]

What do you do with the a/3?. That's a constant. Multiply it by 1. That's your answer, a/3.

 

[pka]

For the first: we have \(\displaystyle xy = 4\quad \Rightarrow \quad y' = \frac{{ - y}}{x}.\)
Therefore, at \(\displaystyle P:\left( {t,\frac{4}{t}} \right)\) the tangent line written in intercept form is \(\displaystyle \L
\frac{x}{{2t}} + \frac{y}{{8/t}} = 1.\)

So we get \(\displaystyle A:\left( {2t,0} \right)\quad \& \quad B:\left( {0,\frac{8}{t}} \right).\)
The area is one-half(OA)(OB) or 8 square units regardless of t.

 

[wind]

ok, thanks for your help but,

Differentiate numerator and denominator

I don't know how to do this and derivatives is the next chapter, I have never even heard of L'Hopital's Rule, is'nt there some other way? why would they put it in this chapter?

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
\frac{{\sqrt[3]{{ax + 1}} - 1}}{x}\left( {\frac{{\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1}}{{\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1}}} \right) & = & \frac{{ax + 1 - 1}}{{x\left( {\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1} \right)}} \\
& = & \frac{{ax}}{{x\left( {\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1} \right)}} \\
& = & \frac{a}{{\left( {\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1} \right)}} \\
& \to & \frac{a}{3} \\
\end{array}\)

 

[galactus]

\(\displaystyle \L
\begin{array}{rcl}
\frac{{\sqrt[3]{{ax + 1}} - 1}}{x}\left( {\frac{{\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1}}{{\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1}}} \right) & = & \frac{{ax + 1 - 1}}{{x\left( {\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1} \right)}} \\
& = & \frac{{ax}}{{x\left( {\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1} \right)}} \\
& = & \frac{a}{{\left( {\sqrt[3]{{\left( {ax + 1} \right)^2 }} + \sqrt[3]{{ax + 1}} + 1} \right)}} \\
& \to & \frac{a}{3} \\
\end{array}\)

That's good, pka. I like that. It take insight to see that.

 

[wind]

Thanks. Sorry but I don't under stand number one y prime = -y/x how did you get that?

 

[galactus]

the tangent line written in intercept form is \(\displaystyle \L
\frac{x}{{2t}} + \frac{y}{{8/t}} = 1.\)

pka, may I ask where you got this from?. It's probably a stupid question, but I fail to see it.

 

[pka]

First comments. I do not know how to work this problem without using the derivative to find the slope. The equation \(\displaystyle xy=4\) is hyperbola with branches in the first and third quadrants and axis-of-symmetry \(\displaystyle y=x\). This problem is about the branch in the first quadrant: all the points look like \(\displaystyle \left( {t,\frac{4}{t}} \right),t > 0.\) The slope at each of these points is \(\displaystyle \frac{{ - 4}}{{t^2 }}.\)

Thus the tangent line is:
\(\displaystyle \L\begin{array}{rcl}
y & = & \frac{{ - 4}}{{t^2 }}\left( {x - t} \right) + \frac{4}{t} \\
t^2 y & = & - 4x + 4t + 4t \\
4x + t^2 y & = & 8t \\
\frac{x}{{2t}} + \frac{y}{{8/t}} & = & 1 \\
\end{array}\)

 

[galactus]

I see it now, pka. Thanks. I got as far as

\(\displaystyle \L\\y-\frac{4}{t}=\frac{-4}{t^{2}}(x-t)\), but didn't follow through.

I knew this was a hyperbola, but didn't put it in the 'hyperbolic' form.

 

[galactus]

Thanks. Sorry but I don't under stand number one y prime = -y/x how did you get that?

That comes from implicitly differentiating xy=4

Product rule:

\(\displaystyle x\frac{dy}{dx}+y=0\)

Solve for dy/dx:

\(\displaystyle \frac{dy}{dx}=\frac{-y}{x}\)

 

[pka]

I don't know how to do this and derivatives is the next chapter

It is clear that the questioner does not know derivatives.

derivatives is the next chapter why would they put it in this chapter?

That is a pet annoyance of mine. Authors are just careless.