Calculus problem find the equation of the line given two points

[O'Fearraigh]

Consider the curve f(x) = 1/x
Consider two points on f(x): Pa and Qa, where the x-coordinate of Pa is a, and the x-coordinate of Qa is a+1.
Let La be the line connecting Pa and Qa

1.) Find the equation for La
2.) Find a formula that expresses A(a) = the area between f(x) and La
3.) Determine lim(a-->∞) A(a) and lim(a-->0) A(a)
4.) Does A(a) have an extremum for 0 < a? If yes, determine if it's a minimum or maximu (or neither?), and what value of A(a) is at the extremum.

Extra: Redo the problem, where the x-coordinate of Qa is a^2Okay, so I know that the coordinate of Pa is (a, 1/a) and I know that the coordinate of Qa is (a+1, 1/a+1). So, you have the equation for the line y=mx+b. So, the slope would be [(1/a+1)-(1/a)] / [(a+1)-(a)], which would simplify down to m = -(1) / (a^2+a). Correct? How do I go on from here?

 

[scurty]

Remember back to your Pre-Calc days... you have the slope and a point on the line. Is there a formula that can combine the two to get the equation of a line?

 

[O'Fearraigh]

Right, well I know the equation y-y1=m(x-x1).
So, I would have: y-(1/a)=(-1 / a^2+a)*(x-a) ---> y=(-x / a^2+a)+(a / a^2+a)+(1/a).
Right? Or... y=(-x / a^2+a)+(2a+1 / a(a+1))
Which would imply that:
y = (2a+1-x)/(a^2+a)
Right?

 

[scurty]

Just be careful with your equations when posting online. Try to remember to use brackets. I know that you intended for your final equation to look like

$y = \displaystyle\frac{-x}{a^{2}+a} + \frac{2a+1}{a^{2}+a}$


And I just saw you updated your post, your final answer is fine (the brackets are good) but it would be better to split up the x term and the constant term (it'll be easier to integrate that way). The equation you had before was good, just be careful about parentheses.

 

[O'Fearraigh]

Okay...
So, when setting up this area equation... would it simply be the the integral of -(x / a^2+a)+(2a+1 / a^2+a) dx - integral of 1/x dx [from x=a to x=a+1]?

Okay, doing that, I get that the function for the area between La and f(x) is:

A(a) = (2a+1 / 2a^2+2a) - ln(a+1) - ln(a)

Is this correct?

 

[scurty]

That is how you set it up, I don't think your final answer is correct though, let me check...

I think you might have forgotten to distribute a negative sign when taking the integral, the natural logs aren't right. Try again and see what you come up with.

 

[Ray Vickson]

Okay...
So, when setting up this area equation... would it simply be the the integral of -(x / a^2+a)+(2a+1 / a^2+a) dx - integral of 1/x dx [from x=a to x=a+1]?

Okay, doing that, I get that the function for the area between La and f(x) is:

A(a) = (2a+1 / 2a^2+2a) - ln(a+1) - ln(a)

Is this correct?

Use brackets! Is A(a) = (2a+1)/(2a^2+2a) - ln(a+1)-ln(a), or is it what you actually *wrote*, which would be [tex]A(a) = 2a + \frac{1}{2a^2} + 2a - \ln(a+1) - \ln(a)[/itex]? <br /> <br /> RGV[/tex]

 

[O'Fearraigh]

Yes, A(a) = (2a+1)/(2a^2+2a) - ln(a+1) - ln(a)

or is it

A(a) = (2a+1)/(2a^2+2a) - ln(a+1) + ln(a) ?

 

[scurty]

The second answer is what I got, and you can simplify that further if you like (properties of logs).

You still seem unsure about the sign of the last logarithm. How come?

 

[O'Fearraigh]

I don't know... I just copied that from my first try.
Anyway, would the reduced equation be:
A(a) = (2a+1)/(2a^2+2a) + ln[(a)/(a+1)] ?

 

[scurty]

Yes, that's the right answer, but if you are unsure, you should go back and redo your work so you know why that is the right answer; i.e. convince yourself it is the right answer.

 

[O'Fearraigh]

No, I see what I did originally: I didn't carry the minus sign over.
Okay, so now that we have a function for A(a), how do the limits work?

We have:

(a) lim[a→∞]
(b) lim[a→0]

I got

lim[a→∞] = 0
lim[a→0] = ∞

Is that correct?