Expert: Paul Klarreich - 10/15/2006

Question
Hi!

My name is Meghan and I am in Grade 12 University level (AP) Calculus and I really need some help. I am currently working on limits and tangents and my teacher well quite frankly sucks and I don't think she taught us any of the stuff I am doing in my homework.

I don't expect you to do it for me, but I really have no idea where to even start with some of these questions. So here they are. If you could explain them to me that would be great.

1)Evaluate lim ?ã2x+3 - ?ãx+3 / ?ã3x=4 - ?ã2x+4
         x->0

I don't know if all that will turn out right, but it says Evaluate the lim when x approaches o, (the square root of 2x+3 subract the square root of x+3) divided by (the square root of 3x+4 subtract the square root of 2x+4)

I tried multiplying it by it's conjugate but I got some really messed up answer, which could mean I screwed up which is likely, but I still don't get it.

2) Find the coordinates of the point(s) on the curve f(x)= x^3 + 2x^2 - 8x where the tangent is parallel to the line y = 12x - 3

This one I just don't understand at all.


3) Use the first principles to determine the equation of the tangent of the curve
y = x^2 - 3x = 4 / x^2
at the point (-1, 8)

I think I understand this one, but could you please walk me through the steps anyways?


Here's the assignment, just in case you didn't understand how I typed things out, or if it changed somehow. It's questions 3, 4 and 5.
http://www.geocities.com/babyjoey717/calculusassignment.jpg

Thanks so much!

*Meghan

Answer
Meghan Asks in Category Calculus ...
 
Subject:  Calculus Help
Private:  no
 
Question:  Hi!

My name is Meghan and I am in Grade 12 University level (AP) Calculus and I really need some help.

I am currently working on limits and tangents and my teacher well quite frankly sucks and I don't think she taught us any of the stuff I am doing in my homework.

I don't expect you to do it for me, but I really have no idea where to even start with some of these questions. So here they are. If you could explain them to me that would be great.

1)Evaluate

lim   sqrt(2x+3) - sqrt(x+3) / sqrt(3x+4) - sqrt(2x+4)
  x->0

I don't know if all that will turn out right, but it says Evaluate the lim when x approaches o, (the square root of 2x+3 subract the square root of x+3) divided by (the square root of 3x+4 subtract the square root of 2x+4)

I tried multiplying it by its conjugate but I got some really messed up answer, which could mean I screwed up which is likely, but I still don't get it.

2) Find the coordinates of the point(s) on the curve f(x)= x^3 + 2x^2 - 8x where the tangent is parallel to the line y = 12x - 3

This one I just don't understand at all.


3) Use the first principles to determine the equation of the tangent of the curve
y = x^2 - 3x = 4 / x^2
at the point (-1, 8)

I think I understand this one, but could you please walk me through the steps anyways?


Here's the assignment, just in case you didn't understand how I typed things out, or if it changed somehow. It's questions 3, 4 and 5.
http://www.geocities.com/babyjoey717/calculusassignment.jpg

Thanks so much!

*Meghan
..........................................
Hi, Meghan,

Good idea -- sending a link to the originals.  The interface at this site is crude and lots of stuff gets messed up.  I have the examples.
**********************************************************
WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.
*********************************************************

Problem 3:[NUMBERED ACCORDING TO THE SHEET.]

      sqrt(2x+3) - sqrt(x+3)
lim   ------------------------
x->0   sqrt(3x+4) - sqrt(2x+4)

Since putting in x = 0 gives you:
sqrt(3) - sqrt(3)    0
----------------- = ---
sqrt(4) - sqrt(4)    0

we have a chance.  Yes, you are right to rationalize, so let's try that:

(sqrt(2x+3) - sqrt(x+3))( sqrt(3x+4) + sqrt(2x+4))
----------------------- --------------------------
(sqrt(3x+4) - sqrt(2x+4))(sqrt(3x+4) + sqrt(2x+4))

The bottom becomes  3x + 4 - (2x + 4) = 3x + 4 - 2x - 4 = x, which is becoming zero at the limit, so this factor of x WILL HAVE TO BE CANCELED SOMEHOW.

But the top will be messy, and even us old math teachers don't like that, either.  [Mathematicians are basically lazy people.]

So we use an old trick: You were taught in high school (yes, yes, I know -- you're in high school now.) to rationalize the denominator, but it is also legal to rationalize the numerator.  So do both:

[AND I MEAN IT ABOUT THE COURIER FONT!]

           A                        B'                        A'
(sqrt(2x+3) - sqrt(x+3))( sqrt(3x+4) + sqrt(2x+4)) (sqrt(2x+3) + sqrt(x+3))
----------------------- ---------------------------------------------------
(sqrt(3x+4) - sqrt(2x+4))(sqrt(3x+4) + sqrt(2x+4)) (sqrt(2x+3) + sqrt(x+3))
           B                        B'                        A'

We already computed  BB' = x.  Now
AA' = 2x + 3 - (x + 3) = 2x + 3 - x - 3 = x.  (Hooray!)

So the expression becomes:

x(sqrt(3x+4) + sqrt(2x+4))
----------------------------
x(sqrt(2x+3) + sqrt(x+3))

and the x cancels, so we have:

(sqrt(3x+4) + sqrt(2x+4))
--------------------------
(sqrt(2x+3) + sqrt(x+3))

NOW we can put  x = 0, and get:

sqrt(4) + sqrt(4)
------------------
sqrt(3) + sqrt(3)
 2 + 2          2
= --------- = -------
 2 sqrt(3)   sqrt(3)

Cute.
...........................................................
Problem 4) Find the coordinates of the point(s) on the curve f(x)= x^3 + 2x^2 - 8x where the tangent is parallel to the line y = 12x - 3

The line  y = 12x - 3  has a slope equal to 12, so any line parallel to it has slope = 12.  So the problem becomes:

Find the coordinates of the point(s) on the curve f(x)= x^3 + 2x^2 - 8x where the tangent has slope = 12.

The slope is the value of the derivative, so:

f'(x) = 3x^2 + 4x - 8

Now we set that equal to 12 and solve.  [Not equal to zero, mind you; we are not looking for horizontal tangents.]

3x^2 + 4x - 8 = 12

3x^2 + 4x - 20 = 0

(3x + 10)(x - 2) = 0

[You used to be good a factoring, right?]

x = -10/3,  and  x = 2.

Follow this up by substituting each into the original  f(x) to get the y-coordinates.  I'll leave that to you.
...........................................................

3) Use first principles to determine the equation of the tangent of the curve
y = x^2 - 3x + 4 / x^2  at the point (-1, 8)

I hate phrasing like that -- who says what 'first principles' are?  So I'll assume you just mean to (1) find the derivative at  x = -1 and and (2) use (-1,8) as the point and (3) put that into the point-slope form of a line.
    x^2 - 3x + 4
y =  ------------- = 1 - 3x^(-1) + 4x^(-2)
         x^2

[This is a sneaky way to avoid the quotient rule.]

                         3       8
y' = 3x^-2 - 8x^(-3) =  ----- + -----
                        x^2     x^3

At x = -1,

y' = m = 3 - 8 = -5

Now the point-slope form is:

y - y0 = m(x - x0)

Use  m = -5,  x0 = -1,  y0 = 8

y - 8 = -5(x + 1)

y - 8 = -5x - 5
5x + y = 3
or some other form, as you see fit.