Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > University Math Help > Calculus
Reply
 
Thread Tools Display Modes
  #1  
Old July 4th, 2009, 09:18 AM
Newbie
 
Join Date: May 2009
Posts: 6
Thanks: 0
Thanked 0 Times in 0 Posts
banana_banana is on a distinguished road
Thumbs down Finding critical points and equations of tangents.

1. Determine the critical points (Clasify as maxima, minima), Inflection points and trace the y=16x+4x^2-x^4
2. Find the equation of tangent to the circle x^2 + y^2 = 9 and parallel to 2x + y= 10.


Last edited by mr fantastic; July 4th, 2009 at 04:57 PM. Reason: Changed post title
Reply With Quote
Advertisement
 
  #2  
Old July 4th, 2009, 09:51 AM
apcalculus's Avatar
Member
 
Join Date: Apr 2009
Location: Boston
Posts: 218
Country:
Thanks: 1
Thanked 64 Times in 64 Posts
apcalculus will become famous soon enough
Default

Quote:
Originally Posted by banana_banana View Post
1. Determine the critical points (Clasify as maxima, minima), Inflection points and trace the y=16x+4x^2-x^4
2. Find the equation of tangent to the circle x^2 + y^2 = 9 and parallel to 2x + y= 10.

2.
First find a slope form for the given circle by implicitly differentiating with respect to x:

2x + 2y y' = 0

and solve for y':

y' = \frac{-x}{y}

Because the slopes are equal, the derivative form above, when evaluated at the point of tangency, should be equal to the slope of the given line, which is is -2. Note:
y = -2x+10

\frac{-x}{y} = -2

which means
x=2y

Substitute this condition into the equation of the circle to solve for x, then find y by taking half of x.

1.y=16x+4x^2-x^4

Differentiate
y' = 4x^3- 8x + 16 = 4 (x^3 - 2x + 4)= 4 (x+2) (x^2-2x+2)

Now solve for zeros (or critical numbers to f(x)) then study the sign to determine types of extrema.

Good luck!!
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Placement Calculus & Statistics::
Reply With Quote
  #3  
Old July 4th, 2009, 06:23 PM
Newbie
 
Join Date: May 2009
Posts: 6
Thanks: 0
Thanked 0 Times in 0 Posts
banana_banana is on a distinguished road
Question Finding the critical points

Quote:
Originally Posted by apcalculus View Post
2.

1.y=16x+4x^2-x^4

Differentiate
y' = 4x^3- 8x + 16 = 4 (x^3 - 2x + 4)= 4 (x+2) (x^2-2x+2)

Now solve for zeros (or critical numbers to f(x)) then study the sign to determine types of extrema.

Good luck!!
, Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
Completing Square:
0=x²-2x+2

0= x²-2x+1=-2+1
(x-1)²=-1
x-1=±√-1
x=1±√-1

Quadratic Formula:
-1±√(-2)²-4(1)(2)
(2)(1)
-1±√-4
2
Reply With Quote
  #4  
Old July 4th, 2009, 07:46 PM
mr fantastic's Avatar
Flow Master

 
Join Date: Dec 2007
Location: Zeitgeist
Posts: 12,242
Country:
Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
mr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond repute
Default

Quote:
Originally Posted by banana_banana View Post
, Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
Completing Square:
0=x²-2x+2

0= x²-2x+1=-2+1
(x-1)²=-1
x-1=±√-1
x=1±√-1 Mr F says: Do you realise that this value of x is not real?

Quadratic Formula:
-1±√(-2)²-4(1)(2) Mr F says: b = -2, not -1 (which is what you substituted in the bit of have highlighted).
(2)(1)
-1±√-4
2
Your equations would look much better and be easier to post if you learnt some basic latex: http://www.mathhelpforum.com/math-help/latex-help/
__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
  #5  
Old July 4th, 2009, 07:48 PM
apcalculus's Avatar
Member
 
Join Date: Apr 2009
Location: Boston
Posts: 218
Country:
Thanks: 1
Thanked 64 Times in 64 Posts
apcalculus will become famous soon enough
Default

Quote:
Originally Posted by banana_banana View Post
, Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
Completing Square:
0=x²-2x+2

0= x²-2x+1=-2+1
(x-1)²=-1
x-1=±√-1
x=1±√-1

Quadratic Formula:
-1±√(-2)²-4(1)(2)
(2)(1)
-1±√-4
2
When you apply the quadratic formula, b = -2, so -b is 2. For some reason you start with a -1 in the numerator in your work.

Pull out the 4 from the square root as a 2, and you get:

\frac{2 \pm 2 \sqrt{-1}}{2} = 1 \pm i

Did you need the complex zeros?
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Placement Calculus & Statistics::

Last edited by apcalculus; July 4th, 2009 at 07:52 PM. Reason: missing a 2 in the numerator
Reply With Quote
  #6  
Old July 4th, 2009, 11:09 PM
Newbie
 
Join Date: May 2009
Posts: 6
Thanks: 0
Thanked 0 Times in 0 Posts
banana_banana is on a distinguished road
Question Critical Points and Equation of tangent Circle

so from that I can conclude that from X+2=0, i can have X= -2, and for 1±√-1, how should I have this If it has plus and minus sign, how could i fit it in the parenthesis because both will yield a different value of y.

First Critical Point (-2, -48)
SEcond critical point (1±√-1,?)

TANGENT EQUATION OF CIRCLE
Quote:
Originally Posted by apcalculus
First find a slope form for the given circle by implicitly differentiating with respect to x:

2x + 2y y' = 0

and solve for y':

y' = \frac{-x}{y}

Because the slopes are equal, the derivative form above, when evaluated at the point of tangency, should be equal to the slope of the given line, which is is -2. Note:
y = -2x+10

\frac{-x}{y} = -2

which means
x=2y

Substitute this condition into the equation of the circle to solve for x, then find y by taking half of x.
I dont understand what you mean about this? If i substitute X=2y to the X^2+Y^2=9 , i will get a y=(√9/5). then what will be the next?

Last edited by mr fantastic; July 5th, 2009 at 12:42 AM. Reason: Added quote tags. And replaced the latex tags that the OP removed from this quote.
Reply With Quote
  #7  
Old July 4th, 2009, 11:25 PM
mr fantastic's Avatar
Flow Master

 
Join Date: Dec 2007
Location: Zeitgeist
Posts: 12,242
Country:
Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
mr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond repute
Default

Quote:
Originally Posted by banana_banana View Post
so from that I can conclude that from X+2=0, i can have X= -2, and for 1±√-1, how should I have this If it has plus and minus sign, how could i fit it in the parenthesis because both will yield a different value of y.

First Critical Point (-2, -48)
SEcond critical point (1±√-1,?)
Have you realised yet that 1 \pm \sqrt{-1} = 1 \pm i are not real (I did point this out in my earlier post)? Surely the implied domain of y=16x+4x^2-x^4 is real numbers.
__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
  #8  
Old July 5th, 2009, 12:48 AM
mr fantastic's Avatar
Flow Master

 
Join Date: Dec 2007
Location: Zeitgeist
Posts: 12,242
Country:
Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
mr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond repute
Default

Quote:
Originally Posted by banana_banana View Post
[snip]TANGENT EQUATION OF CIRCLE
I dont understand what you mean about this? If i substitute X=2y to the X^2+Y^2=9 , i will get a y=(√9/5). then what will be the next?
What you do next is recall that x=2y and substitute the value of y ....

(By the way, please don't edit posts that have been replied to by adding slabs of new stuff. Make a new post in the thread).
__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 11:41 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.