| 
July 4th, 2009, 09:18 AM
| | Newbie | | Join Date: May 2009
Posts: 6
Thanks: 0
Thanked 0 Times in 0 Posts
| | 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
| 
July 4th, 2009, 09:51 AM
|  | Member | | Join Date: Apr 2009 Location: Boston
Posts: 218
Country: Thanks: 1
Thanked 64 Times in 64 Posts
| | Quote:
Originally Posted by banana_banana 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:
and solve for 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:
which means
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
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:: | 
July 4th, 2009, 06:23 PM
| | Newbie | | Join Date: May 2009
Posts: 6
Thanks: 0
Thanked 0 Times in 0 Posts
| | Finding the critical points Quote:
Originally Posted by apcalculus 2.
1.y=16x+4x^2-x^4
Differentiate
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 | 
July 4th, 2009, 07:46 PM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
Posts: 12,242
Country: Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
| | Quote:
Originally Posted by banana_banana , 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.
| 
July 4th, 2009, 07:48 PM
|  | Member | | Join Date: Apr 2009 Location: Boston
Posts: 218
Country: Thanks: 1
Thanked 64 Times in 64 Posts
| | Quote:
Originally Posted by banana_banana , 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:
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
| 
July 4th, 2009, 11:09 PM
| | Newbie | | Join Date: May 2009
Posts: 6
Thanks: 0
Thanked 0 Times in 0 Posts
| | 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:
and solve for 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:
which means
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.
| 
July 4th, 2009, 11:25 PM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
Posts: 12,242
Country: Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
| | Quote:
Originally Posted by banana_banana 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  are not real (I did point this out in my earlier post)? Surely the implied domain of  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.
| 
July 5th, 2009, 12:48 AM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
Posts: 12,242
Country: Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
| | Quote:
Originally Posted by banana_banana [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.
| | Thread Tools | | | | Display Modes | Linear Mode |
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts HTML code is Off | | | All times are GMT -7. The time now is 11:41 PM. | | |