My Math Forum Help-Finding the slope of a tangent

 Trigonometry Trigonometry Math Forum

 October 29th, 2007, 12:25 PM #1 Newbie   Joined: Oct 2007 Posts: 3 Thanks: 0 Help-Finding the slope of a tangent I'm Canadian so I'm taking "advanced functions and relations" which is a mandatory course first semester before calc. second. I need help with a couple questions but I'll make a different thread for each one. Anyway lets get started. I cannot understand how to come to the final answer on this question. Find the slope of the tangent to each curve at the given point. Remember i have not taken calc. so i will not know anything about derivatives, so don't use that in your answer. y= square root (cant find symbol on keyboard) of 25-x^2; (3,4) Answer = -3/4 If anyone out there can show me how to do this, it would be much appreciated.
 October 29th, 2007, 12:30 PM #2 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Do try to make your post titles more descriptive and use better language. Thanks, Moderator.
October 29th, 2007, 12:37 PM   #3
Newbie

Joined: Oct 2007

Posts: 3
Thanks: 0

Quote:
 Originally Posted by Infinity Do try to make your post titles more descriptive and use better language. Thanks, Moderator.

fine, but she really does suck...

October 29th, 2007, 12:40 PM   #4
Senior Member

Joined: Dec 2006

Posts: 1,111
Thanks: 0

Quote:
 fine, but she really does suck...
That may be true, but keep in mind it's your job to learn the material whether or not there is any teacher there at all, let alone whether they can actually teach.

October 29th, 2007, 02:54 PM   #5
Newbie

Joined: Oct 2007

Posts: 3
Thanks: 0

Quote:
Originally Posted by Infinity
Quote:
 fine, but she really does suck...
That may be true, but keep in mind it's your job to learn the material whether or not there is any teacher there at all, let alone whether they can actually teach.
thats why I'm here, I've only been living in the city I'm currently in for a year, (gr.11 and now) i really dont have many friends and any in that class.

So this is how i'ma learn this.

If you have any FAQ's, self help, sticky's, etc.. i would be glad to read em'

 October 29th, 2007, 04:49 PM #6 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Good choice! Learning is great fun, especially when you can manage your education by yourself. A teacher can greatly help, but letting the teacher manage everything you do doesn't always work.
 October 29th, 2007, 07:24 PM #7 Member   Joined: Oct 2007 From: CA, USA Posts: 46 Thanks: 0 Alright, you need to use derivatives. Granted, that it doesn't have to be calculus derivatives, but nonetheless, a derivative is the slope of the line tangent to a point on the graph. There are three equations you can used called difference quotients. Forward DQ: (f(x) - f(x-.001))/.001 Backwards DQ (f(x+.001) - f(x))/.001 Symmetrical DQ: (f(x+.001) - f(x-.001))/.002 I plugged in the equation into all three, and I got -0.75 all three times.
 October 30th, 2007, 02:20 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 You would need calculus for graphs in general, but simple geometry suffices in certain cases, including the specific case you are asking about. The curve y = √(25-x^2) is a semicircle. The slope of the radius to the point (3, 4) is 4/3. The tangent through (3, 4) is perpendicular to that radius and hence has slope -1/(4/3), i.e., -3/4. Note: some graphs (including this semicircle) have tangents which are perpendicular to the x-axis and therefore don't have a slope.

 Tags helpfinding, slope, tangent

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post texaslonghorn15 Calculus 2 June 21st, 2013 07:52 PM pomazebog Calculus 5 January 27th, 2012 10:33 PM Striker2 Trigonometry 1 March 11th, 2010 03:49 AM Emily14:) Calculus 12 November 8th, 2009 06:42 AM AdamL Calculus 2 September 17th, 2007 05:25 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top