
Trigonometry Trigonometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 29th, 2007, 12:25 PM  #1 
Newbie Joined: Oct 2007 Posts: 3 Thanks: 0  HelpFinding 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 25x^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:
fine, but she really does suck...  
October 29th, 2007, 12:40 PM  #4  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
 
October 29th, 2007, 02:54 PM  #5  
Newbie Joined: Oct 2007 Posts: 3 Thanks: 0  Quote:
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 = √(25x^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 xaxis and therefore don't have a slope. 

Tags 
helpfinding, slope, tangent 
Thread Tools  
Display Modes  

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