My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 10th, 2013, 01:18 PM   #1
Newbie
 
Joined: Mar 2013

Posts: 4
Thanks: 0

what is the differential of a trigonometric function

I have a calculus question I can't figure out how to solve,

The function y = tan(3x + 6); what is the dy when x=3 and dx=0.03

I tried these steps:

f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx, (need to calculate in radians)

so, for f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

= (1/[(1+cos2(15))/2])

= (2/(1+cos30))

= {2/[1+ sqrt(3)/2]}

= [2/[(2+sqrt(3))/2]]

= 4/((2+sqrt(3)),

so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3)) = and would be ..... am I doing this right?
Thank you.
dokrbb is offline  
 
March 10th, 2013, 03:15 PM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 20,931
Thanks: 2205

You are incorrect. You mentioned "need to calculate in radians", but didn't clarify whether x is already in radians. When you evaluated cos(30), you seemed to be working in degrees, but that would mean that everything was already in degrees and should have been converted to radians from the outset to make the differentiation valid.
skipjack is online now  
March 10th, 2013, 03:50 PM   #3
Newbie
 
Joined: Mar 2013

Posts: 4
Thanks: 0

Re: what is the differential of a trigonometric function

My mistake, yes worked as everything is in degrees, what I mentioned about radians it's because of the transformations I needed to do in order to use the calculator, what would you think about if you see the problem in this way:

The function y = tan(3x + 6); what is the dy when x=3 and dx=0.03

I tried these steps:

f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx,

so, for f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

= (1/[(1+cos2(15))/2])

= (2/(1+cos30))

= {2/[1+ sqrt(3)/2]}

= [2/[(2+sqrt(3))/2]]

= 4/((2+sqrt(3)),

so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3)) =
dokrbb is offline  
March 10th, 2013, 04:39 PM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 20,931
Thanks: 2205

Why have you repeated three lines for no apparent reason? The trigonometric manipulation is valid, but the calculus isn't. You can't use sec(x) as the derivative of tan(x) unless x is in radians.
skipjack is online now  
March 10th, 2013, 04:45 PM   #5
Newbie
 
Joined: Mar 2013

Posts: 4
Thanks: 0

Re: what is the differential of a trigonometric function

so I have to transform y = tan(3x + 6) in radians and after that to obtain the derivative, or I just have to assume that it was from the beginning in radians? sorry, but the problem doesn't state whether x is in radians or in degrees and I assumed from the beginning that it is in degrees, I'm completely confused now?
dokrbb is offline  
March 11th, 2013, 04:35 PM   #6
Math Team
 
Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: what is the differential of a trigonometric function

ALL derivatives of trig functions are based on radians (Technically, the trig functions can be defined without reference to angles at all. "Radian" measure for angles happens to give the same derivatives.) Unless a problem specifically refers to angle measured in degrees, 'radians' are to be assumed.
HallsofIvy is offline  
March 11th, 2013, 05:05 PM   #7
Newbie
 
Joined: Mar 2013

Posts: 4
Thanks: 0

Re: what is the differential of a trigonometric function

So, am I finally right in my calculations or not, and if not can you show where is my mistake

f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx,

for f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

= (1/[(1+cos2(15))/2])

= (2/(1+cos30))

= {2/[1+ sqrt(3)/2]}

= [2/[(2+sqrt(3))/2]]

= 4/((2+sqrt(3)),

so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3))...
dokrbb is offline  
March 13th, 2013, 09:29 AM   #8
Math Team
 
Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: what is the differential of a trigonometric function

Quote:
Originally Posted by dokrbb
So, am I finally right in my calculations or not, and if not can you show where is my mistake

f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx,

for f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

= (1/[(1+cos2(15))/2])

= (2/(1+cos30))

= {2/[1+ sqrt(3)/2]}

= [2/[(2+sqrt(3))/2]]

= 4/((2+sqrt(3)),

so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3))...
Where did that "(3)" just before "(0.03)" come from?
HallsofIvy is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
differential, function, trigonometric



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Trigonometric Function hatchelhoff Trigonometry 3 November 4th, 2011 04:45 PM
trigonometric function brumby3 Algebra 4 April 2nd, 2011 07:54 PM
Trigonometric Function Tetra Algebra 1 December 5th, 2009 08:28 PM
Trigonometric Function srlavos Algebra 1 April 16th, 2009 03:32 AM
Trigonometric equation / Differential inequality (12/18/06) julien Differential Equations 0 April 30th, 2007 12:05 AM





Copyright © 2019 My Math Forum. All rights reserved.