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 October 10th, 2016, 01:11 PM #1 Member   Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 Height of a water tower problem This question is at the end of a section on limits and differentiation, A landmark on a distant hill is 'x' metres from a water tower. The angle of elevation from the top of the tower is observed to be '0' (0 is theta) degrees whereas the angle from the foot of the tower is observed to be 0+h degrees a)how high is the water tower b)show that if 'h' is small then the height of the water tower is approx (Pi)xh / 180(cosn)^2 I've managed to do the first part. If the height of the landmark is 't' and the height of the water tower is 'n' then, tan (0+h) = t/x and tan 0 = (t - n) / x Eliminating t, t = xtan(0+h) = xtan0 +n So the height of the water tower 'n' is x(tan(0+h) - tan0) Can anyone show me how to do the second part. I don't understand how (Pi) and 180 can become involved?? October 11th, 2016, 01:06 PM #2 Global Moderator   Joined: May 2007 Posts: 6,763 Thanks: 697 Statement if b) doesn't look right. How can a dumber with dimension of length appear as an argument for cosn? Pi and 180 become involved, since trig functions use radians as a "natural" argument - for example sin(x) ~ x when x is in radians. Thanks from wirewolf October 11th, 2016, 01:53 PM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 if $h$ is small then $\tan(\theta+h) \approx \tan(\theta)+\left. \dfrac{d}{d\theta}\tan(\theta)\right|_{\theta=h}$ $\tan(x+h) \approx \tan(\theta)+\sec^2(h)$ the rest follows Thanks from wirewolf October 12th, 2016, 01:45 PM   #4
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 Originally Posted by romsek if $h$ is small then $\tan(\theta+h) \approx \tan(\theta)+\left. \dfrac{d}{d\theta}\tan(\theta)\right|_{\theta=h}$ $\tan(x+h) \approx \tan(\theta)+\sec^2(h)$ the rest follows
This assumes h is in radians, not in degrees. h in degrees is where $\displaystyle \frac{\pi}{180}$ comes into play. October 20th, 2016, 02:29 AM #5 Member   Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 Thanks for the replies. Having looked at it some more I think this is how it is done n=x(tan(0+h) - tan0) now dy/dx is defined as limit as h tends to zero of [f(x+h) - f(x)] / h therefore n=xh [lim (h tends to zero) (tan(0+h) - tan0 / h)] so n=xh (d/d0 tan0) = xh / (cos0)^2 degrees or (pi)xh / 180(cos0)^2 degrees Tags height, problem, tower, water Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ganesh Ujwal Physics 2 January 3rd, 2015 06:47 AM mr. jenkins Calculus 1 May 26th, 2014 01:09 AM rakmo Algebra 3 March 28th, 2013 05:20 AM thearae Calculus 7 September 26th, 2012 03:51 AM manich44 Algebra 9 June 23rd, 2009 09:23 AM

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