My Math Forum Application of differentiation ! HELLP !!

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 September 2nd, 2009, 02:28 AM #1 Newbie   Joined: Sep 2009 Posts: 3 Thanks: 0 Application of differentiation ! HELLP !! A bush walker can walk at 5km/h through clear land and 3km/h through bushland. If she has to get from point A to point B following a route indicated at right, find the value of x so that the route is covered in a minimum time. (note: time=distance/speed) im sorry i didnt know how to upload a picture on to this post , so i have posted a link with my question ! http://i188.photobucket.com/albums/z106 ... tled-5.jpg thank you all so much in advance
 September 2nd, 2009, 07:31 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: Application of differentiation ! HELLP !! Time = distance/speed, let $T$be time, so $T=\frac{\sqrt{4+x^2}}{3}+\frac{3-x}{5}=\frac{5\sqrt{4+x^2}+9-3x}{15}$ Derivative: $\left(\frac{1}{15}\right)\frac{d}{dx}\left(5\sqrt{ 4+x^2}+9-3x\right)=\left(\frac{1}{15}\right)\left(\frac{5x} {\sqrt{4+x^2}}-3\right)=\left(\frac{1}{15}\right)\frac{5x-3\sqrt{4+x^2}}{\sqrt{4+x^2}$ To minimize time we set the numerator of the above equal to zero: $5x-3\sqrt{4+x^2}=0$ $5x=3\sqrt{4+x^2}$ $25x^2=9(4+x^2)$ $16x^2=36$ $x^2=\frac{9}{4}$, since distance is always positive we do not consider the negative root. $x=\frac{3}{2}$ Thanks from KekkoMenny
 September 3rd, 2009, 02:16 AM #3 Newbie   Joined: Sep 2009 Posts: 3 Thanks: 0 Re: Application of differentiation ! HELLP !! thankk you ! i was so close xD
 September 3rd, 2009, 08:06 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,809 Thanks: 2150 Below the text box, there's an "Upload attachment" button! If you were so close, how come you didn't originally post the work you'd already done?
 March 3rd, 2017, 10:27 PM #5 Newbie   Joined: May 2015 From: Australia Posts: 10 Thanks: 0 Sorry for posting on a 8 years old post, but I had the same question in my textbook, and I get how you find $T=\frac{\sqrt{4+x^2}}{3}+\frac{3-x}{5}=\frac{5\sqrt{4+x^2}+9-3x}{15}$ , but how did you derived? Thanks.
March 3rd, 2017, 10:33 PM   #6
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Quote:
 Originally Posted by KekkoMenny Sorry for posting on a 8 years old post, but I had the same question in my textbook, and I get how you find $T=\frac{\sqrt{4+x^2}}{3}+\frac{3-x}{5}=\frac{5\sqrt{4+x^2}+9-3x}{15}$ , but how did you derived? Thanks.
$T=\dfrac{\sqrt{4+x^2}}{3}+\dfrac{3-x}{5} =$

$\dfrac 5 5 \dfrac{\sqrt{4+x^2}}{3} + \dfrac 3 3 \dfrac{3-x}{5} =$

$\dfrac{5\sqrt{4+x^2}}{15} + \dfrac{3(3-x)}{15} =$

$\dfrac{5\sqrt{4+x^2} + 9-3x}{15}$

 March 4th, 2017, 08:34 PM #7 Newbie   Joined: May 2015 From: Australia Posts: 10 Thanks: 0 Maybe I didn't explain my self correctly, how did you get from $T=\frac{\sqrt{4+x^2}}{3}+\frac{3-x}{5}=\frac{5\sqrt{4+x^2}+9-3x}{15}$ to $5x-3\sqrt{4+x^2}=0$ Thanks.
March 4th, 2017, 08:52 PM   #8
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Quote:
 Originally Posted by KekkoMenny Maybe I didn't explain my self correctly, how did you get from $T=\frac{\sqrt{4+x^2}}{3}+\frac{3-x}{5}=\frac{5\sqrt{4+x^2}+9-3x}{15}$ to $5x-3\sqrt{4+x^2}=0$ Thanks.
he took the derivative of this expression, set it equal to zero, and simplified it a bit so just this piece is set equal to zero.

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