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 December 10th, 2016, 10:56 PM #1 Newbie   Joined: Dec 2016 From: Pakistan Posts: 14 Thanks: 0 How condition will change when changing variable value in this function? The function is: H(x) = 0.002x^2 + 0.005 - 0.029 , 10<=x<=100 in this function x is speed in mph when converting it to kph I multiply x with 6.209 and get this (see if its correct): H(1.6x) = 0.002(1.6x)^2 + 0.005 - 0.029 , 6.215 <= x <= 62.15 Now if I plot graph it is like this: Blue line is for Kph and red for mph. Kph is more for same value of mph so its value on graph should be more on x axis then mph but as you can see in graph its reverse. Mph is more for same horse power as kph is for same horse power. Should Kph be more? Did I made mistake in changing condition?
 December 11th, 2016, 08:50 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,052 Thanks: 431 Several things. First and foremost what is H(x) supposed to represent? If H(x) is speed, then x is not speed. If x is speed, then H(x) is not speed. ALWAYS identify your variables clearly, or confusion is likely to result. I am going to guess that x is horsepower (or perhaps time), and H(x) is speed. Is that guess correct? I suggest in any case giving different names to the functions involving different units. Second, whatever H(x) means, should it be $H(x) = 0.002x^2 + 0.005x - 0.029?$
December 11th, 2016, 09:32 AM   #3
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 Originally Posted by JeffM1 Several things. First and foremost what is H(x) supposed to represent? If H(x) is speed, then x is not speed. If x is speed, then H(x) is not speed. ALWAYS identify your variables clearly, or confusion is likely to result. I am going to guess that x is horsepower (or perhaps time), and H(x) is speed. Is that guess correct? I suggest in any case giving different names to the functions involving different units. Second, whatever H(x) means, should it be $H(x) = 0.002x^2 + 0.005x - 0.029?$
Sorry for confusion. H is horsepower and x is speed in miles per hour. Here is the complete question:

 December 12th, 2016, 01:38 AM #4 Newbie   Joined: Dec 2016 From: Pakistan Posts: 14 Thanks: 0 I replied yesterday but it's under moderation review.... H(x) is horse power and x is speed in Mph. And not function is written correct I have triple checked. [Moderator note: the earlier reply is now approved.] Last edited by skipjack; December 13th, 2016 at 06:14 AM.
 December 12th, 2016, 07:41 AM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,392 Thanks: 100 $\displaystyle H(x)\approx .002x^{2}$ in the speed range given. The other two terms don't make sense (.005-.029=-.024) unless each is experiment dependent. $\displaystyle H(x)=.002x^{2}, Hhp,xmph\\ H(x)=.002\frac{hp}{mph^{2}}(xmph)^{2}\\ h(x)=.002\frac{hp}{mph^{2}}\left (xkmh\frac{1mph}{1.6kmh} \right )^{2}\\ h(x)=\frac{.002}{(1.6)^{2}}x^2, hhp, xkmh$ kmh=km/hr If you use the same x scale for plotting mph and kmh, then for 100 interpreted as 100mph you get 20hp, for 100 interpreted as kmh, you get .002(100)$\displaystyle ^2$/(1.6)$\displaystyle ^2$=7.8hp Thanks from shahbaz200
 December 13th, 2016, 05:59 AM #6 Senior Member   Joined: May 2016 From: USA Posts: 1,052 Thanks: 431 I do not find post # 4 at all clear. Assuming that horsepower is measured in the same units and that the function relating horsepower to miles per hour is correct, we have: $H_m(m) = 0.002m^2 + 0.005 - 0.029 = 0.002m^2 - 0.024,$ $where\ m = speed\ in\ miles\ per\ hour\ and$ $H_m(m) = horsepower\ measured\ in\ some\ unknown\ units.$ $k = speed\ in\ kilometers\ per\ hour.$ We are asked to find the comparable function where the function's argument is expressed in terms of kilometers per hour. That is, we need to find $H_k(k).$ $k = a \implies m = \dfrac{a}{1.6}.$ $\therefore H_k(a) = H_m (0.625a) = 0.002 \left ( \dfrac{a}{1.6} \right ) ^2 - 0.024 = 0.002 * \dfrac{a^2}{2.56} - 0.024 = 0.00078125a^2 - 0.024.$ What is the relevant domain? $m = b \implies k = 1.6b.\ \therefore m = 10 \implies k = 1.6\ and\ m = 100 \implies k = 160.$ $THUS\ 16 \le k \le 160 \implies H_k(k) = 7.8125 * 10^{-4}k^2 - 2.4 * 10^{-2}.$ This is, I believe, what zylo intended to say. Thanks from shahbaz200 Last edited by JeffM1; December 13th, 2016 at 06:04 AM.
December 13th, 2016, 06:51 AM   #7
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 Originally Posted by zylo $\displaystyle H(x)\approx .002x^{2}$ in the speed range given. The other two terms don't make sense (.005-.029=-.024) unless each is experiment dependent. $\displaystyle H(x)=.002x^{2}, Hhp,xmph\\ H(x)=.002\frac{hp}{mph^{2}}(xmph)^{2}\\ h(x)=.002\frac{hp}{mph^{2}}\left (xkmh\frac{1mph}{1.6kmh} \right )^{2}\\ h(x)=\frac{.002}{(1.6)^{2}}x^2, hhp, xkmh$ kmh=km/hr If you use the same x scale for plotting mph and kmh, then for 100 interpreted as 100mph you get 20hp, for 100 interpreted as kmh, you get .002(100)$\displaystyle ^2$/(1.6)$\displaystyle ^2$=7.8hp
That is exactly what I meant to say, post #5. All units are clearly spelled out, as they should be.

EDIT: It's not significant but you can add -.024hp to equations. -.029hp could indicate AC turned off and .005hp could be some other accessory.

Last edited by zylo; December 13th, 2016 at 07:39 AM. Reason: add -.024

 December 14th, 2016, 07:16 AM #8 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,392 Thanks: 100 H(x) = .002x$\displaystyle ^{2}$, xmph, Hhp xmph = (x'kmh)(.61mph/kmh) x=x'/1.6 H(x'/1.6) = (.002/1.6$\displaystyle ^{2}$)x'$\displaystyle ^{2}$, x'kmh, Hhp
 December 16th, 2016, 09:01 AM #9 Newbie   Joined: Dec 2016 From: Pakistan Posts: 14 Thanks: 0 Thanks zylo and JeffM1. I got my answer and graph correct, thank you for your help both!

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