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December 29th, 2016, 08:38 AM  #1 
Newbie Joined: Dec 2016 From: United Kingdom Posts: 5 Thanks: 0  3 Problems I am stuck on
Hi all, never had the chance to learn, so addressing the issue later in life! Been doing OK but stuck on three questions on variation and none of the literature I have is helping me at all. If anyone can advise on the following, I would be very appreciative. 1. If the distances of an object and of its image, formed by a mirror, are measured from a certain point, it is found that the sum of the distance varies as their product. If the image distance is 120mm when the object distance is 300mm, calculate the image distance when the object is 540mm. 2. The square of the speed of a particle varies as the cube of its distance from a fixed point. If the distance is increased by 1.2 percent, what is the approximate percentage in the increase in the speed. 3. A clock keeps accurate time at 10c but gains as the temperature falls and viceversa, the rate of gain or loss varying and the square of the number of nC between the actual temp and 10c. If it gains 2 seconds per day when the temp is 5c, how much does it lose to the nearest second when the temp is 42c? I am sure these are fairly stock questions, but I am stuck and mind is numb! Appreciate your time. P Last edited by skipjack; December 30th, 2016 at 09:01 AM. 
December 29th, 2016, 09:12 AM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 1,862 Thanks: 968  Quote:
So here we have $120+300 = \alpha (120)(300) \Rightarrow \alpha = \dfrac{420}{36000} = \dfrac{7}{600}$ $540 + d_o = \alpha (540)d_o = \dfrac{7}{600}(540)d_o$ and I leave the remaining bit of algebra to you. Quote:
$s_2^2 = \alpha (1.012 d)^3$ $s_2^2 = (1.012)^3 \alpha d^3 = (1.012)^3 s^2$ $\dfrac{s_2^2}{s^2} = (1.012)^3$ $\left(\dfrac{s_2}{s}\right)^2 = (1.012)^3$ $\dfrac{s_2}{s} = \sqrt{(1.012)^3} = (1.012)^{3/2}$ Quote:
Last edited by skipjack; December 30th, 2016 at 09:03 AM.  
December 29th, 2016, 09:40 AM  #3 
Newbie Joined: Dec 2016 From: United Kingdom Posts: 5 Thanks: 0 
Hi romsek  thanks for a prompt reply. I see the patterns yes (I got the other 11 questions right in the exercise with little issue), but it is not providing the answers the book does still and this is where I am stuck. (Is the book wrong on 3 consecutive answers? I doubted it, hence reaching out but your answer for 1 is one of those I could achieve but confused why book is so different? If I read you formatting correctly, the final answer for 1 is 6.3, yet the book I am working from states 102mm? Again for 2 a different answer I can not resolve book = 1.8% Maybe I have just had mind freeze! Been at this all day and none the wiser now!! Appreciate the response. P Last edited by skipjack; December 30th, 2016 at 09:06 AM. 
December 29th, 2016, 10:07 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,862 Thanks: 968  Quote:
If you solve for $d_o$ you get $101.887~mm$ For #3 $(1.012)^{3/2} = 1.01805$ this is an increase of $0.01805$ or $1.805\%$ Last edited by skipjack; December 30th, 2016 at 09:07 AM.  
December 29th, 2016, 10:29 AM  #5 
Newbie Joined: Dec 2016 From: United Kingdom Posts: 5 Thanks: 0  Thanks you
Sorry  yes 1. I see now I missed the do part on the right. 2. Yes tired and frustrated I see now I have 1.01805 on paper, but it's a percentage! I will work on three tomorrow when I have reviewed this thoroughly. Would be really useful if they put worked examples in the book, nothing of this sort in there at all! Your help is greatly appreciated. P Last edited by skipjack; December 30th, 2016 at 09:07 AM. 
December 31st, 2016, 08:30 AM  #6 
Newbie Joined: Dec 2016 From: United Kingdom Posts: 5 Thanks: 0 
OK so i Have looked at 3 and got the right answer. Could someone advise if I have taken a typical approach etc The question is A clock keeps accurate time at 10c (c) but gains as the temperature falls and viceversa, the rate of gain (g) or loss varying and the square of the number of nC between the actual temp and 10c. If it gains 2 seconds per day when the temp is 5c, how much does it lose to the nearest second when the temp is 42c? k = unknown element So I used logic that g=c^2/k. So 2=5^2/k 2=25k k=25/2 k=12.5 Back into the original formula g=32^2/12.5 g=1024/12.5 g=81.92 seconds per day. *4 days = 328' seconds. Just want to know if my approach is the best (it answers the question but limited ideas on if I have used a typical approach). Thanks P 
December 31st, 2016, 12:29 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 998 Thanks: 410 
3. A clock keeps accurate time at 10c but gains as the temperature falls and viceversa, the rate of gain or loss varying and [WITH ?] the square of the number of nC ?? between the actual temp and 10c. If it gains 2 seconds per day when the temp is 5c, how much does it lose to the nearest second when the temp is 42c? If I understand the problem, your formula is not quite right, but your logic is solid. I think a better formula would be $g = k * sgn(10  t) * (10  t)^2.$ Obviously t stands for temperature. Usually constants of proportionality are shown as multiplications. You took the change of sign into account logically, but the formula should take it into account directly. The sgn, standing for the signum function, is one way to do this. $\therefore t = 5\ and\ g = +2 \implies 2 = k * (+\ 1) * (10  5)^2 \implies 2 = 25k \implies k = 0.08.$ This value of k means the same thing as your value of course. Thus $g = 0.08 * sgn(10  42) * (10  42)^2 = 0.08(\ 1)(1024) = \ 81.92.$ What is the relevance of the four days? EDIT: More information on the signum function https://en.wikipedia.org/wiki/Sign_function Last edited by JeffM1; December 31st, 2016 at 12:37 PM. Reason: Added link 
January 2nd, 2017, 01:51 PM  #8 
Newbie Joined: Dec 2016 From: United Kingdom Posts: 5 Thanks: 0 
Thanks for response Jeff will read up on the link. Answers are right which is key and process is now proven as repeatable. 4 days was just in the question so *4 at the end 

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