Practical applications of max and min... (1 Viewer)

[ThandraCityRail]

...Why must they be so hard?

the actual volume/area q's etc i can do but when it comes to those speed, cost etc questions i find them almost impossible to do. fitzpatrick splits them into two seperate sections, the first one is doable and the second section i can't even follow along the worked answers... what's wrong with me?

/rant

 

[taeyang]

haha, there is nothing wrong with you.. If you are having too much trouble with them, why are you starting with Fitzpatrick? What does your teacher give you, do them!

Besides.. the process is basically problem solving, just stop and think. Put the pen down and try to think of ideas that lead to what you want rather than jumping straight in and doing what you did on the last question. Yes 2u and Ext1 are usually just repetition but Max and Min are usually q9 or 10 of the exam which require some sort of practical solving.

Just because you can't see the answer straight away doesn't mean to say you suck, do you know how many people have a problem with Circle Geometry in Ext mathematics? This is generally the case of "you see it, or you don't" but my close mate just worked his way up from the very basic and now he is pro brah.

Just don't get caught up in thinking that if you can't do questions out of, arguably, the hardest textbook, then you suck at math. Keep at it man you can't be pro at every topic.

 

[ThandraCityRail]

haha, there is nothing wrong with you.. If you are having too much trouble with them, why are you starting with Fitzpatrick? What does your teacher give you, do them!

Besides.. the process is basically problem solving, just stop and think. Put the pen down and try to think of ideas that lead to what you want rather than jumping straight in and doing what you did on the last question. Yes 2u and Ext1 are usually just repetition but Max and Min are usually q9 or 10 of the exam which require some sort of practical solving.

Just because you can't see the answer straight away doesn't mean to say you suck, do you know how many people have a problem with Circle Geometry in Ext mathematics? This is generally the case of "you see it, or you don't" but my close mate just worked his way up from the very basic and now he is pro brah.

Just don't get caught up in thinking that if you can't do questions out of, arguably, the hardest textbook, then you suck at math. Keep at it man you can't be pro at every topic.

thanks for the advice!

yeah i'm gonna give it another go tomorrow really can't be stuffed doing anymore max and min today

i'm usually good at problem solving... i can do basically most q 9's and 10's that don't involve max and min but when they do... :/ (see 2003 q 9c)

 

[taeyang]

well, all you do is differentiate the equation in the first part with respect to E in terms of v, now, you should get v = 0 and v = 3u/2.. Now, to check whether v = 3u/2 is a minimum, we can either find the second derivative OR test the slope of the gradient of the tangent to the curve on either side of v = 3u/2 (so basically, if the gradient is - on the left of v = 3u/2 and then changes to + on the right of v = 3u/2 the point is a minimum)

SO... since we don't want to derive it again ( I mean look at the damned thing! ) we will just test points to the left and right of v = 3u/2 (we don't worry about v = 0 because that gives us the gradient being 0) and see if the sign changes...

Since we know v > u > 0 ... we know that u has to be greater than 0.. so what I did was sub in values such as v = 5u/2 (Becuase it is MORE than 3u/2 for the right side) and v = u/2 (because it is LESS than 3u/2 for the left side) ... turns out the sign changes from negative to positive meaning it is a minimum. It's as simple as that!

The whole point of maxima and minima is to get you to find where given things are a maximum or a minimum value and the way we are taught is with those two methods... Just keep in mind it almost CERTAINLY wants you to differentiate and test points (like we did) or differentiate again.. They will also try to confuse you (like in this question) with other variables to make you get confused (a lot of people will ) so just keep in mind it is a 2 Unit exam, we are not bloody Euler.

 

[taeyang]

By the way, if you mean the first part ^^ ... They assume you know the formula v = d/t

So, just rearrange the formula for t ... so.... t = d/v where d = L and v = (v-u) (given in the question)

so you just plug in these values into the formula given

 

[UnrealAnchovies]

hahahahaha all the time in the world and you still suck!

 

[taeyang]

hahahahaha all the time in the world and you still suck!

??

 

[ThandraCityRail]

well, all you do is differentiate the equation in the first part with respect to E in terms of v, now, you should get v = 0 and v = 3u/2.. Now, to check whether v = 3u/2 is a minimum, we can either find the second derivative OR test the slope of the gradient of the tangent to the curve on either side of v = 3u/2 (so basically, if the gradient is - on the left of v = 3u/2 and then changes to + on the right of v = 3u/2 the point is a minimum)

SO... since we don't want to derive it again ( I mean look at the damned thing! ) we will just test points to the left and right of v = 3u/2 (we don't worry about v = 0 because that gives us the gradient being 0) and see if the sign changes...

Since we know v > u > 0 ... we know that u has to be greater than 0.. so what I did was sub in values such as v = 5u/2 (Becuase it is MORE than 3u/2 for the right side) and v = u/2 (because it is LESS than 3u/2 for the left side) ... turns out the sign changes from negative to positive meaning it is a minimum. It's as simple as that!

The whole point of maxima and minima is to get you to find where given things are a maximum or a minimum value and the way we are taught is with those two methods... Just keep in mind it almost CERTAINLY wants you to differentiate and test points (like we did) or differentiate again.. They will also try to confuse you (like in this question) with other variables to make you get confused (a lot of people will ) so just keep in mind it is a 2 Unit exam, we are not bloody Euler.

thanks for the great response! yeah i'm kinda getting my head around it now it's much more doable than what ti was before... i'll keep at it

thanks!

 

[ThandraCityRail]

By the way, if you mean the first part ^^ ... They assume you know the formula v = d/t

So, just rearrange the formula for t ... so.... t = d/v where d = L and v = (v-u) (given in the question)

so you just plug in these values into the formula given

yeah i figured, thanks again!

 

[ThandraCityRail]

hahahahaha all the time in the world and you still suck!

ease down mr 1/1