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#1 Re: Help Me ! » Problem in implicit differentiation » 2018-12-07 21:24:06

hi Malubia

On the page about partial differentiation f is a function of y and x.  So if you wanted to make a graph of this function a sheet of 2 dimensional graph paper wouldn't be enough.  You need a 3 dimensional graph with f having its values plotted in the 'z' direction.  Imagine you have the x and y axes on a flat surface and the z axis goes up in the air.  If you choose values of x and y, there will be values of z making a (wavy) surface above (and below) the x-y plane.

For partial differentiation we assume that one of the variables (let's say y) is fixed.  That means we are only considering a part of the surface where x and z can vary but y stay constant.  This is now a 2 dimensional problem once more so we can use the usual rules of differentiation. Let's imagine a possible situation where we could use this.  If the surface has a dip down to a lowest point, taking y as constant will be like taking a slice through the surface and making a curve which has a lowest point.  If we now re-do the differentiation with x constant there will similarly be a lowest point.  For each there will still be a variable (the one that we've been pretending is constant) in the equations.  But with a pair of such equations we can find the single point where both x and y are lowest together, in other words the bottom of the dip. 

In the same way we can find maximum points on the surface and also saddle points where one variable is lowest and the other highest.  If you are walking towards a mountain pass, that is an example of a saddle point.  As you go through the pass the ground slopes down in front and behind so you are at a maximum in that direction; but to either side the mountain sides slope up, so in that direction you are at a minimum.  In both directions the partial derivatives are zero.

Now back to the 2 dimensional cases.  If the function is expressed as y = function of x, then the differentiation is relatively straight forward.  But sometimes we don't have that; we may have an equation such as x^2 + y^2 = R where neither variable is the subject of the equation.  In this case it is possible to re-arrange the equation so that y is the subject, but it's not an easy differentiation then, and sometimes you cannot do the re-arrangement at all.

That's when implicit differentiation becomes useful.  Whenever you are differentiating it is always with respect to a variable, let's say x.  Do the x containing components as usual but you have to remember that y is a variable too.  So it must be differentiated with respect to x as well.  And that's the process of implicit differentiation.

It's worth noting that even with a simple function like y = x^2 you are actually using implicit differentiation.  The right hand side becomes 2x.  The left hand side becomes dy/dx.  So it has also been differentiated with respect to x.

Hope that helps,


Thanks for the good wishes in the other post.  Let me know if you discover how to achieve it smile Living for a 1000 years that is.

#2 Re: Help Me ! » Problem in calculus. Please help!!!! » 2018-12-06 20:43:31

hi Malubia

Welcome to the forum.

That had me puzzled for a while.

Draw a horizontal line  AB across from point   A = ( a, f(a) ) to  B = ( b, f(a) ) with a point X = (x , f(a) ) anywhere between A and B.

Define t to be the fraction XB/AB.


Hope that helps,


#3 Re: Help Me ! » Stuck on previous exam questions! » 2018-11-30 21:23:03

hi Monox D. I-Fly

The express train starts off later.  See post #18.


#5 Re: Exercises » triangle » 2018-11-23 22:12:23

Hi Tony123

I used the cosine rule thus:


#6 Re: Help Me ! » Physics: Help with understanding Centripetal and Centrifugal forces. » 2018-11-23 03:46:35


I've managed to do A level maths and physics and then later teach it, without ever needing to refer to centrifugal force.  Why do we need something that's made up ?  Mathematicians would never consider something that's imaginary would they ? smile

According to Newton, an object such as your yo yo, will go in a straight line unless acted upon by an external force.  To make it go in a circle a centrally acting force is needed and that's what we call centripetal force.  We can feel the string tugging but that's just the tension in the string.

If you could cut the string whilst the yo yo is whizzing round, it would continue in a straight line, tangential to the circle. If you're sitting on the back seat of a car as it is driven round a sharp bend, there's nothing to stop you from carrying on in a straight line.  You might think you're being pushed sideways but actually you're going straight and the car (which can go in a circle due to friction between the tyres and the road) is curving towards you.  When you make contact with the side of the car, the car can exert a force on you, to make you go in a circle too.

The genius of Newton was he could see past the individual's self centred perspective and understand what is really going on. I regard avoiding the term centrifugal as part of my homage to the great man.


#7 Re: Exercises » Differentiability » 2018-11-21 21:21:15

Correct.  Here's the graph: … )+abs(x-1)

So just show that the left and right limits are not the same.


#8 Re: Exercises » Domain and Range » 2018-11-17 20:49:45

The domain is the set of possible 'x' values and the range is the set of possible 'y' values.


You are correct that the curve tends to zero as x tends to infinity but that doesn't tell us the range.  You cannot have the value x=3; it is excluded from the domain, but you can get as close to 3 as you like and as you do the function has larger and larger values.  There is no value of 'y' that cannot be attained so the range is ( -∞  , ∞ )  Note y = zero is attained at x = -2


#9 Re: Exercises » Domain and Range » 2018-11-14 21:36:33


There's a lot of brackets there so here's how I analysed the function:


(SQRT(x+2))      /      ((x^2)-9)

Drop unnecessary brackets

SQRT(x+2)        /       (x^2-9)

Question 1.  Is this the correct function ?

Question 2.  Have you looked at the graph linked in post 2 ?

Please answer these questions.  If I don't get both answers I shall not be posting again on this thread.


#10 Re: Exercises » Domain and Range » 2018-11-14 03:43:51

x                           F(x)
     0                     -0.15713484
  100                0.00101086
   -1                    -0.125
    2                    -0.4
2.99999            -37267.82447
2.999999999    -372677965.4
3.00000001     37267799.89
3.000000001     372677965.5

Using Excel to calculate =SQRT(x+2)/(x^2-9)

As x approaches 3 from below the values of F are getting increasingly big in the negative direction.
As x approaches 3 from above the values of F are getting increasingly big and positive.

Have you looked at the graph?

Are we using the same function ?


#11 Re: Exercises » Domain and Range » 2018-11-12 20:07:10

hi Zeeshan 01

I'm not sure if there is another way.  Have a look at this page: … omain.html

Somehow you have to find all the possible values that F(x) can have.

You cannot list them in a set as this is (except at x=3) a continuous function.  So you have to write the answer in this way:

[lowest values, highest value] although sometimes a square bracket may be replaced with a round bracket to indicate that an endpoint is not included.

I think that considering the graph helps a lot with this.  Did you follow the link?  It shows that the function is continuous in two sections separated by x=3.  It also shows that the function tends to both + and - infinity.  Thus we know that the range is all real values of x.


#12 Re: Exercises » Domain and Range » 2018-11-11 20:34:00

Domain f(x)= [-2,+infinity)

Yes, but you must also add x  ≠ 3 as there is no value for F(x) at this point. infinity is not regarded as a number.

The range is all the values that F(x) can take.  (x-3)(x+3) determines this because when the denominator approaches 3, the denominator tends to zero so the function tends  to either + or - infinity.  This means that all values occur for F(x) so the range is (-∞ , +∞) If you look at the graph, choose any value for F and draw a horizontal line to try to cut the curve.  It is always possible.


#13 Re: Exercises » Domain and Range » 2018-11-10 20:52:46

hi Zeeshan 01

Haven't heard from you for a while.  How are you doing?

To answer your question, I made a sketch of the graph.

Firstly, if x < -2 then the square root cannot be computed so that sets a lower limit and determines the domain.

At x = -2 , F(x) = 0

x^2 - 9 = (x-3)(x+3) so there will be vertical asymptotes at -3 and +3, meaning that x tends to either + or minus infinity. The negative one is outside the domain so we needn't consider it.   Consider what sign F(x) has as x approaches 3 from above and below.  When x > 3 and approaching 3 both the numerator and denominator are + so the curve will tend to + infinity.  When x < 3 and approaching 3 the denominator switches to negative so that part of the curve tends to - infinity.  That's enough to decide the range.

If you want to view the graph go to … 2)/(x^2-9)

Hope that helps,


#14 Re: Help Me ! » Law of sines and Trig » 2018-11-05 20:11:20

hi careless25

Angles PQS and SQR are shown with two arcs.  I think this can be interpreted as meaning that they are equal.

So in triangle PQS that angle can be found and then in triangle SQR you know two (and therefore all three) angles and a side.  So x can be found.


#15 Re: Help Me ! » How do you get v dv/dt = d/dt (v^2/2) » 2018-11-01 23:48:55

hi kpstatic

Welcome to the forum.

In mechanics, to switch between acceleration, a, velocity, v and distance, s it is necessary to differentiate and integrate as

I seem to remember this bit of algebraic manipulation arises when dealing with simple harmonic motion, because a is defined as a function of s rather than t.  So a way is needed to express a in terms that can be integrated.

If you post back more precisely where you've got to in your course and why you came searching here, I'll try to provide more help.


ps.  s is used for distance as d is likely to be used for differentiating. x may also be used.

#16 Re: Help Me ! » Volume of irregular polyhedrons » 2018-10-20 19:51:35

Hi Mathegocart,

Thanks for the help.

I'd forgotten I'm on someone else's wi fi at the moment. It's their content blocker.  Seems to be a common problem with Vodafone. So I'll wait and see if the OP wants my diagram before tinkering with the settings.


#17 Re: Help Me ! » Volume of irregular polyhedrons » 2018-10-19 21:34:57

hi thehay95

Welcome to the forum.

I've had a go at sketching this solid.  Trouble is that just having the letters PQRSTUV doesn't exactly tell me which points to join to which.  It looks like there's a prism PQRSTV, which has a triangular cross section QRV, and sitting on this is a rectangular based pyramid TVRSU, where the base is TVRS and the vertex U.

Volume of the prism should be fairly straight forward.  Calculate the area of the triangle QRV and multiply by the length, PQ.

The pyramid is more tricky.  You'll need the area of the base, TVRS.  This is a rectangle.  You need Pythag. to calculate VR.  Then the perpendicular height from TVRS to U.  If you draw the triangle UZY, you'll see that UZ is not perpendicular to ZY so that length won't do.  But you can calculate UY and the angle UYZ and so treating UY as a hypotenuse you can get the required height by opposite (height) = hyp x sin(UYZ).  Once you have this the volume of the pyramid is  one third of the area of base times the perpendicular height.

Then just add together the two volumes.  smile

Hope that helps,

image upload problem.  I have a diagram but I cannot log in to imgur to upload it.  Some problem with a Vodafone security certificate ??? I'm on BT broadband.  Anyone got any suggestions.  It happens with both IE11 and Chrome. My account still exists as old images are still showing on the forum


#18 Re: Introductions » Introduction » 2018-10-08 19:23:24

hi iamaditya

There's a thread on the forum that includes the discussion and the link to the download.  It was a while back and I've forgotten the details; sorry.


#19 Re: Introductions » Introduction » 2018-10-07 19:53:06

hi iamaditya

No Spam. Spam includes messages that have no relevance to the topic, that are annoying, repetitious or promotional in nature.

I remove adverts because they are (1) sometimes pornographic; (2) they 'clog' up the site with irrelevant rubbish; and (3) it's not fair to the paying sponsors to allow free advertising. If they stop supporting us the forum's very existence  is at risk.

Mostly, these posters join the forum; create one or more posts which contain an advert, either in the post or the signature, then we never hear from them again.

Just once, we had a genuine mathematician who posted a link to the sale of his maths book.  Bobbym had a word with him, and he removed the advert and replaced it with a link to a free download of the book.  That seemed like a good compromise.


#20 Re: Help Me ! » Algebra: Polynomial Question » 2018-10-01 19:34:49

hi math9maniac

The 'simplification' was not valid.  √ (A-B) is not generally equal to √ A - √ B

I squared both sides of the equation to get

I'm still not seeing any real solutions.  See … ^8 - x + 2


#21 Re: Help Me ! » 3d plane in a cube » 2018-10-01 01:28:18

hi n0tjyxgsgs

Welcome to the forum.

This problem came up ages ago so I've mostly forgotten it.  Some other posters provided quicker solutions than mine; that's fine; I never said mine was the only one nor that it was quick.  The OP was happy to be taught some new maths along the way which I was happy to do.

PQRC is a sloping plane.  So QR isn't a continuation of PQ.  Not sure what you mean here.


#23 Re: Help Me ! » Question » 2018-09-29 21:04:05

Hi Sterling94,

Welcome to the forum.

When two dice are thrown there are 36 possible outcomes but we can limit ourselves to just (5,x) 6 possibilities plus (x,5) another 6 possibilities less the one we've counted twice (5,5).  So 11 in total.  Of these two give a total of 8 so 2/11


#24 Re: Computer Math » Software Solution 2019 Latest Updates » 2018-09-27 19:19:50

hi joseph

Welcome to the forum.

Thanks for the supportive comments.  MIF is funded by carefully selected advertisers.  Please don't use the forum for free advertising.


#25 Re: Help Me ! » Trigonometric Confusion? » 2018-09-27 19:16:47

It's very odd.  You'll have to raise this with your teacher.


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