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#1 Re: Help Me ! » Catenary » Yesterday 01:06:48


Further analysis:

It looks like the water makes no difference.  see … y-and-sag/

I've played about with those numbers but no solution has appeared so far.  I'm trying to take the riverbed as the x axis and then find the lowest point on the curve ( made symmetrical so x = 0), which is the number that mathematicians usually call a in the equation.

But I also need to know the height above the bed of the wire fixing points.


#2 Re: Help Me ! » Write A Linear Equation » 2021-05-07 20:04:21


Does it?

the dollar value V of the product in terms of the year t

That's what you have done.


#3 Re: Help Me ! » Derivative of Composite Functions » 2021-05-07 20:02:04


Please post your answer as far as you have got.


#4 Re: Help Me ! » Chain Rule » 2021-05-07 06:13:57


θ(t) = (pi/3)cos[(1/2)•sqrt{(2k/5}t]

Unpick the separate functions from the inside to the outside:

u = (2k/5) t

v = u^(0.5)  [ do you recognise the use of a power for a square root?]

w = v/2

θ = (pi/3)cos(w).

A four step chain is a tough one if you haven't done any two step ones first.


#5 Re: Help Me ! » Derivative of Composite Functions » 2021-05-07 06:07:45


Start by doing the three differentiations.

'Chain' then together to create an expression for dy/dx.

Maybe you could stop at that, but I expect you should also put the final answer in terms of x only (if possible).

So eliminate 'u' by writing any u bits in terms of v.  Simplify if poss.

Then eliminate v by writing any v bits in terms of x.


#6 Re: Help Me ! » Catenary » 2021-05-07 06:02:27


hi vellawitting

Welcome to the forum.

I see you think the wire will follow a catenary curve.  If you used a rope then the weight of water in the rope would probably throw the calculations.  As it's a wire, maybe we're ok.

The catenary is usually used for a suspension bridge between two fixed points. Because the barge is under tow that may upset the symmetry of the curve.  Also the formula that is usually used gives the 'sag' using the distance between the fixed points.  You've given the length of the wire.

So I've got some homework to do looking up the underlying theory that is used to develop the equation.

I'll come back to this when I've made some progress.

Can I assume that the height of the fixed points is close enough to the water level so that we can disregard it?  If not I'll need more measurements.

Are you concerned that the wire will scrape the bottom in shallow water?


#7 Re: Help Me ! » Equidistance » 2021-05-07 05:53:19


That's the correct place to start.

But we need all the points that are equidistant.  So you need the perpendicular bisector of AB … isect.html

Steps: 1.  work out the gradient of AB

2. Hence the gradient of a line that is perpendicular to AB.

3. Using that and the midpoint, work out the equation of that line.


#8 Re: Help Me ! » Write A Linear Equation » 2021-05-07 05:45:18


First is correct.

second: The drop in value is 24000 - 2000 over ten years, so divide that by 10 to get the rate of decrease, is the negative m value.  You have an intercept (24000) so the equation follows from that.


#9 Re: Help Me ! » Write A Linear Equation » 2021-05-07 05:40:43



The first is a decrease so the gradient should be negative.

Also I would include the explanation of what t represents as it is unusual.


#10 Re: Help Me ! » Road Grade » 2021-05-05 20:33:18


6 isn't that amount.

Imagine a triangle ABC with a right angle at B and angle A = 3.43 degrees.  So BC = 6 and AC = 100.

Now a second, larger triangle DEF with a right angle at E and angle D = 3.43 degrees.  DF = 200 and you are trying to find EF.

So EF = 200 * sine(3.43)

That isn't 6.


#11 Re: Help Me ! » Sales » 2021-05-05 20:26:56


Yes, very strange and you're better off doing something more challenging I think smile


#12 Re: Help Me ! » Sales » 2021-05-05 19:55:21


That fits with what I said.  First is 135% increase.

It takes longer to figure out what they're asking than to do it.


#13 Re: Help Me ! » Road Grade » 2021-05-05 19:54:02


OK. That's the angle of the slope in degrees.

Now use Opp = Hyp x sin(angle)


#15 Re: Help Me ! » Electronics » 2021-05-05 19:48:40


For part (a), I must evaluate for y given all the values of x.  Yes.

For part (b), I must graph the points formed by the values
if and y. How do I use the graph to estimate the resistance when x is
85.5, a value for x not given in the table?

It's called 'interpolation'.  Literally 'between the points'. Once you have the points plotted you'll see they lie on a curve.  If you try to connect them with a smooth curve, you can then read off the y value for x = 85.5

The formula will give you the actual y value.  It should be about the same as your estimate using interpolation.

(d) I'm not sure either.  If the formula was x^2 we would say it's a square law.  As it's 1/x^2 I guess they want you to say it's an inverse square law.  That's like Newton's law for gravity.


#16 Re: Help Me ! » Sales » 2021-05-05 19:39:12


The wording is rather unclear.  I'm assuming that for a single year the line representing the increase (or decrease) is given by the value of m.

So in the first case, if the actual sales figure at the start was 100, then at the end it is 100 + 135/1 = 235.  So the percentage increase is (235-100)/100 %

If you use a different start value (not 100) , you'll find the result is the same.


#17 Re: Help Me ! » Road Grade » 2021-05-05 19:33:07


The speed is not a part of this question.  It is just a trigonometry problem.

Be careful about slope questions on roads.  It's a lot easier to calculate the slope using the gain in height (Opp) and the distance along the road (Hyp).  A theodolite will give Opp and there are plenty of ways to find Hyp.

But this question clearly says the slope is 6/100, which means 6 = Opp and 100 = Adj(cent) 

If you drive 200 feet that's a Hyp distance.  So you need to calculate Adj.  6/100 is the tangent not the sine, so you'll have either find the sine or calculate  Adj using Pythag.


#18 Re: Help Me ! » Using Intercept Form » 2021-05-05 00:12:12


Therefore d = 1 and you're done.

You know already that a = b = d so you can plug into the intercept equation.


#19 Re: Help Me ! » Speed of An Object » 2021-05-04 05:44:12


Firstly, note the use of the word speed, not velocity.  Velocity is a vector variable meaning that direction is important as well as magnitude.  Speed is a scalar, meaning that only magnitude matters.

So the maximum speed occurs when the ds/dt term has a maximum or a minimum.  It might seem odd that the minimum also counts, but on the graph a minimum means a highest but negative value.

So we are looking for the values of t that make |(-1/8)sin(t)| maximum.  The || lines here mean the 'absolute value of' ignoring the sign of the answer.

We cannot do much about -1/8 as its fixed so the only chance to vary the speed comes from the sin(t) term.

Look at … raphs.html

This shows the graph of a sine function.  It has a maximum positive value at t = pi/2 (measured in radians**) and a minimum at t = 3pi/2.  And it is a periodic function, meaning the values repeat every 2pi.

So there are many answers; every time a max or min occurs.  We could write the answer like this:

pi/2;  3pi/2;  5pi/2;  7pi/2 etc etc.

A neater way to say this is to give the first value, pi/2, and then give a term that describes all the repeats in one go.  The repeats are pi apart so n.pi does this (where n is any integer).  Writing the answer in this way is called the general answer taking full account or the periodic nature of the sine curve.

** I'm hoping you know about radians.  If not read on:

When you first meet angles you are taught that there are 360 degrees in a full turn.  This way of writing angles is not the only way.  In calculus d(cosx)dx = -sinx only if the angles are measured in radians. The conversion is:

pi radians = 180 degrees.

so sin(90°) = sin(pi/2) which is why the answer is given with t in  radians not degrees.

More here:


#20 Re: Help Me ! » Restaurant » 2021-05-04 05:19:46


Yes those are correct.


#21 Re: Help Me ! » Parallel, Perpendicular, or Neither. » 2021-05-04 05:09:16


Ok.  I'll try again.

The slope or gradient of a line is found by noting the coordinates of two points and working out [difference in ys]/[difference in xs]

So I choose a line, AC, and then boxed it in with a rectangle ABCD, making the sides horizontal and vertical.

The gradient of AC is CD/AD which in my diagram is 2/3

Then I turned the rectangle around 90 degrees to make IJKL.  The diagonal IK is at right angles to AC.  Its gradient is IL/LK.

In going from L to K we are going in the positive direction so we can record that distance as +2.  But we have to go down to get from I to L, so rather than recording this distance as +3, it should be written as -3.  That makes the gradient -3/2.  This is what you should expect as it slopes the other way, so it should have a negative gradient.

Note that 2/3 multiplied by -3/2 gives -1.

Of course my diagram only shows this for those measurements but you should be able to see that it will always work out like this whatever the size of the rectangle because the up in one rectangle becomes the across in the other.

So the general test for perpendicular lines is to ask if the product of the gradients is -1. 

My third rectangle has HF parallel to AC.  I've made the rectangle twice the size but, as both the up and the across have got bigger by the same factor, the gradient comes out the same as AC.

Again this is a general rule: same gradient means parallel.

Hope that helps,


#22 Re: Help Me ! » Speed of An Object » 2021-05-03 19:26:15


You're right that domain is used for the x values and range for the ys.

But range is also used, as here, as a general term for any variable., x or y.


#23 Re: Help Me ! » Restaurant » 2021-05-03 19:24:05


where t = 0 corresponds to November 1, 2014; t = 1 corresponds
to January 1, 2015; t = 2 corresponds to March 1, 2015;
and so on.

So each time t increases by 1, add two months to the date.


#25 Re: Help Me ! » Arch Bridge » 2021-05-03 10:39:49


I'm viewing the forum using a Lenovo ideapad laptop.  The screen is roughly 14 inches wide.  In the top left hand corner of every post is a number: #1 for the first post; #2 for the next and so on.  If you cannot see this, what device are you using?  A mobile screen is probably too small for you to make much sense of many of my replies.

In the third post of this thread I explained exactly how to work out the width.  Try graphing the equation.

It's too late at night for me to look now at any of your other posts.


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