Calculus! For beginners, with a ‘no theorems’ approach!

Quote: Originally posted by aga

If expulsion does get discussed again, may i remind the teaching staff of the modulus of those negatives i still have in safe keeping from the 1992 school cultural visit to Paris. You know, the ones of yourselves with 'Monsieur Fouetter la Derriere' at madame Fifi's place on the Rue de La Huchette ...

Quote: Originally posted by Darkstar

I was only four years old in 1992! Shame on you! P.S. - Blackmailer!

Simple Differential Equations:

Perhaps Robo-Integrator suddenly regains some appeal?

Next up: some real world examples...

[Edited on 14-4-2016 by blogfast25]

Quote: Originally posted by aga

OK. All looks fine, right up to the integration factor example, apart from the logs. Are these general Rules ? : $$e^{\ln x^n}=x^n$$ $$n\ln x=\ln x^n$$ This step i do not follow at all:- $$\int \frac{3}{x}dx=3\ln x$$ I was expecting it to be this infinte horror $$\int \frac 3x dx = 3\int x^{-1}dx = \frac 30$$ Probably best to 'fess up now that that i do not know what logarithms are.

Quote: Originally posted by blogfast25

Exception to the power rule of integration: $$\int x^n dx=\frac{x^{n+1}}{n+1}+C$$ EXCEPT for: $$n=-1$$ Then: $$\int x^{-1}dx=\ln x+C$$

Quote: Originally posted by aga

Does the 'normal' rule hold true for all other -ve values of n ?

Train stopping problem:

Although this problem can also be solved in a quicker way, using calculus is a nice, simple demonstration of calculus in kinematic problems, so here goes...

A 150-car freight train is approaching Carbondale, Yorkshire at a constant speed of 15 meter per second (approximately 34 mph). As the train nears a railroad crossing, the locomotive engineer sees a car stalled on the tracks. It takes him 4 seconds to react before he applies the brakes. It then takes the train another 1.25 minutes to come to a full stop. Assume that the deceleration of the train is constant while the brakes are being applied.

How far does the train travel from the moment the engineer sees the car until the time the train comes to a full stop?

Schematic:


Let’s express the basic data in ‘mathspeak’:
$$v_0=15\;\mathrm{m/s}, t_0=4\:\mathrm{s}, t_s-t_0=1.25\:\mathrm{min}=75\:\mathrm{s}$$
We imagine the train to run along a x-axis, pointing in the direction of motion.

Now we take a quick short cut, to calculate the deceleration during the braking period.
$$a=\frac{\Delta v}{\Delta t}=\frac{0-v_0}{t_s-t_0}=-\frac{15}{75}=-\frac15=-0.2\:\mathrm{m/s^2}$$

So the train's velocity is reduced by 0.2 m/s, per second of braking.

The velocity – time function is not smooth and continuous because v suddenly changes at t=4 s. For that reason we’ll ignore what went on before t=4 s, at least for now.

From higher up in the thread we know that:
$$a=\frac{dv}{dt}$$
$$dv=adt$$
$$\int dv=\int adt=a\int dt$$
$$v=-\frac15 t+C$$
Using the boundary condition:
$$t=4, v=15$$
$$\implies 15=-\frac15 \times 4+C$$
$$C=\frac{79}{5}$$
$$\implies v=-\frac15 t+\frac{79}{5}$$
So the v,t function (for t > 4 s) is:
$$v=\frac15(79-t)$$
We also know that
$$v=\frac{dx}{dt}$$
$$dx=vdt$$
To calculate the travelled distance during braking we can apply:
$$\Delta x=\int_4^{79} vdt=\int_4^{79} \frac15 (79-t)dt$$
$$=\frac15\big[79t-\frac12 t^2\big]_4^{79}=562.5\:\mathrm{m}$$
Now we need to add the distance travelled during the 4 seconds before the driver puts on the brakes:
$$4 \times 15=60\:\mathrm{m}$$
So the total distance travelled between the driver noticing the car and full stop is 622.5 m.

[Edited on 15-4-2016 by blogfast25]

Quote: Originally posted by aga

In the missing steps after this step : $$\frac15\big[79t-\frac12 t^2\big]_4^{79}$$ does the 1/5 apply to both bits or to the overall result ?

Let's nip that bit of confusion in the bud immediately: 1/5 is a factor and applies to everything between the brackets. So calculate the bit between [ ] brackets first, then multiply by the factor.

Learning to read calculus problems is a already a huge step forward (seriously!)

[Edited on 15-4-2016 by blogfast25]

Quote: Originally posted by aga

$$= 562.5 + 60 = 622.5m$$ ... exactly the same answer ! In't maffs amazing ?

It's totally maff!

Your way is how most would do it, BTW. But I'm an old oddball!

One minor comment: when calculating:

$$\Big[\int f(x)dx\Big]_a^b=\Big[F(x)\Big]_a^b=F(b)-F(a)$$

... you don't need to invoke C because we know it will drop out. But what you wrote was mathematically correct nonetheless.

Hot Coffee: no need to blow on it!

So you've just bought yourself a cup of hot, steaming coffee in a flimsy throw away cup with lid. Now the annoyingly inevitable starts: it cools down fairly quickly. In this exercise we'll try and model that cooling.

A few definitions and assumptions:

The coffee of mass m contains heat energy (Enthalpy) Q (J). Coffee has a specific heat capacity c_p (J/kg K).

Heat loss (cooling) is through convective loss to the surrounding air (radiative losses are considered small at < 100 C).

Temperature of the coffee is more or less uniform throughout the cup.

In those circumstances Newton's cooling law tells us that:
$$-\frac{dQ}{dt}\propto A(T-T_a)$$
... where dQ/dt is the rate of heat loss energy, A is the total surface area of the cup, T is the temperature of the coffee and T_a the ambient temperature (which is presumed constant). The negative sign is needed because dQ < 0 (it's a LOSS).

We can turn this into an identity by means of a proportionality constant h:
$$ \frac{dQ}{dt}=-hA(T-T_a)$$
h (W/K m²) is known as the Heat Transfer Coefficient. It's specific to each heating/cooling problem (it's not a universal constant). (in a different context, double glazing people often refer to it as the "K-factor").

Thermodynamics also tells that when an object of mass m with specific heat capacity c_p changes in temperature it's heat content also changes acc.:

$$\Delta Q=mc_p\Delta T$$
By limit taking we can convert this into a differential equation:
$$\lim_{\Delta T \to 0}\frac{\Delta Q}{\Delta T}=\frac{dQ}{dT}=mc_p$$
$$\implies dQ=mc_pdT$$
Inserting this into the second equation above we get:
$$mc_p\frac{dT}{dt}=-hA(T-T_a)$$
Or:
$$\frac{dT}{dt}=-\frac{hA}{mc_p}(T-T_a)$$
Now we call ('tau'):
$$\tau=\frac{hA}{mc_p}$$
$$\frac{dT}{dt}=-\tau (T-T_a)$$
$$\frac{dT}{T-T_a}=-\tau dt$$
Because:
$$d(T-T_a)=dT-0=dT$$
$$\implies \frac{d(T-T_a)}{T-T_a}=-\tau dt$$
We can now integrate both sides directly, using the boundary conditions:
$$(t=0, T=T_i),(t, T)$$
... where T_i is the initial temperature of the coffee and T the temperature after some time t, so:
$$\int_{T_i}^{T}\frac{d(T-T_a)}{T-T_a}=-\int_0^t\tau dt$$
Remembering the properties of logs ( ), we get:
$$\big[\ln(T-T_a)\big]_{T_i}^{T}=-\tau t$$
$$\ln \Big[\frac{T-T_a}{T_i-T_a}\Big]=-\tau t$$
$$\frac{T-T_a}{T_i-T_a}=e^{-\tau t}$$
$$\large{T=T_a+(T_i-T_a)e^{-\tau t}}$$
The temperature thus evolves as follows (schematic):



Evaluate (qualitatively):

1. The time needed to reach T_a.
2. The influence of h, A, m and c_p on the cooling process.


[Edited on 16-4-2016 by blogfast25]

Quote: Originally posted by blogfast25

... you don't need to invoke C because we know it will drop out.

Quote: Originally posted by aga

You might know that, but for me, only having seen C a matter of minutes ago, i felt it important to remember that it does appear then drop out. That working-out/post is also intended to be useful for me to refer back to.

I am a maff-nerd and I endorse those messages!


[Edited on 16-4-2016 by blogfast25]

Quote: Originally posted by blogfast25

Evaluate (qualitatively): 1. The time needed to reach Ta. 2. The influence of h, A, m and cp on the cooling process.

Quote: Originally posted by aga

Erm, the 'qualitatively' part knocked me sideways (my fall being broken by a rather large pile of empty beer cans, so no lasting damage) What does the question mean ? Come up with an equation that :- 1. predicts the time needed to reach Ta starting from an abitrary time T ? 2. re-arrange said equation into other equations so h, A, m and cp respectively are on the other side ? Edit: You ain't fooling nobody : tau is u in disguise, trying to sneak in via the back door.

Quote: Originally posted by blogfast25

Qualitatively means: without actual numbers or calculation. How long for the coffee to completely cool to Ta?

Quote: Originally posted by blogfast25

Evaluate (qualitatively): 1. The time needed to reach Ta. 2. The influence of h, A, m and cp on the cooling process.

@aga:

Well, well.

Decided to inject some experimentalism into it? Good!

I agree with your value of V but for A I get 0.0568 m². I need to check that value, though. Did you take into account the surface area of the top and bottom?

I'm curious to see your data and how you will derive your estimate of h from them. I know how I would.

[Edited on 17-4-2016 by blogfast25]

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   Evaluate (qualitatively): 1. The time needed to reach Ta. 2. The influence of h, A, m and cp on the cooling process. 1.Mathematically the coffee will never reach ambient temperature (i.e. inifinite time). In practice it reaches room temperature shortly before you finish doing whatever distracted you from drinking it hot (see above !) 2. less mass (m) = quicker coooling less specific heat capacity (cp) = quicker cooling less area (A) = slower cooling less heat transfer coefficient (h) = slower cooling [Edited on 17-4-2016 by aga]

All present and correct. 100/100.

Quote: Originally posted by blogfast25

Did you take into account the surface area of the top and bottom? I'm curious to see your data and how you will derive your estimate of h from them. I know how I would.

Quote: Originally posted by aga

I just set the equation up in a spreadsheet, used the calculated values (also the s.h.c. of water @ 4.186) and tried 'h' numbers until the result roughly agreed with the experimental result. Quicker for me that way.

The scientific way of doing this would be as follows.

Remember that:
$$\ln \Big[\frac{T-T_a}{T_i-T_a}\Big]=-\tau t$$

You would plot in Excell:

$$\ln \Big[\frac{T-T_a}{T_i-T_a}\Big]$$

versus:

$$t$$

That should give something looking like:



The + are the data points, they should roughly be on a straight line. Then make Excell fit a line to the data (linear regression).

The slope of the line is:

$$-\tau$$

Then calculate h from tau.

Checking surface area of the cup, by calculus.

For a revolution body:



... the surface area A (end bits not counted) is given by:
$$A=2\pi\int_a^bf(x)\sqrt{1+[f'(x)]^2}dx$$
For the cup, with an x-axis running through it horizontal centre axis:
$$f(x)=\frac{5}{17}x+70$$
$$f'(x)=\frac{5}{17}$$
$$A=2\pi\int_0^{85}\sqrt{1+\frac{5^2}{17^2}}(\frac{5}{17}x+70)dx$$
$$A=2.085\pi\Big[\frac{5}{34}x^2+70x\Big]_0^{85}=45927$$
Lid:
$$\frac{\pi}{4}95^2=7088$$
Cone + lid:
$$53015\:\mathrm{mm^2}=0.053\:\mathrm{m^2}$$

Quote: Originally posted by aga

All day reviving algebra cells did not leave any brain capacity to learn what a linear regression was. What does a second or third order integrally equation look like ?

Quote: Originally posted by blogfast25

$$m\frac{d^2y}{dt^2}=mg-ky$$

Is 'g' gravitational acceleration ?

Must be, seeing as it popped out of nowhere

You do realise that i'll soon be able to calculate proof for agaspace ...

Well, maybe if i can finally master those nasty fractions, do 'u' substitutions, drink less, think straight etc.

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   $$m\frac{d^2y}{dt^2}=mg-ky$$ Is 'g' gravitational acceleration ?

@aga:

I happen to know that for agaspace you need partial derivatives:

For a function f of x AND y:
$$f(x,y)$$
Partial derivative to x:
$$\frac{\partial f}{\partial x}$$
... and to y:
$$\frac{\partial f}{\partial y}$$

You can't calculate agaspace astral mass time differential tensors w/o them!

But we don't need them for agasyphon^TM, so we'll stick to the latter, tomorrow.

[Edited on 17-4-2016 by blogfast25]

Here's the plot for:

$$\ln \Big[\frac{T-T_a}{T_i-T_a}\Big]=-\tau t$$

$$\Big[\frac{T-T_a}{T_i-T_a}\Big]$$

... is also called the "Reduced Temperature", a dimensionless number between 1 and 0.

I used T_a=20 C.



The linear regression equation:

$$y=-0.0006x-0.021$$

... gives us an estimate for:

$$\tau=0.0006\:\mathrm{s^{-1}}$$
$$\tau=\frac{hA}{mc_p}$$

With A=0.053 m², m=0.458 kg, c_p=4181 J/kg K then:

$$h=\frac{\tau mc_p}{A}=22\:\mathrm{Wm^{-2}K^{-1}}$$

Considering there's some noise on the data, this is a crude estimate, of course.

This source puts up a value range of 5 to 37 Wm^-2K^-1 for free convection in gas (air) or dry vapour, so we're in the right ball park.

[Edited on 18-4-2016 by blogfast25]

Quote: Originally posted by aga

22 ? Doh. I was very uncertain of the Units, so perhaps i got metres/millimetres mixed up somewhere. $$f(x,y,z)$$ now that would definitely come in handy. Quote: astral mass time differential tensors They sound very cool ! What colour are they ?

Quote: Originally posted by blogfast25

Perhaps the most common f(x,y,z) type function is the volume of a right angled box: $$V=lhw$$

Syphon: tank emptying time

The tank's shape is of no consequence as long as its cross-section A is constant.

The outflow speed v₂ can be determined by applying Bernoulli's Principle between points 1 and 2:
$$\frac12 v_2^2+gy_2+\frac{p_2}{\rho}=\frac12 v_1^2+gy_1+\frac{p_1}{\rho}$$
The derivation is a bit lengthy, so I've posted it below the second page break.

The result is:
$$v_2=c\sqrt{2gy}$$
So the volumetric throughput Q_v is:
$$Q_v=\frac{\pi}{4}D^2v_2=\frac{\pi}{4}D^2c\sqrt{2gy}$$

During an infinitesimal amount of time dt, the water level decreases by dy (<0), so:
$$dV=-Ady=Q_vdt$$
$$-Ady=\frac{\pi}{4}D^2c\sqrt{2hy}dt=\frac{\pi}{4}D^2c\sqrt{2g}\sqrt{y}dt$$
Let's call:
$$\alpha=\frac{\pi D^2c\sqrt{2g}}{4A}$$
$$-y^{-1/2}dy=\alpha dt$$
Integrate between relevant boundaries:
$$-\int_H^{h_0}y^{-1/2}dy=\int_0^t \alpha dt$$
$$2\big[\sqrt{y}\big]_{h_0}^H=\alpha t$$
$$t=\large{\frac{2}{\alpha}(\sqrt{H}-\sqrt{h_0})}$$
Note that the higher up the tank is placed with respect to 0, the smaller t becomes.
<hr>
But. There's a big but.

The model above doesn't take into account any viscous losses in the syphon's pipe.

To take them into account, with Bernoulli, we get:
$$\frac{1}{2c^2} v_2^2= gy-\frac{\Delta p_s}{\rho}$$
For laminar flow (turbulent flow makes it more complicated) through the pipe:
$$\Delta p_s=\frac{32 \mu L v_2}{D^2}$$
Where:
$$L \approx 2(H-h_0)+h_0$$
Substituting we get:
$$\frac{1}{2c^2} v_2^2= gy-\frac{32 \mu L v_2}{\rho D^2}$$
$$\frac12 v_2^2+\beta v_2-c^2gy=0$$
With:
$$\beta=\frac{32 \mu L c^2}{\rho D^2}$$
This quadratic equation has one positive root:
$$v_2=-\beta+\sqrt{\beta^2+2c^2gy}$$
So:
$$Q_v=\frac{\pi}{4}D^2v_2=\frac{\pi}{4}D^2(\sqrt{\beta^2+2c^2gy}-\beta)$$
Inserting into the DV above the break:
$$-\int_H^{h_0}\frac{dy}{\sqrt{\beta^2+2c^2gy}-\beta}=\int_0^t \alpha dt$$
Where:
$$\alpha=\frac{\pi D^2c}{4A}$$
Although that integral is computable, the result gets very bulky and messy, so I won't go there.
<hr>
Full derivation:
$$\frac12 v_2^2+gy_1+\frac{p_2}{\rho}=\frac12 v_1^2+gy_2+\frac{p_1}{\rho}$$
$$\frac12 (v_2^2-v_1^2)+\frac{p_0}{\rho}=gh_0+\frac{p_1}{\rho}$$
$$A_2v_2=Av_1$$
$$\implies v_2^2-v_1^2=v_2^2\Big[1-\Big(\frac{A_2}{A}\Big)^2\Big]$$
$$A_2=\frac{\pi}{4}D^2$$
$$\frac{1}{c^2}=[1-\Big(\frac{A_2}{A}\Big)^2\Big]$$
$$\frac{1}{2c^2}v_2^2+\frac{p_0}{\rho}=gh_0+\frac{p_1}{\rho}$$
$$p_1=\rho g(y-h_0)+p_0$$
$$\frac{1}{2c^2} v_2^2= gy$$
$$v_2=c\sqrt{2gy}$$
$$Q_v=\frac{\pi}{4}D^2v_2=\frac{\pi}{4}D^2c\sqrt{2gy}$$
$$dV=-Ady=Q_vdt$$
$$-Ady=\frac{\pi}{4}D^2c\sqrt{2hy}dt=\frac{\pi}{4}D^2c\sqrt{2g}\sqrt{y}dt$$
$$\alpha=\frac{\pi D^2c\sqrt{2g}}{4A}$$
$$-y^{-1/2}dy=\alpha dt$$
$$-\int_H^{h_0}y^{-1/2}dy=\int_0^t \alpha dt$$
$$2\big[\sqrt{y}\big]_{h_0}^H=\alpha t$$
$$t=\large{\frac{2}{\alpha}(\sqrt{H}-\sqrt{h_0})}$$




But.
$$\frac{1}{2c^2} v_2^2= gy-\frac{\Delta p_s}{\rho}$$
$$\Delta p_s=\frac{32 \mu L v_2}{D^2}$$
$$L \approx 2(H-h_0)+h_0$$
$$\frac{1}{2c^2} v_2^2= gy-\frac{32 \mu L v_2}{\rho D^2}$$
$$\frac12 v_2^2+\beta v_2-c^2gy=0$$
$$\beta=\frac{32 \mu L c^2}{\rho D^2}$$
$$v_2=-\beta+\sqrt{\beta^2+2c^2gy}$$
$$Q_v=\frac{\pi}{4}D^2v_2=\frac{\pi}{4}D^2(\sqrt{\beta^2+2c^2gy}-\beta)$$
$$-\int_H^{h_0}\frac{dy}{\sqrt{\beta^2+2c^2gy}-\beta}=\int_0^t \alpha dt$$
$$\alpha=\frac{\pi D^2c}{4A}$$

[Edited on 19-4-2016 by blogfast25]

Quote: Originally posted by aga

Cor Blimey ! I think a bit of commentary is going to be required to follow that properly !

Jeez. All that algebra is seriously heavy stuff.

Are all the students going to draw lots to decide who does the experiment(s) ?

Quote: Originally posted by aga

Are all the students going to draw lots to decide who does the experiment(s) ?

Of course. But I've got some Darwins on a dead cert!

Re. the algebra, the next one is much lighter on it.

[Edited on 19-4-2016 by blogfast25]

Hydrogen generator: (student problem)

In the hydrogen gas generator above, a (constant) volume V of dilute hydrochloric acid of initial concentration c₀ reacts with zinc granules acc.:

Zn(s) + 2 HCl(aq) === > ZnCl₂(aq) + H₂(g)

We assume the amount of Zn (number of moles) present to be much, much higher than that of HCl and the temperature to be constant. In those circumstances kinetics allows us to write:
$$\frac{dN}{dt}=kc$$
N is the number of moles of hydrogen formed, t is time, k the reaction rate constant and c the hydrogen chloride concentration.

The concentration of HCl is obviously NOT constant, as each mol of hydrogen generated requires 2 mol of HCl to react away. Stoichiometrically we can write:
$$c=\frac{c_0V-2N}{V}$$



We start off at t=0 with N=0.

Develop expressions for:

1. N(t): the number of moles hydrogen generated as a function of time.

2. The rate of hydrogen generation dN(t)/dt as a function of time.

Heigh-ho, heigh-ho!
Off to work you go!

[Edited on 19-4-2016 by blogfast25]

Mini-recap on substitutions and differentiation

Quote: Originally posted by aga

i'm assuming the C generated in each integration is the same C, so cancel out.

Quote: Originally posted by aga

I guess the rate of gas generation is the first derivative of that, which i think is $$-\frac 12 e^{-\frac {2k}{V}}$$ [Edited on 22-4-2016 by aga] Hang on, surely the Rate will depend on the concentration as well ?

Quote: Originally posted by aga

Oh bollocks. Big fat fecking C bollocks ! Ah well, did i get into the top 30% for effort, or do i remain in the bottom 20% ? [Edited on 22-4-2016 by aga]

Top 40 %!

One more problem coming up tonite, similar to this one (mathematically speaking). Hold beer till about 8 to 8.30 UTC.

After that: partial derivatives!

[Edited on 22-4-2016 by blogfast25]

Quote: Originally posted by blogfast25

Hold beer till about 8 to 8.30 UTC.

Projectile velocity problem:

A 50 kg mass is shot from a cannon straight up with an initial velocity of 10m/s off a bridge that is 100 meters above the ground. If air resistance (drag force Fd) is given by 5v determine the velocity of the mass when it hits the ground.

Solution: First, notice that when we say straight up, we really mean straight up, but in such a way that it will miss the bridge on the way back down. Here is a sketch of the situation.



According to Newton’s Second Law, the projectile will experience acceleration acc.:
$$ma=mg-F_{drag}$$
$$a=\frac{dv}{dt}=v’$$
$$v'=g-\frac{F_{drag}}{m}$$
$$v’=9.8-\frac{5v}{50}=9.8-\frac{v}{10}$$
Initial condition:
$$t=0,v(0)=-10$$
This is a two-step problem. Firstly find an expression for v(t) from the above.


[Edited on 22-4-2016 by blogfast25]

Quote: Originally posted by aga

While we're at it, is bloggers the consumate Calculus devil or an angel on a deep-cover operation ?

Quote: Originally posted by aga

OK. So before i miss another C, here's the Plan : Assuming that the cannon is on a retractable arm that removes the cannon immediately after firing (so we got true straight up/straight down motion), we have to calculate distance travelled from the firing point to the 0m/s point at the top (of the arc in the sketch). Add that to 100m then calculate the velocity of the mass when it hits the ground. It will accelerate down from there by 9.8 m/s/s as per the question data, adjusted by the drag factor. The 5v is giving me some concern, as in 'what units ?', although i suspect that is because i have not done the calculations yet.

Quote: Originally posted by aga

A direction-changing two-dimensional parabola (time is one ..) OK. I'll have to invoke the Beer Ammendment and attempt it on the 'morrow. Side-note: Do you find these bits of maths Easy ? Currently i find them really really hard (as you may have noticed).

Strictly speaking we should do the whole thing in vector calculus:

$$\frac{d\vec{v}}{dt}=\vec{g}-5\vec{v}$$

Does that make it any clearer? As luck has it, here we can deal with the scalars (magnitude and sense of the vectors) only, as above.

Easy? Maybe not but it really is all about practice. I've been using calculus for over 30 years now.

Quote: Originally posted by aga

Can i at least write a letter to my mum before all the Whipping starts ?

So sorry I missed this for some reason.

Your solution is:

{Drum roll}...

{another drum roll...}

{final, longer drum roll!}

100 % correct!

A slightly simpler form (but equal to yours) is:

$$v(t)=98-108e^{-0.1t}$$
<hr>

Next step: calculate total flight time. For this we need an expression for x(t):

$$v(t)=\frac{dx(t)}{dt}$$
$$\implies x(t)=\int v(t)dt$$
Initial condition:
$$t=0, x(0)=0$$

Chop, chop!

[Edited on 25-4-2016 by blogfast25]

Quote: Originally posted by aga

Will your solution work out easier than mine in the next step ? I'm guessing it will.

Chain rule, g-dammit!

Looks like there might be some whipping today, kids...

I spy with my little eye, something like this:







[Edited on 26-4-2016 by blogfast25]

$$C=-1080$$
... is correct!

So we get:
$$x(t)=98t+1080e^{-0.1t}-1080$$
And no, you can't isolate t from that. To find the flight time, we set x(t) = 100:
$$100=98t+1080e^{-0.1t}-1080$$
This is a transcendental equation that has no analytical solution. But it does have a numerical solution, which I give you:
$$t=5.98\:\mathrm{s}$$

Now plug that into v(t) to get final velocity.

PS. That 'cheat-sheet' has been posted here at least twice before!

Quote: Originally posted by aga

25.51 m/s I was thinking that the total distance travelled would be the Up part (approx 9m) plus the 100m drop. Cheat sheet bookmarked, thanks.

Quote: Originally posted by aga

Doh ! After all that i pressed the wrong calculator buttons at the easy step ! How did you get a number out of the 'transcendental equation' ? Is there a faster method than Meditiation or Reincarnation ?

Partial Derivatives:

Earlier we saw that the first derivative of a function f(x):
$$f’(x)=\frac{df(x)}{dx}$$
... is the gradient (slope) of that function. So how does that work for a function of two variables, f(x,y), or even three or four? Will there be 2, 3 or 4 gradients?

Let’s look at a simple function f(x,y):

$$z=ax+by+c$$
This is the description of a (flat) plane in three dimensions:



z is the black plane in the pic.

Now we add another plane, this one red:

$$y=B$$

The intersection of the black and red planes is the green line with formula:

$$z=ax+bB+c$$
The first derivative to x of this line is:
$$\frac{dz}{dx}=a+0+0=a$$

This is the gradient of the black plane in the x-direction.

Analogously, if we add another plane, this one light blue:
$$x=A$$

The intersection of the black and light blue planes is the yellow line with formula:

$$z=aA+by+c$$
The first derivative to y of this line is:
$$\frac{dz}{dy}=0+b+0=b$$

This is the gradient of the black plane in the y-direction.

We call these gradients partial derivatives:
$$\frac{\partial z}{\partial x}\:\text{and}\:\frac{\partial z}{\partial y}$$
The partial derivative of z to x is computed by ‘pretending’ y is a constant:
$$\frac{\partial z}{\partial x}=a+0+0=a$$
The partial derivative of z to y is computed by ‘pretending’ x is a constant:
$$\frac{\partial z}{\partial y}=0+b+0=b$$
But this isn’t limited to simple functions like z. It works for ALL functions f(x,y), as a few examples will show:

1.
$$z=3x^2-5y^3$$
$$\frac{\partial z}{\partial x}=6x$$
$$\frac{\partial z}{\partial y}=-15y^2$$
2.
$$u=x^2y+3xy^3-xy$$
$$\frac{\partial u}{\partial x}=2xy+3y^3-y$$
$$\frac{\partial u}{\partial y}=x^2+9xy^2-x$$
3.
$$v=\frac{x}{y} -2\frac{y}{x}$$
$$\frac{\partial v}{\partial x}=\frac1y+2\frac{y}{x^2}$$
$$\frac{\partial v}{\partial y}=-\frac{x}{y^2}-\frac2x$$
4.
$$y=ze^{2x}$$
$$\frac{\partial y}{\partial x}=2ze^{2x}$$
$$\frac{\partial y}{\partial z}=e^{2x}$$
Rules of Partial Derivation:

The rules of partial derivation are exactly the same as for ‘normal’ derivation, including the chain rule. Two examples of the latter:

1.
$$z=\sqrt{x^2-3xy}$$
$$\frac{\partial z}{\partial x}=\frac12 (x^2-3xy)^{-\frac12}\frac{\partial}{\partial x}(x^2-3xy)$$
$$=\frac12 (x^2-3xy)^{-\frac12}(2x-3y)$$
and:
$$\frac{\partial z}{\partial y}=\frac12 (x^2-3xy)^{-\frac12}\frac{\partial}{\partial y}(x^2-3xy)$$
$$=\frac12 (x^2-3xy)^{-\frac12}(-3x)=-\frac32 x(x^2-3xy)^{-\frac12}$$
2.
$$z=e^{x^3-4xy^2}$$
$$\frac{\partial z}{\partial x}=e^{x^3-4xy^2}\frac{\partial}{\partial x}(x^3-4xy^2)$$
$$=e^{x^3-4xy^2}(3x^2-4y^2)$$
and:
$$\frac{\partial z}{\partial y}=e^{x^3-4xy^2}\frac{\partial}{\partial y}(x^3-4xy^2)$$
$$=e^{x^3-4xy^2}(-8xy)=-8xye^{x^3-4xy^2}$$

Partial derivatives of functions dependent on more than two independent variables:

Such a function can generically by represented by:
$$f(x_1,x_2,x_3,...,x_i,...,x_n)$$
For each variable x_i a partial derivative:
$$\frac{\partial f}{\partial x_i}$$
... can be computed. It's done by considering all other variables (other than x_i as constants.

Consider the following example:

$$u(x,y,z)=x^3y+x^2z^2-3zy^2+5xyz$$
$$\frac{\partial u}{\partial x}=3x^2y+2xz^2+5yz$$
$$\frac{\partial u}{\partial y}=x^3-6zy+5xz$$
$$\frac{\partial u}{\partial z}=2x^2z-3y^2+5xy$$
<hr>
I'll let that 'sink in' for a bit and take questions if there are any. Then some simple exercises will be posted.



[Edited on 28-4-2016 by blogfast25]

1. Correct

2. Incorrect.
$$\cos x+\sin y$$

If y is a constant, then so is siny, so:
$$f_x=-\sin x$$
If x is a constant, then so is cosx, so:
$$f_y=\cos y$$
3. Correct

4. (product rule)

Incorrect, you're applying the sum rule, not the product rule. Try:
$$(x^2-y)(1+y^2)$$
$$f_x=f_x(x^2-y) \times (1+y^2)+(x^2-y) \times f_x(1+y^2)$$
Similar for f_y.

5. fx is correct but fy should be:
$$x^3y^2-6xy+3x$$
$$f_y=2x^3y-6x$$
6. (chain rule)
Almost but not quite:
$$\cos(2x-y)$$
$$f_x=-\sin(2x-y)f_x(2x-y)=2\sin(2x-y)$$
$$f_y=-\sin(2x-y)f_y(2x-y)=-\sin(2x-y)$$

Ok, so we've got some gremlins to deal with.

[Edited on 28-4-2016 by blogfast25]

Quote: Originally posted by blogfast25

So definitely some gremlins, not to mention a few missiles going down rather than up!

Quote: Originally posted by aga

It's pretty much apparent that i was not designed to be a gifted theoretical mathematician.

For most (like me), math is hard work (but rewarding, IMO). So there's few with an excuse ("whoohoo, look at me, I'm sooooo gifted!"), so what?

In characteristic human style we'll now just stick our heads in the sand and plow on!

Tomorrow: simple multi-variate optima.

Don't get me wrong :

i actually Like this algebra stuff, just that it's pretty clear that i'm not careful or thoughtful enough for such a Precise discipline.

Chemistry should be fine as i have Buckets to mix stuff up in

Multi-variate optima:

Higher up we saw that for a simple function f(x), an optimum exists, if:
$$\frac{df(x)}{dx}=0$$



Similarly, a multivariate function:
$$f(x_1,x_2,...,x_i,...,x_n)$$
... has an optimum if:
$$f_{x_1}=0,f_{x_n}=0,...,f_{x_n}=0,$$
The latter condition forms a system of simultaneous equations which when solved yields the numerical values x_i of the optimum.

We won't concern ourselves here with the formal theory of multi-variate optima (which includes detection of absolute and relative extremes and saddle points), instead we'll use common sense to determine whether a found optimum is a maximum or minimum.

First example:

$$f(x,y)=x^2+y^2$$
$$f_x=2x$$
$$f_y=2y$$
An optimum exists for:
$$2x=0 \implies x=0$$
$$2y=0 \implies y=0$$
The optimum value is:
$$f(0,0)=0$$
For slightly larger values of x and y, e.g. (1,1), we get:
$$f(1,1)=2$$
... which indicates (0,0) is a minimum.

Second example:

$$x^2+2y^2-xy+14y$$
$$f_x=2x-y=0 \implies y=2x$$
$$f_y=4y-x+14=0 \implies 8x-x=-14$$
$$x=-2, y=-4$$

3D plot:



Third example:
$$f(x,y)=xy+\frac8x+\frac8y$$
$$f_x=y-\frac{8}{x^2}=0$$
$$f_y=x-\frac{8}{y^2}=0$$
Some reworking and substituting:
$$y=\frac{8}{x^2}$$
$$x-\frac{x^4}{8}=0$$
$$x(1-\frac{x^3}{8})=0$$
$$1-\frac{x^3}{8}=0$$
$$x=2$$
$$y=2$$
$$f(2,2)=4+4+4=12$$
$$f(4,4)=16+2+2=18$$
(2,2) is a minimum.

Exercise:
$$f(x,y)=(x-1)^2+y^3-3y^2-9y+5$$


[Edited on 30-4-2016 by blogfast25]

aga:

Both partial derivatives were correct.

So is x = 1.

But your attempt at solving:

$$y^2-2y-3=0$$

... was very '#creative#' but not very correct!

No, it does not imply that if:
$$y(y-2)=3$$
... then:
$$y=3$$

For a general quadratic equation:

$$ax^2+bx+c=0$$
$$\implies x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
So we have two roots:
$$y=3,y=-1$$
So we have two optima, (1,-1), a maximum and (1,3), a minimum, as the 3D plot shows:




[Edited on 30-4-2016 by blogfast25]