# Differentiation assignment

Hi!

I'm new to the forum so first I would like to say 'Hi' to everyone.

I would like to ask you to check my assignment. It is already finished (at least I hope it is). Please do not get me wrong. I don't ask anyone to do any work for me. I'm only asking if it's correct. If any part isn't I only want to know which part and I'll try to figure out the correct answer myself.

I've read the rules of the forum and I understand that you don't particularly like work presented in pictures however it would be very difficult for me to type everything into the post and I could make many mistakes. It's in PDF format, about 1.2MB. It's properly scanned not a mobile phone picture. The only trouble could be my handwriting but I swear that I've done my best.

I don't think that the questions them self are necessary as all the data is stated at the beginning of every question.

Thank you in advance for any possible interest.
mathi85

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Mark44
Mentor
I doubt that you'll get many responses. It's best to ask about one problem at a time, not 15 or so that you have in the PDF. One reason we ask people to post the problems in text, rather than photos or PDFs is that we can pinpoint exactly where someone has gone astray, and don't have to type in the whole equation for context.

And your writing is a little difficult to read, which you noted.

Hi!

Thank you for your answer. I will try to put it into text as soon as I'll find some time.

mathi85

Rate of change

σ(x,y)=e1/2yln(x+2y)

When x = 2.5cm and y = 4cm
x is increasing @ 5cm/s
y is increasing @ 3cm/s

The rate of change of σ is:
dσ/dt=(δσ/δx*dx/dt)+(δσ/δy*dy/dt)

Partial derivatives: u=e1/2y v=ln(x+2y)
δσ/δx=e1/2y*1/x+2y*1=e1/2y*1/x+2y

Using product rule:
δσ/δy=e1/2y*1/x+2y*2+1/2e1/2y*ln(x+2y)
=2e1/2y*1/x+2y+1/2e1/2y*ln(x+2y)

Furthermore:
dx/dt=+5cm/s
dy/dt=+3cm/s

dσ/dt=[(e1/2y*1/x+2y)(5)]+[(2e1/2y*(1/x+2y)+1/2e1/2y*ln(x+2y)(3)]
=5e1/2y*6e1/2y*(1/x+2y)+1.5e1/2y*ln(x+2y)
=11e1/2y*(1/x+2y)+1.5e1/2y*ln(x+2y)

When x=2.5 and y=4
dσ/dt=11e2*(1/2.5+8)+1.5e2*ln(2.5+8)= 33.80258144cm2/s

This is the task that worries me most.
If anyone has some spare time I'd be very glad for help.

mathi85

The stream function, ψ(x,y) is related to the velocity components u and v of the fluid flow by the following:
u=δψ/δy and v=-δψ/δx

Find u and v if:
ψ(x,y)=3ln(x2+y2)

If ψ satisfies Laplace's equation then the flow is irrotational. Given taht Laplace's equation is:
δ2ψ/δx22ψ/δy2=0

Determine wheter the flow is irrotational.

Solution:

δψ/δx=6x/(x2+y2)
δψ/δy=6y/(x2+y2)

Therefor:
u=6y/(x2+y2)
v=-6x/(x2+y2)

δ2ψ/δx2=(x2+y2)6-6x2x/(x2+y2)2=-6x2+6y2/(x2+y2)2

δ2ψ/δy2=(x2+y2)6-6y2y/(x2+y2)2=6x2-6y2/(x2+y2)2

Hence:
δ2ψ/δx22ψ/δy2=0

Many thanks for any interest!

mathi85

Your work for task 2 looks correct to me. Can you please state what u and v are in #4?

u and v are just a side note for me so I don't get confused while using a product/quotient rule etc. It wont be included in a submission version.

u and v are just a side note for me so I don't get confused while using a product/quotient rule etc. It wont be included in a submission version.

Your derivatives look good to me. I haven't checked the arithmetic.

Just a side note, its a good practice to use parentheses in your working. For instance,

δσ/δx=e1/2y*1/x+2y*1=e1/2y*1/x+2y

Without the use of parentheses, someone might interpret it as
δσ/δx=e1/(2y)*(1/x)+2y*1
which is not what you mean. The correct way to write the above is

δσ/δx=e(1/2)y*(1/(x+2y))*1=e(1/2)y*(1/(x+2y))

Ray Vickson
Homework Helper
Dearly Missed
σ(x,y)=e1/2yln(x+2y)

When x = 2.5cm and y = 4cm
x is increasing @ 5cm/s
y is increasing @ 3cm/s

The rate of change of σ is:
dσ/dt=(δσ/δx*dx/dt)+(δσ/δy*dy/dt)

Partial derivatives: u=e1/2y v=ln(x+2y)
δσ/δx=e1/2y*1/x+2y*1=e1/2y*1/x+2y

Using product rule:
δσ/δy=e1/2y*1/x+2y*2+1/2e1/2y*ln(x+2y)
=2e1/2y*1/x+2y+1/2e1/2y*ln(x+2y)

Furthermore:
dx/dt=+5cm/s
dy/dt=+3cm/s

dσ/dt=[(e1/2y*1/x+2y)(5)]+[(2e1/2y*(1/x+2y)+1/2e1/2y*ln(x+2y)(3)]
=5e1/2y*6e1/2y*(1/x+2y)+1.5e1/2y*ln(x+2y)
=11e1/2y*(1/x+2y)+1.5e1/2y*ln(x+2y)

When x=2.5 and y=4
dσ/dt=11e2*(1/2.5+8)+1.5e2*ln(2.5+8)= 33.80258144cm2/s

A minor(?) quibble: it makes no sense to have units (cm) attached to x and y in expressions like e^y and log(x+2y). Arguments of functions like exp, log, sin, cos, etc., must be dimensionless. However, if that is how the questions were written out, you just have to hold your nose and accept the questions as given.

A minor(?) quibble: it makes no sense to have units (cm) attached to x and y in expressions like e^y and log(x+2y). Arguments of functions like exp, log, sin, cos, etc., must be dimensionless. However, if that is how the questions were written out, you just have to hold your nose and accept the questions as given.

I'm not sure if it makes any difference but I didn't give you a full question. Here it is:

"The stress of a particular material, σ, is a function of x and y, where x denotes stress in the x-direction and y denotes stress in the y-direction. The stress function is given by:

σ(x,y)=e(1/2)yln(x+2y)

Determine the rate of increase of σ given that x=2.5 cm, y=4 cm and that x is increasing at the rate of 5 cm/s and y is increasing at the rate of 3 cm/s."

It might make more sense now, however my maths tutor admits that she doesn't know much about an engineering and she only tries to make the questions more relevant.

Pranav-Arora thank you for your advice. I didn't think about it before but now I can see that without the parentheses it might be confusing.

Ray Vickson thank you for your time. I do appreciate it.

I wasn't going to ask about any more tasks, mainly because I didn't expect any interest but as it's going so well I'm asking you for a little more help:) I hope I'm not abusing your good will.

It is Task No. 3 and here it is:

"The volume, V, of a liquid of viscosity coefficient η delivered after time t when passed through a tube of length l and diameter d by a pressure p is given by:

V=(pd4t)/(128ηl)

If the errors in V, p, and l are 1.2%, 2.2% and 3.2% respectively, determine the error in η."

Solution:

η=(pd4t)/(128Vl)

Approx change in η:

δη≈∂η/∂V*δV+∂η/∂p*δp+∂η/∂l*δl

Partial derivatives:

∂η/∂V=(-pd4t)/(128V2l)
∂η/∂p=(d4t)/(128Vl)
∂η/∂l=(-pd4t)/(128Vl2)

Furthermore we have:

δV=+0.012V
δp=+0.022p
δl=+0.032l

Hence:

δη≈(-pd4t128-1V-2l-1)(0.012V)+(d4t120-1V-1l-1)(0.022p)+(-pd4t128-1V-1l-2)(0.032l)
≈-0.022[(pd4t)/(128Vl)]

Hence

≈-0.022η since η=(pd4t)/(128Vl)

I was hoping for Ray or someone else to help you out as I myself had a very short introduction to partial derivatives.

That looks right to me but please post one problem per thread.

Thank you very much Pranav-Arora!

I definitely will post one problem per thread when I get another assignment.

Thanks everyone for your help. Much appreciate.