Question

please help!

Answer 1

all you have to do is fill in the numbers in the equation(:

Answer 2

it didnt work!

Answer 3

oh then im not sure what to do sorry

Answer 4

Why would you expect that to work? Did you use differentials?
Why would you even try that? It makes no sense at all.

The problem statement tells you how to do it. Use Differentials! Find the complete differential of the viscosity equation. It's a couple of Partial Derivatives.

Answer 5

i thought i need to do differentials.

Answer 6

in this equation, which one is the constant?

Answer 7

For the partial derivative with respect to v, everything but 'v' is a constant.
For the partial derivative with respect to r, everything but 'r' is a constant.
For the partial derivative with respect to p, everything but 'p' is a constant.

\(\pi\), 8 and 4 are always constants.

Answer 8

I see. thank you for the explanation.
how do i deffertiate this equation to find out maxmium error?

Answer 9

You find the complete differential - essentially the sum of all possible partial differentials.

Answer 10

mmmm Show e steps? or guide me

Answer 11

Let's see how far we get...

If \(Y = r^{2}\), find \(dY/dr\)

Answer 12

2r. ?

Answer 13

Yes. Why the "?"

If \(Y = 3s^{2}\) find \(dY/ds\).

Answer 14

it should be 6s ?

Answer 15

Perfect. A couple more...

If \(Y = \pi s^4r\), find \(dY/dr\)

Answer 16

hmm... pis^4?

Answer 17

Perfect. Now, use the same definition and find \(dY/ds\)

Answer 18

from what equation?

Answer 19

\(Y = \pi s^{4}r\). Find \(dY/ds\).

Answer 20

pis^4??

Answer 21

No, that's \(dY/dr\).

Find \(dY/ds\). r is constant.

Answer 22

just pi?

Answer 23

@tkhunny

Answer 24

No. Let's back up.

If \(S = r^{4}\), find \(dS/dr\)

Answer 25

I have to go now, but thank you for guide me.
i will be back but i am not sure if you are still here or not.

Answer 26

@tkhunny are you here?

Answer 27

Let's go. Answer the last one.

Answer 28

Ok,
S=r^3.

Answer 29

Did you remember the problem? \(S = r^{4}\)

Find \(dS/dr\)

Answer 30

S=4r^3?

Answer 31

@tkhunny

Answer 32

Now \(S = \pi r^{4}\cdot 3\)

Find \(dS/dr\)

Answer 33

mmmm chain rule?
12pir^3

Answer 34

No, there was no Chain Rule. Why do you think there was a Chain Rule.

\(\dfrac{dr^{4}}{dr} = 4r^{3}\) -- No chain rule

\(\dfrac{d \pi r^{4}}{dr} = \pi4r^{3}\) -- No chain rule

\(\dfrac{d \pi r^{4}3}{dr} = \pi4r^{3}3\) -- No chain rule

\(\dfrac{d \pi r^{4}3q}{dr} = \pi4r^{3}3q\) -- No chain rule

\(\dfrac{d \pi r^{4}3qp}{dr} = \pi4r^{3}3qp\) -- No chain rule

No constant add anything to the complexity of the derivative operation. Just carry them along.

Answer 35

thanks for your effort to show me steps,
I see so in the end I get Pi4r^33pq?

Answer 36

Then, if you wish to simplify, you can make it \(12\pi r^{3}pq\). The simplification has nothing to do with the derivative.

Answer 37

aw, I see.
so either 12 pir^3pq or pi4r^33pq?

Answer 38

That's enough of that. Find these derivatives:

\(\dfrac{dp}{dp} = \)

\(\dfrac{dpr^{4}}{dp} = \)

\(\dfrac{d\pi pr^{4}}{dp} = \)

\(\dfrac{d\dfrac{\pi}{8}pr^{4}}{dp} = \)

\(\dfrac{d\dfrac{\pi}{8}\dfrac{pr^{4}}{v}}{dp} = \)

Answer 39

\[\frac{ dp }{ dp }=1\]

Answer 40

Perfect. Next?

Answer 41

\[\frac{ dpr^4 }{ dp }=dpr^3\]

Answer 42

@tkhunny

Answer 43

Why did you do ANYTHING with the 'r'? It's dp. Ignore the 'r and work with the p. It can be rewritten:

\(\dfrac{dpr^{4}}{dp} = \left(\dfrac{dp}{dp}\right)r^{4} = r^{4}\dfrac{dp}{dp}\)

Answer 44

Ok,
\[\frac{ d \pi pr^4 }{ dp }=\(\frac{ dp }{dp })\pi pr^4\]

Answer 45

@tkhunny

Answer 46

Forget that. You haven't done the second one, yet. Please try again on #2.

Answer 47

R\[r^4\frac{ dp }{ dp }p\]

Answer 48

Where did you get the extra "p"? Look at mine, again and provide the derivative.

Answer 49

Pay close attention. When the derivative is with respect to 'p'. EVERYTHING else is a constant. Everything.