Question

Need help determining the missing steps in my professor's notes. Coded steps soon to come

Answer 1

\[\large{\tau_{rz}={-\mu \over g_c}({du \over dr})_{r=r_i}}\]where mu, gc, and ri are constants\[u=u_{\max}[1-({r \over r_i})^2]\]\[\large\tau_{rz}={-\mu \over g_c} \times {u_\max \over r_i} \times -2r; \text{for } {r=r_i}\]\[\large \tau_{rz}={2\mu u_\max \over g_c r_i}\]

Answer 2

i'm confused after he substitutes 'u' into equation 1 to achieve the new Tau equation in line 3

Answer 3

I'm guessing he takes the derivative of u then substitutes in for du

Answer 4

u(max) is also a constant

Answer 5

\(\large \mathbb{u=u_{\max}[1-({r \over r_i})^2]}\)

Answer 6

differentiate both sides with.respect.to r

Answer 7

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}[1-({r \over r_i})^2])}\)

Answer 8

yes

Answer 9

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}[1-({r \over r_i})^2])}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-u_{\max}({r \over r_i})^2)}\)

Answer 10

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}[1-({r \over r_i})^2])}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-u_{\max}({r \over r_i})^2))}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-\frac{u_{\max}}{r_i^2}(r^2)})\)

Answer 11

derivative of 1 is 0

Answer 12

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}[1-({r \over r_i})^2])}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-u_{\max}({r \over r_i})^2))}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-\frac{u_{\max}}{r_i^2}(r^2)})\)

\(\large \mathbb{\frac{du}{dr}= 0-\frac{u_{\max}}{r_i^2}(2r)}\)

\(\large \mathbb{\frac{du}{dr}= \frac{u_{\max}}{r_i^2}(2r)}\)

Answer 13

^^you should get that

Answer 14

\[\large du={2u_{\max}r \over r_i^2}\]

Answer 15

dont ignore the dr in the bottom of du ok ?

Answer 16

its du/dr
not du

Answer 17

Can you help me explain the terminolgy a little bit. You are leaving it as du/dr because it is with respect to r

Answer 18

both are different things
one is called derivative, the other is called differential

Answer 19

d/dr just means the derivative with respect to r as du/dr means the derivative of u with respect to r

Answer 20

yup !

Answer 21

du/dr would be the differential

Answer 22

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}[1-({r \over r_i})^2])}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-u_{\max}({r \over r_i})^2))}\)

\(\large \mathbb{\frac{du}{dr}= \frac{d}{dr} (u_{\max}-\frac{u_{\max}}{r_i^2}(r^2)})\)

\(\large \mathbb{\frac{du}{dr}= 0-\frac{u_{\max}}{r_i^2}(2r)}\)

\(\large \mathbb{\frac{du}{dr}= \frac{u_{\max}}{r_i^2}(2r)}\)

\(\large \mathbb{\frac{du}{dr}\Big|_{r=r_i} = \frac{u_{\max}}{r_i^2}(2r_i) = \frac{u_{\max}}{r_i}(2) }\)

Answer 23

so now the professor probably moved the dr to the other side and substiuted du then reduced to the final equeation

Answer 24

differentials : \(du, dr\)

derivatives : \(\frac{du}{dr}, \frac{d}{dr}\)

Answer 25

ur professor simply substituted the value of \(\frac{du}{dr}\)

Answer 26

Thanks for your help. Makes sense now.

Answer 27

good, u wlc =)