Question

Calculus II Question (in terms of Physics II):

How did the d become "1" in the next line?

(See photo)

Answer 1

Just curious, is this dealing with circuits? :)

Answer 2

@Jhannybean : Nah, it's currents through wires. :)

Answer 3

\[\frac{\mu_0 I I'}{2 \pi}\int\limits_{0}^{\frac{a}{2}}\frac{1}{d+\sqrt{3}x} dx \\ \text{ \let } u=d+\sqrt{3}x \\ \text{ so } du=\sqrt{3} dx \\ \text{ so } \frac{1}{\sqrt{3}} du=dx \\ \frac{\mu_0 I I'}{2\pi} \int\limits_{d+\sqrt{3} \cdot 0}^{d+\sqrt{3} \cdot \frac{a}{2}} \frac{1}{\sqrt{3}} \frac{du}{u} \\ \frac{ \mu_0 I I'}{2 \pi} \frac{1}{ \sqrt{3}} \ln|u| \\ \frac{ \mu_0 I I'}{2\sqrt{3} \pi} [\ln|d+\sqrt{3}\cdot \frac{a}{2}|-\ln|d+\sqrt{3}\cdot 0|] \\ \frac{\mu_0 I I' }{2 \sqrt{3} \pi} \ln |\frac{d+\sqrt{3} \cdot \frac{a}{2}}{d}|\]

Answer 4

\[\frac{\mu_0 I I'}{2 \sqrt{3} \pi} \ln |1+\frac{\sqrt{3} \frac{a}{2}}{d}|\]

Answer 5

\[\frac{\mu_0 I I'}{2 \sqrt{3} \pi} \ln|1+\frac{\sqrt{3} a}{2d}|\]

Answer 6

so the d didn't turn into 1
the d/d turned into 1

Answer 7

that is what happened between those two steps is they plugged in the limits
and use this property of log
log(a/b)=log(a)-log(b)

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @freckles
\[\frac{\mu_0 I I'}{2 \pi}\int\limits_{0}^{\frac{a}{2}}\frac{1}{d+\sqrt{3}x} dx \\ \text{ \let } u=d+\sqrt{3}x \\ \text{ so } du=\sqrt{3} dx \\ \text{ so } \frac{1}{\sqrt{3}} du=dx \\ \frac{\mu_0 I I'}{2\pi} \int\limits_{\color{red}{d+\sqrt{3} \cdot 0}}^{\color{red}{d+\sqrt{3} \cdot \frac{a}{2}}} \frac{1}{\sqrt{3}} \frac{du}{u} \\ \frac{ \mu_0 I I'}{2 \pi} \frac{1}{ \sqrt{3}} \ln|u| \\ \frac{ \mu_0 I I'}{2\sqrt{3} \pi} [\ln|d+\sqrt{3}\cdot \frac{a}{2}|-\ln|d+\sqrt{3}\cdot 0|] \\ \frac{\mu_0 I I' }{2 \sqrt{3} \pi} \ln |\frac{d+\sqrt{3} \cdot \frac{a}{2}}{d}|\]
\(\color{blue}{\text{End of Quote}}\)

Can you explain the integral? How did you find that?

Answer 9

I can't explain circuits
It was the problem in the pdf

Answer 10

@freckles : Ahhh it's U-sub that I failed to see, in solving this problem! Thank You very much! Your answer was just what I wanted!

Answer 11

I can do that calculus part . :p

Answer 12

Darn it.I just wanted to know how you got the limits :P

Answer 13

oh well I let u=d+sqrt(3) x

Answer 14

so if x=0 then u=d+sqrt(3)*0 or just d
and if x=a/2 then u=d+sqrt(3)*a/2

Answer 15

Ohh, and the x part was x=0 , x= a/2 .... HAhaha Nvm~!

Answer 16

Yeah it all makes sense now lol

Answer 17

let me put one more thing okay
in my thingy

Answer 18

\[\frac{\mu_0 I I'}{2 \pi}\int\limits\limits_{0}^{\frac{a}{2}}\frac{1}{d+\sqrt{3}x} dx \\ \text{ \let } u=d+\sqrt{3}x \\ \text{ so } du=\sqrt{3} dx \\ \text{ so } \frac{1}{\sqrt{3}} du=dx \\ \frac{\mu_0 I I'}{2\pi} \int\limits\limits_{d+\sqrt{3} \cdot 0}^{d+\sqrt{3} \cdot \frac{a}{2}} \frac{1}{\sqrt{3}} \frac{du}{u} \\ \frac{ \mu_0 I I'}{2 \pi} \frac{1}{ \sqrt{3}} \ln|u||_{d+\sqrt{3} \cdot 0}^{d+\sqrt{3} \cdot \frac{a}{2}} \\ \frac{ \mu_0 I I'}{2\sqrt{3} \pi} [\ln|d+\sqrt{3}\cdot \frac{a}{2}|-\ln|d+\sqrt{3}\cdot 0|] \\ \frac{\mu_0 I I' }{2 \sqrt{3} \pi} \ln |\frac{d+\sqrt{3} \cdot \frac{a}{2}}{d}| \]

Answer 19

I just wanted to put those limits on that 6th line

Answer 20

so everyone cool
and everyone can follow that?

Answer 21

Im not too sure about the last line. Maybe cause I'm out of practice lol

Answer 22

do you know that ln(a)-ln(b)=ln(a/b)

Answer 23

Yeah

Answer 24

\[\ln|d+\frac{\sqrt{3} a}{2}|-\ln|d| \\ =\ln|\frac{d+\frac{\sqrt{3}a}{2}}{d}| \\ =\ln|\frac{d}{d}+\frac{\sqrt{3}a}{2d}|\]

Answer 25

Ohh, I got it.

Answer 26

Then you just simplify.

Answer 27

ya