Question

find dr/d(theta): theta^1/2+r^1/2=1

Answer 1

differentiate wrt to theta the whole equation
can you proceed ?

Answer 2

what is wrt

Answer 3

wrt = with respect to
i think i should have written it as w.r.t

Answer 4

1/2theta^-1/2+1/2r^-1/2

Answer 5

alright there are little mistakes
given a function f(r) = r
tell me what is the derivative of f(r) w.r.t theta?

Answer 6

does theta represent x and r represent y

Answer 7

just think it this way that there are two variables r and theta
if you really want x and y
then say f(y) = y
tell me derivative of f(y) w.r.t x?

Answer 8

1/2y^-1/2 dy/dx

Answer 9

why 1/2 y^-1/2 ???

what is derivative of y wrt y ?

Answer 10

when substracting 1/2-1 =-1/2

Answer 11

derivative of y wrt y is 1 i think isn`t it?

Answer 12

i believe so

Answer 13

yes

Answer 14

then from where did 1/2 come in ?

Answer 15

when doing the derivative you have to move the exponent in front of the variable then subtract the exponent from one

so...

you have y^1/2 the derivative would be 1/2y^-1/2

Answer 16

because 1/2-1/1=-1/2

Answer 17

that is fine but i never asked for derivative of y^1/2 , i asked you for derivative of y

Answer 18

1

Answer 19

alright now switch back to the above eq (in the question) and differentiate it wrt to theta and tell me what you get?

Answer 20

y represents theta for this equation

Answer 21

I don`t know what the confusion is all about earlier you had x and y and the two variable now it`s r and theta the steps to solve still remain the same

Answer 22

Find \(\frac{dr}{d\theta}: \) for
\[ \theta^\frac{1}{2}+r^\frac{1}{2}=1\]
If you are more comfortable with x and y, then consider our usual practice, we just let r=y and x=θ here.

So, we get
\[ x^\frac{1}{2}+y^\frac{1}{2}=1\]
Now, differentiate both sides with respect to x. What do you get?

Answer 23

1/2theta^-1/2+1/2r^-1/2dr/dtheta

(1/1/2theta^1/2)+(1/1/2r^1/2dr/dtheta)=0

Answer 24

The first step looks fine, except there should be an "=0" after dr/d theta
\[\frac{1}{2}\theta^{-\frac{1}{2}}+\frac{1}{2}r^{-\frac{1}{2}}\frac{dr}{d\theta}=0\]

For the second step, we need to use the property that \(\large a^{-x}=\frac{1}{a^x}\).
It should be like this:
\[\frac{1}{2\theta^{\frac{1}{2}}}+\frac{1}{2r^{\frac{1}{2}}}\frac{dr}{d\theta}=0\]

Answer 25

why isn't it 1/1/2(theta)^1/2 if 1/2 is the number in front of the variable

Answer 26

\[\frac{1}{2}\theta^{-\frac{1}{2}}=\frac{1}{2}\times \frac{1}{\theta^{\frac{1}{2}}}=\frac{1}{2\theta^{\frac{1}{2}}}\]

Answer 27

oh ok

Answer 28

\[1/\sqrt{2\theta}+1/\sqrt{2r}\]

Answer 29

well technically \[1/\sqrt{2\theta}+1/\sqrt{2r}dy/dx\]

Answer 30

my guess is to add the fractions

Answer 31

No?!
No square root for 2s.

Answer 32

so it would just be \[1/2 \theta ^{1/2} +1/2r ^{1/2}dr/d \theta \]

Answer 33

then add the fractions

Answer 34

Hmmm
\[\frac{1}{2\theta^{\frac{1}{2}}}+\frac{1}{2r^{\frac{1}{2}}}\frac{dr}{d\theta}=0\]And then, you should isolate dr/d theta

Answer 35

\[1/2r ^{1/2} dr/d \theta =-1/2\theta ^{1/2}\]

Answer 36

then multiply 2r^1/2/1 on both sides i think

Answer 37

(2r^1/2)/1 - Yes :)

Answer 38

\[-1/2 \theta ^{1/2} (2\theta ^{1/2})=-2\theta ^{1/2}/2\theta ^{1/2}=-1\]

Answer 39

ooops wrong

Answer 40

\[-2r ^{1/2}/2\theta ^{1/2}=-r ^{1/2}/\theta ^{1/2}=-\sqrt{r}/\sqrt{\theta}\]

Answer 41

\[\frac{1}{2\theta^{\frac{1}{2}}}+\frac{1}{2r^{\frac{1}{2}}}\frac{dr}{d\theta}=0\]\[\frac{1}{2r^{\frac{1}{2}}}\frac{dr}{d\theta}=-\frac{1}{2\theta^{\frac{1}{2}}}\]\[\frac{2r^{\frac{1}{2}}}{1}\times \frac{1}{2r^{\frac{1}{2}}}\frac{dr}{d\theta}=-\frac{2r^{\frac{1}{2}}}{1}\times \frac{1}{2\theta^{\frac{1}{2}}}\]\[\frac{dr}{d\theta}=\frac{-2r^{\frac{1}{2}}}{2\theta^{\frac{1}{2}}}=...\]
Yes!!~

Answer 42

Thank you again =^^=

Answer 43

Welcome :)