Question

Find the rate of change of F with respect to theta.

Answer 1

\[F=\frac{ \mu W }{ \mu \sin \theta +\cos \theta }\]

Answer 2

Are you struggling to differentiate this?

Answer 3

yes :(

Answer 4

It might help to rewrite it:\[F=\frac{ \mu W }{ \mu \sin \theta +\cos \theta }= \mu W(\mu \sin(\theta)+\cos(\theta))^-1\]
Do you know how to use the chain rule?

Answer 5

kinda, I just learned it recently but I am not to confident using it it

Answer 6

Okay, no problem. Firstly, we perform a substitution: \(z=\mu \sin(\theta)+\cos(\theta)\) which gives us \(F=\mu W(z)^{-1}\). Using this substitution, we can find \(\frac{dF}{d \theta }=\frac{dF}{dz}\times \frac{dz}{d \theta}\). Does this make sense so far?

Answer 7

yep

Answer 8

Great. Reckon you can find dF/dz and dz/do?

Answer 9

i think so, but i think i might need help getting started on dF/dz

Answer 10

That's the easy one :) \[F=\mu Wz^{-1}\]\[\frac{dF}{dz}=-\mu Wz^-2\]

Answer 11

ahhh i was over thinking it haha

Answer 12

Yeah it can be a bit confusing with a load of variables, but all you essentially had was az^-1.

Answer 13

is to second one
\[-\mu \cos \theta -\sin \theta\]

Answer 14

The first term shouldn't be negative. d/dx(sin(x))=cos(x).

Apart from that, it's perfect.

Now multiply the two together and undo our z substitution for your answer.

Answer 15

\[-\mu W(\mu \cos \theta - \sin \theta)-2\]

Answer 16

oh i meant for that -2 to be an exponent

Answer 17

You've done the substitution, but you haven't multiplied it by the derivative you found.

Answer 18

\[\frac{ -\mu W }{ \mu \cos^2 \theta - 2 \mu \sin \theta \cos \theta +\sin^2 \theta }\]

Answer 19

\[\frac{-\mu W(\mu \cos \theta -\sin \theta)}{(\mu \cos\theta +\sin\theta)^2}\]

Answer 20

Didn't notice before but you'd substituted z back in incorrectly, it's + not -. You just had to multiply by dz/do to get your final answer.

Answer 21

ohh so i multiply the derivative by the original equation?

Answer 22

So to summarise:\[\frac{dF}{d \theta }=\frac{dF}{dz}\times \frac{dz}{d \theta}\]
\[F=\mu Wz^{-1}\]\[\frac{dF}{dz}=-\mu Wz^-2=-\mu W(\mu \sin(\theta)+ \cos(\theta))^{-2}\]
\[z=\mu \sin(\theta)+\cos(\theta)\]\[\frac{dz}{d \theta} = \mu \cos(\theta) -\sin(\theta)\]
\[\frac{dF}{d \theta} = \frac{-\mu W(\mu \cos \theta -\sin \theta)}{(\mu \cos\theta +\sin\theta)^2}\]

Answer 23

ohhh

Answer 24

Hope that clears it up a bit, should be easier to see when it's all in the same place.

Answer 25

yes :)

Answer 26

thank you so much for helping me and for being so patient!

Answer 27

Great! No problem at all, happy to help. You use the chain rule a lot so get used to it!