Question

Can I have some help with this differentiation question please? *question attached* will give medal

Answer 1

Can I have some steps as well as an explanation of the steps of how to do this question, please?

Answer 2

dy/dt
----
dx/dt

Answer 3

what are dx/dt and dy/dt?

Answer 4

Let me work it out :)

Answer 5

Do I need to apply the product rule for dx/dt and dy/dt?

Answer 6

since x(t) is a product and so is y(t) then yes

Answer 7

Hmm, this is giving me a lot of trouble. Won't I have to apply the chain rule twice since sin2t is a composite function?

Answer 8

what is the derivative of sin 2t ?

Answer 9

sin(u) to u' cos(u) , the chain rule applies yes

Answer 10

2t is a product, and we can use the product rule ....

[2t]' = 2't + 2t'

2' = 0 and t' = 1

[2t]' = 0t + 2 = 2

Answer 11

So, we end up with 2cos2t

Answer 12

yes :) for that portion of it

Answer 13

And now, for dx/dt, I ended up with \[2e ^{2t}\sin2t + 2e ^{2t}\cos2t\]

Answer 14

that worked our nice .... almost as if someone planned it that way :)

dy/dt is a similar process

Answer 15

:) And for the dy/dt I ended up with:
\[2e ^{2t}\cos2t-2e ^{2t}\sin2t\]

Answer 16

now the rest is trig identites right?

Answer 17

Haha, yup! Now, dy/dx = \[\frac{ 2e ^{2t}\cos2t-2e ^{2t}\sin2t }{ 2e ^{2t}\sin2t+2e ^{2t}\cos2t }\]. And the two \[2e ^{2t}\cos2t\] will cancel out each other, right? So then how are we going to end up with a tan when there will be only sin remaining? :S

Answer 18

cos(2t) doesnt factor out does it?

Answer 19

and we dont always need to factor everything out ... but rather use it to our advantage as the need arises

Answer 20

I don't understand :S

Answer 21

when we convert one thing into another, the most useful properties we have are to add 0 or multiply by 1

factoring everything that factors and simplifying is not always the best course of action is all

does a cos(2t) actually factor out tho? the e^(..) might be useless to us so we can get rid of them i agree

Answer 22

No it doesn't factor, but it's both in the numerator and denominator, so isn't it cancelled out?

Answer 23

If we divide by it

Answer 24

perhaps: multiply by 1/cos(2t) top and bottom is multiplying by 1

\[\frac{ 1-\tan2t }{1+\tan2t}\]
is what you get with your idea so far