Parametric calculus i think (1 Viewer)

[enigma_1]

Hey, can someone give me worked solutions to 12b?

thanks

 

[braintic]

I've attached solutions. Note two things:

(1) The question is poorly asked - y' could mean dy/dx or dy/dt. This is why you shouldn't use that notation.

(2) I've provided two methods for each derivative. To understand the second method for the 2nd derivative, you will need to have done implicit differentiation. OR you might still understand it in the context of the chain rule.

View attachment BOS Q.pdf

 

[enigma_1]

Thanks so much man

 

[bokat]

Make the t the subject in both equations you get: t=x/2 and t=1/y then
equate the right hand sides to get: x/2=1/y cross multiply to get:
xy=2 then make y the subject to have y=2/x or y=2x^-1
Differentiate to get the 1st derivative dy/dx=-2x^-2
differentiate again to get the second derivative: dy^2/(dx)^2=4x^-3

Hope it helps,
Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au

 

[enigma_1]

For question 12d, how do I find the SECOND derivative?

My answer is coming out sorta different.. it's just deriving it again right?

The answer for 12 d second derivative is 4t^3 /(1+t^2 )^3

 

[bokat]

dy/dt=1-t^-2

dx/dt=1+t^-2

dy/dx=(dy/dt)/(dx/dt)

so dy/dx=(1-t^-2)/(1+t^-2) Multiply the top and the bottom by t^2 to get:

or dy/dx=(t^2-1)/(t^2+1) differentiate RHS with respect to t then multiply by dt/dx (the chain rule)

The second derivative= (2t(t^2+1)-2t(t^2-1))/(t^2+1)^2x t^2/(t^2+1) , t^2/(t^2+1) is dt/dx which can be found by flipping dx/dt .
This after simplifying will give you the 4t^3/(t^2+1)^3 which is the answer.

Hope it is useful.

Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au