help me guys can anyone help me

[zenith wyvern]

find the second derivative of the given parametric equation: x = sin(2t), y = -cos(t) at the point t= π/6

[flat monolith]

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[old flare]

Second derivative with respect to?

[zenith wyvern]

this is just the question

[old flare]

Then most likely with respect to t

So far what have you worked out?

[zenith wyvern]

ive already did the derivative of both then turned it to dy/dx

but im confused if i should enter the value of t or derive the dy again then input the value of t

[zenith wyvern]

aight bro whats next

[old flare]

I thought we just needed to take second derivatives of each component with respect to t

But then you proceeded to look for d^2y/dx^2

So I thought you knew better than I do

[zenith wyvern]

i dont know what to do next bro

[old flare]

I don't even know what a derivative of an equation is

[zenith wyvern]

aight

[lone carbon]

Can you post the actual question?

[zenith wyvern]

this is the actual question

[lone carbon]

Aaaah then I agree with rafain in that they want you to give the 2nd derivative wrt t

[zenith wyvern]

i dont get the question too

i mean i followed the rules then inserted the t after solving the parametric

i got .250 then shit is wrong

hahahaha

[lone carbon]

How have you arrived at the first derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$?

[oblique thorn]

Lumberdude #LeaveWolfAsHeIs

[zenith wyvern]

this is the part where im confused at

[lone carbon]

So you took the first derivatives of both wrt t and then did the $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$?

[oblique thorn]

Lumberdude #LeaveWolfAsHeIs

[lone carbon]

And then you took the derivative of that monster wrt time?

Why?

[zenith wyvern]

thats parametric

the format is d^2y/dx^2

[lone carbon]

I know, but you're taking the derivative of $\frac{dy}{dx}$ wrt to time, why not do it the same way you got $\frac{dy}{dx}$ in order to get $\frac{d^2y}{dx^2}$?

[oblique thorn]

Lumberdude #LeaveWolfAsHeIs

[lone carbon]

I don't think what you're doing is entirely correct tbh

If you've got $\frac{dy}{dx}$ the way to the second derivative would be to take the derivative wrt to x of this expression

[oblique thorn]

Lumberdude #LeaveWolfAsHeIs

[old flare]

It's basically $\frac{dy}{dx}=\frac{y'}{x'}$ and $\frac{d^2y}{dx^2} = \frac{\left(\frac{dy}{dx}\right)'}{x'}$

[oblique thorn]

Rafain | #WhoIsWolf

[old flare]

where the "primes" denote differentiation with respect to t

[lone carbon]

This is correct

[old flare]

I was mainly disapproving of the problem statement for being unclear

[lone carbon]

It is very unclear yeah. You could easily read it as give the second derivative wrt time of both x and y

[old flare]

I didn't even want to assume to understand it until derivatives of an equation are introduced