Question

For the curve defined by the parametric equations x = 10sin(2t), y = 16cos(2t), with 0 ≤ t ≤ π, evaluate dy/dx at the point (5√3, 8)

Answer 1

\[\large \frac{dy}{dx}=\frac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \qquad = \qquad \frac{\left(\dfrac{d}{dt}16\cos(2t)\right)}{\left(\dfrac{d}{dt}10\sin(2t)\right)}\]Understand the process we'll be taking? :)

Answer 2

Hmm lemme make sure I'm remembering this correctly :D

Answer 3

Ok ok ya we're on the right track.
Before we take a derivative, let's find out what `t` value corresponds to the given (x,y) value.

Given \(\large x=5\sqrt{3}\),
\[\large x=10\sin(2t) \qquad \rightarrow \qquad 5\sqrt3=10\sin(2t)\]

Hmm I don't think that's going to give us a nice `t` value... hmmmmmm :C

Answer 4

Oh yes it will :D I'm being silly...

Answer 5

Dividing both sides by 10 gives us,\[\large \frac{\sqrt3}{2}=\sin(2t)\]This is one of our special angles.
So we've determined that the angle \(2t\) is equal to \(\dfrac{\pi}{3}\).
Solving for t gives us,\[\large 2t=\dfrac{\pi}{3} \qquad \rightarrow \qquad t=\dfrac{\pi}{6}\]

Answer 6

so do we still need to use chain rule

Answer 7

For what part? for taking the derivatives? :o yes

Answer 8

\[\large \frac{d}{dt} 16\cos(2t) \qquad = \qquad -16\sin(2t)\color{royalblue}{\frac{d}{dt}(2t)}\]There is the chain rule being applied in the blue, which will give us,\[\large -16\sin(2t)(2) \qquad = \qquad -32\sin(2t)\]

Answer 9

i am confused the question says to evaluate dy/dx at the point \[(5\sqrt{3},8)\]

Answer 10

We'll end up with an answer of `dy/dx` equal to some function of `t`.
If you took the derivative of your y and x and plugged them into that formula I posted before you would see the equation we get.

You look at that and go, hmm we don't have any x's or y's. What am I suppose to plug into this derivative function?

Look back at what I posted earlier.
I plugged in the x value and solved for `t`.

They gave us a specific (x,y). Those coordinates correspond to a specific `t` value.
That's the `t` value you want to plug into the derivative.