00:01
The curve 4y cubed is equal to a square times x plus 3y can be parameterized as x is equal to a cosine of 3 theta, and y is equal to a cosine of theta.
00:18
Obtained the expressions for dydx by implicit differentiation and in parameterized form, verify that they are equivalent.
00:26
Part b showed that the only point of inflection occurs at the origin.
00:30
It a stationary point of inflection.
00:33
Part c is the information to sketch the graph paying particular attention near the points negative a over 2 and positive a negative a over 2 and the slope at its end points.
00:47
Okay, so first part a by implicit differentiation, this is going to be 12y squared dy is equal to a squared dx plus 3a squared dx, so this is 12 y squared minus 3a squared, dy, is a squared dx.
01:30
And then we get dydx is a squared over 12 y squared minus 3a squared, d squared.
01:40
So that is by implicit differentiation, and then we can use the parameterized form.
01:45
If we're looking at the parameterized form, then dx, d theta is negative 3a sine of 3 theta, and dyd theta is negative a sine of theta, and dydx is dyd theta times d theta dx d ydx.
02:15
Which is, so we're going to pop this up here.
02:25
This is a sine of theta over 3a sine of 3 theta.
02:37
And then you can use trig properties to substitute this through...