00:01
In this question, given the parametric curve xy, we are asked to find the values, we are asked to find dy or dx, d double y over dx squared, and determine the intervals of concavity of the curve.
00:19
First, let's find dy over dx.
00:22
For parameter curves, d .y over d x equals to d .y over dt, divide by dx over dt.
00:33
D .y over dt is simply the derivative of t squared plus 9t and equals to 2t plus 9.
00:44
Dx over dt is the derivative of t squared plus 6 and equals to 2t.
00:51
Therefore, dy over dx equals to 2t plus 9 over 2t.
00:56
To calculate d double y over dx squared, we need to differentiate to find d over dx of dy over dx.
01:15
If we call d -y or dx by some other function, as by, let's say, as g of t, right? by g, then we are basically asked to find the derivative dg over dx.
01:37
Note that it's similar to finding the derivative of d -y over dx.
01:42
But instead of why we have g now.
01:46
And we know that the formula for d -y over d -x is d -y -d -t, divide by d -x over d -d -t.
01:54
Similarly, we can rewrite dg over dx as dg over dt, divide by dx over dt.
02:09
We recall that g equals to this expression and equals to dy or dx basically.
02:15
So we're replacing dy over dx by g, where g equals to 2t plus 9 over 2t, and now we need to calculate the derivative of dg over dt to get dyy over dx squared.
02:31
Dg over dt equals dg over dt equals, to the derivative of 2t plus 9 over 2t.
02:39
And to find this, we will use the quotient rule.
02:43
By the quotient rule, we need to differentiate the numerator.
02:49
The derivative of the numerator equals to 2, multiplied by the denominator, minus the numerator, multiplied by the derivative of the denominator, which is 2, divided by the square of the denominator, which is 4t squared.
03:05
This simplifies to 4 t minus 2...