00:01
In this problem, we're given these parametric equations and we want to find the slope of the curve defined by these parametric equations at t equal to 2 and the value of its second derivative at t equal to 2.
00:14
Now for the first part, we note that the slope is equal to dy over dx and if it's in terms of parametric equation, then this is equal to dy over dt.
00:33
This is equal to the derivative with respect to t of 1 over t minus 1, which we can rewrite as t minus 1 raised to negative 1.
00:44
It's going to be negative of t minus 1 raised to negative 2, or that's negative of 1 over t minus 1 squared, while dx over dt equals the derivative with respect to t of t plus 1 raised to negative 1, which is the same process as dy dt.
01:07
That's negative of t plus 1 raised to negative 2 equal to negative of 1 over the square of t plus 1.
01:16
So our dy over dx equals negative of 1 over t minus 1 squared times the reciprocal of negative of 1 over t plus 1 squared.
01:30
That's times the negative of t plus 1 squared over 1 equal to square of t plus 1 all over the square of t minus 1 at t which equals to dy over dx at t equals 2 equals square of 2 plus 1 over the square of 2 minus 1...