00:01
So we have x is equal to sine of t, y is equal to cosine of t.
00:09
And we want to find the equation of tangent at t is equal to pi over four.
00:22
So we can start by finding d x by d t, which is equal to cosine of t, and d y by d t, which is equal to negative sign of t.
00:35
So, d, y by d x would be equal to d, d, y, over dt divided by dx over dt, which is negative sine of t divided by cosine of t, which is negative tangent of t.
00:57
So now we can find the derivative at t is equal to pi over four, so d over, d, y, over d x, where t is equal to pi over four would be equal to negative tangent of pi over four and tangent of pi over four is equal to one so this would be negative one so we have our slope which is equal to negative one next we need a point so x when t is equal to pi over four is equal to uh sign of pi over four which is a square root of two over 2.
01:41
And in the same way, why, when t is equal to pi over 4, would be equal to cosine of pi over 4, and that would also be equal to square root of 2 over 2.
01:54
So we have a point...