00:01
In this question, we're given this parametric equation and we want to find the horizontal and vertical tangents.
00:07
So that is the tangent has to do with the gradient of the tangent which is dy dx and that will be dy over dt divided by dx over dt.
00:21
So looks like we need to do a differentiation on both x and y with respect to t.
00:32
So dx dt differentiate x with respect to t.
00:35
For t cubed bring down the power 3, repeat the t, subtract 1 to the power w square minus 3.
00:44
Now t when differentiate respect to t is just 1.
00:48
Dy dt for t square bring down the power 2, repeat the t and subtract 1 to the power which is power 1 and minus 8 is a constant so when you differentiate just get this would be dy dt is 2t dx dt is 3t square minus 3.
01:13
For horizontal tangent, now horizontal tangent is when the dy dx or the slope is 0.
01:32
So we're gonna set dy dx equals to 0 so that would be 2t over the denominator equals to 0.
01:45
Denominator cannot be 0 so only numerator is 0 so therefore t equals to 0.
01:49
So when t equals to 0, sub into the x, my x is 0, my y will be minus 8.
02:02
So the point with horizontal tangent would be 0 minus 8...