00:01
In this question, we will be finding the equation of the tangent line.
00:07
The line is tangent to the curve defined by these parametric equations, and tangent at the point corresponding to t equals 1 over 4.
00:28
For the equation of the tangent line, we want to find the derivative of the function, sorry, the derivative of y with respect to x at the point, multiply that by x minus the x coordinate and equate that to y minus the y coordinate of the point of tangency, that is.
00:54
So we need three things.
00:56
The slope, the x coordinate, and the y coordinate.
01:02
Since we're given that this must be at t equals 1 over 4, we can easily find the x and y coordinates.
01:11
X is just t, which is 1 over 4, and y is the square root of t, which in this case is 1 .5.
01:23
So the positive square root, that is, to be specific and precise.
01:34
So we found the x and y coordinates, and so what do we have left? just the slope, the tangent line.
01:48
This is going to be given by the derivative of y, one of the parametric variables, with respect to.
01:55
To x, the other of the two parametric variables.
02:01
So how do we find d, y, d, d, x if we don't have y in terms of x? well, there are two things we can do.
02:09
We could either express y in terms of x...