00:01
In this problem, we have been given the equation of a curve, we have x squared plus y squared is equal to 25, and we want to find the tangent and normal lines to the curve at the point negative 4 comma negative 3.
00:12
So for that purpose, first of all we're going to require dy dx, so we can obtain this using implicit differentiation.
00:19
So we differentiate both sides of the equation with respect to x.
00:22
So the derivative of x squared will be 2x, and using the chain rule, the derivative of y squared with respect to x will be the derivative of y squared with respect to y times dy dx.
00:33
And on the right hand side we have the derivative of 25 that will be 0 because the derivative of a constant is 0.
00:39
So we obtain that dy dx times 2y is equal to negative 2x which will imply that dy dx is equal to negative 2x divided by 2y.
00:51
So this is negative x over y.
00:53
This is dy dx.
00:55
So now what we're going to do is determine the slope of the tangent line.
01:00
And this will be equal to the value of the derivative at the point negative 4 comma negative 3.
01:08
So we have negative x over y, that is dy dx, we calculated this.
01:13
So instead of x, we write negative 4, instead of y, we write negative 3.
01:17
So we end up with negative of negative 4 divided by negative 3.
01:22
And so we end up with negative 4 over 3.
01:25
Then what is the equation of the tangent line? so what is the equation of a line with a slope of m and passing through the point x1, y1? that will be y minus y1 is equal to m times x minus x1...