00:01
The question we are given the x equal to the second square t plus 1, and now the y equal to the tension t and t equals to minus pi out of 4.
00:16
From here we can get the d x over d t equal to 2 and then second of the t and derivative of the second equal to the second tension so we have a second square tension.
00:30
T and now the d -y over d -t equal to the second square t and from here we get the y prime were equal to the d -y over dt divided by the d x over dt then we get equal to the second square t dividing by two times second square t t, cosso, this one and then we get equal to a half of the one divided tension equal to the core tension of the t.
01:08
This will be the form of the yt, y, prime, therefore the y -prem and the value of the t, minus pi to 4, get equal to one half, co -tension of minus pi out of four, we get equal to the minus 1, therefore we get the minus 1 out of 2.
01:25
And for the equation of the tangent line, we have the formula y equal to m x plus b.
01:31
And this is an example equal to m, therefore we get equal to minus a half x plus b.
01:37
To find the b here, notice that for this value for the t, we get the x now equal to the second square of the t...