2. Let S be the surface of revolution obtained by rotating...

2. Let $S$ be the surface of revolution obtained by rotating the curve $x^2 = z^2 + 1$ in the $xz$-plane around the $z$-axis. (a) Show that $v_p = (1, 2, \sqrt{2})$ is a tangent vector to $S$ at the point $p = (\sqrt{2}, 0, 1)$. (b) Find the equation of the normal section of $S$ at $p$ in $v_p$ direction.

Transcript

00:01 In this problem we are given that s denotes the cone z equals to x squared plus y squared under z equals to 9.

00:16 In subpart a we are asked to parameterize s while considering x and y to be the parameters.

00:30 So we have r of x comma y to be equal to x remains as it is and y also remains as it is.

00:37 And here in the place of z we have x squared plus y squared so the z components becomes x squared plus y squared and since we have under z equals to nine it implies that x squared plus y squared must be less than or equal to nine and therefore this is the required parameterization of the curve s next, in subpart b, we are asked to use y equals to c and to find out the tangent vector 2s at the point 1 to 5.

01:28 So now in the parametric equation, let us replace y by c.

01:34 So we get r of x equals to x c, x squared, plus.

01:41 C squared.

01:42 Since we require the tangent vector, let us differentiate this with respect to x.

01:47 We get 1, 0, 2 times x.

01:53 In this point, the value of x is 1.

01:56 So let us substitute the value of x as 1.

01:59 We have 1 0 2 times 1 which is 2 and therefore this is the required tangent vector.

02:09 Next, in subpart c, we are asked to consider x equals to c and we are asked to find out the tangent vector at the same point that is 1 to 5.

02:23 So in this case we have r of y to be equal to c y c squared plus y squared.

02:33 So differentiating with respect to y we have 0 1 2 times y.

02:39 So in the given point the value of y is 2...

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