2. Consider the parametric curve defined by x = ln(2t+1), y...

2. Consider the parametric curve defined by x = ln(2t+1), y = t^2 + 3t + 1. a) Evaluate dy/dx and d^2y/dx^2 at t = 1. b) Set up but do not evaluate the integral for the arc length of the curve for 0 ? t ? 1. c) Set up but do not evaluate the integral for the area of the surface obtained by revolving the curve about the x axis for 0 ? t ? 1.

Transcript

00:01 Okay, so let's get started by noticing that here, from this equation, we get 2t plus 1 equals exp of x, which implies t equals e to the x, which is the same as xp of x, minus 1 over 2.

00:22 So in particular here what do we have? we have that y as a function of x is given by t squared, t squared which is e to the 2x minus 2 e to the x minus plus 1 over 4.

00:49 Then we have plus three halves e to the x minus three halves plus one plus one half here okay now what is the derivative of this function here well y prime of x is given by two e to the x minus two the x minus two or two or two to the two x here minus 2 e to the x over 4 plus 3 halves e to the x now what is the second derivative well the second derivative is going to be 4 e to the 2 x minus 2 e to the x again over 4 plus 3 halves e to the x and so we are done with the first point indeed at t equals 1 we have log of 3 x equals log of 3 so y prime computed okay let's write it like this the y over the x at t equals 1 is what is okay 2 multiplied by e to the 2 log of 3 minus 2 e to the log of 3 over 4 plus 3 halves e to the log of 3.

02:50 Now by using the properties of logarithmic functions and exponential functions, here what do we get? we get 2 multiplied by 3 squared so 9 minus 2 multiplied by 3 which is 6 over 4 plus 3 plus 3 halfs multiplied by 3 so at the end we get negative 3 4ths plus 9 over 2 now let's do the same for the second derivative okay so let's see if we can do it here we have that the second derivative of y with respect to x evaluated at t equals 1 is okay i'm going to do it very quickly okay here we are going to have okay 4 over 4 so 1 and here we are going to have 9 here we're going to have minus 100 half multiplied by 3.

04:03 Okay, so negative 3 halves...

Question

Please give Ace some feedback