PROFESSOR: Calculate the arc length of the parameterized curve r of t equals angle bracket 2t squared plus 1, 2t squared minus 1, t cubed close angle bracket with 0 less than or equal to t less than or equal to 3. The equation for arc length requires us to think of each component of our parameterized curve as a function of t. In the case of our function r of t, we're going to call f of t 2t squared plus 1, g of t will be 2t squared minus 1, and h of t will be equal to t cubed. And the bounds are going to be dealing with are a equals 0 and b equals 3 from the bounds on t. So the arc length function is equal to the integral from a to b of the integrin the square root of f prime of t squared plus g of t squared plus h prime of t squared.

Again, that's the square root of all of that, that sum, multiplied by dt. So given our f, g, and h, we can find the derivatives of those. And so our arc length becomes the integral from 0 to 3 of the square root of 4t squared. So that would be 16t squared plus derivative of g is 4t. And then we square that.

So that would b 16t squared again plus derivative of h of t of t cubed is 3t squared. And then we square that. That's 9t to the 4th. So combining like terms, our integral is from 0 to 3 of the square root of the quantity 32t squared plus 9t to the 4th dt.

Now we might notice we can divide by-- or factor out a t squared so that this integral becomes 3t. I'm actually going to factor it at 9t squared. This becomes 3t [INAUDIBLE] integral. So that would be the integral from 0 to 3 of 3t square root of t squared plus 32 over 9, all in our square root dt.

Now I'm going to make a u substitution. So we'll say u equals t squared plus 32 over 9, which means that du is equal to 2t dt, which means that 1/2du is equal to t dt. This fits the form of the integral we have.

All right. So this integral becomes 3 over 2 becomes integral of 0 to 3. That would be a view to the 1/2 du. And actually our endpoints in the integral will change. They will not be 0 to 3 because we've made a u substitution. We had to change our endpoints.

So using the function u equals t squared plus 32 over 9, our endpoint of 0 becomes 32 over 9. And our endpoint of 3 becomes 113 over 9. So we have the integral-- it's 3 over 2 times the definite integral from 32 over 9 to 113 over 9 of u to the 1/2du. And antiderivative of u to the 1/2 is u to the 3/2 multiplied by 2/3.

And then evaluating that from 32 over 9 to 113 over 9, and also multiplying by the constant we had-- constant factor 3 over 2-- this value is approximately 37.785. So the arc length of this parameterised curve r of t from 0 to 3 is 37.785.