PROFESSOR: Given the vector valued functions r of t equals cosine ti plus sine tj minus e to the 2t k and u of t equal to ti plus sine tj plus cosine tk, calculate the derivative with respect to t of r of t dotted with r prime of t and the derivative with respect to t of u of t crossed with r of t. To begin, we need to find several derivatives of these functions because the product rule for the dot product and the cross product, when you're taking the derivative of those, to begin with, the dot product is the derivative with respect to t of r of t times or dotted with u of t equals r prime of t dotted with u of t plus r of t dotted with u prime of t. And for the cross product, the derivative with respect to t of r of t crossed with u of t equals r prime of t crossed with u of t plus r of t cross with u prime of t. And because one of our functions is r prime, we actually need to find the second derivative of r as well.

So let's go and first find what u prime of t is. And again, these are vector valued functions, although I'm not drawing the arrows over them. That is understood. So the derivative of u of t is 1 i-- so just i-- plus cosine j-- cosine tj-- minus sine of tk.

And the first derivative of r of t-- r prime of t-- would be equal to minus sine ti plus cosine tj minus 2e to the 2t k. And then r double prime the second derivative of r would be equal to minus cosine ti minus sign tj minus 4e to the 2t k.

Now applying our definition of the derivative of a dot product of vector valued functions, the derivative with respect to t of r of t dotted with r prime of t is going to be equal to r prime of t dotted with r prime of t plus r of t dotted with r double prime of t, the second derivative. Now that would be equal to parentheses sine squared t plus cosine squared t-- remembering our definition of the dot product-- plus 4e to the 4t close parentheses plus open parentheses minus cosine squared t minus sine squared t plus 4e to the 4t close parentheses. Now remembering that sine squared plus cosine squared is equal to 1, and negative cosine squared t minus sine squared t would be equal to negative 1, we can combine terms so that this is equal to 8e to the 4t, so that the derivative with respect to t of r of t dotted with r prime if t is equal to 8e to the 4t.

Now for the second part of this question, the derivative with respect to time-- or with respect to t of u of t crossed with r of t-- I think of t is time, but really, it's just a parameter of these functions. So got to correct myself. So the derivative with respect to t of u of t crossed with r of t from our product rule for cross product-- derivative of cross product of vector valued functions, that would be equal to u prime of t crossed with r of t plus u of t crossed with r prime of t. So referring back to those functions, and our definition of the cross product, that would be equal to open parenthesis cosine t times e to the negative-- sorry. Let me step back.

That would be equal to parentheses cosine t times negative e to the 2t plus-- or minus a negative-- so plus sine squared t close parentheses i minus open parentheses 1 times negative e to the 2t plus again, sine of t times cosine t close parentheses j plus parentheses 1 times sine of t minus cosine times cosine. So I'm going to say 1 sine t minus cosine squared t close parentheses k. And that is the u prime of t crossed with r of t plus-- we're going to now calculate u of t crossed with r prime of t. And that will be equal to parentheses sine t times negative 2e to the 2t minus cosine squared t close parentheses i minus open parentheses negative 2t e to the t minus cosine squared t close parentheses j plus open parentheses t cosine t minus sine t times a minus sine t. So that would be plus sine t-- or sine squared t close parentheses k.

Adding these two together, we come up with parentheses negative e to the 2t open parentheses 2 sine t plus cosine t close parentheses plus sine squared t minus cosine squared t close parentheses i plus parentheses e to the t plus sine t cosine t plus 2t e to the 2t plus cosine squared t close parentheses-- that's our j term-- plus parentheses sine t minus cosine squared t plus t cosine t plus sine squared t close parentheses k.

Now, we need to recognize at this point that sine squared minus cosine squared-- so sine squared t minus cosine squared t-- is equal to cosine of 2t so that the simplifies just a bit. So our first term parentheses negative e to the 2t parentheses cosine t plus 2 sine t close parentheses plus cosine of 2t. And again, that is coming from sine squared t minus cosine squared t. And that close parentheses makes for our i term plus parentheses e to the 2t open parentheses 2t plus 1 close parentheses minus sine of 2t close parentheses j plus parentheses t cosine t plus sine t minus cosine of 2t close parentheses k.

So the derivative with respect to t of u of t cross r of t for our given functions is equal to negative e to the 2t multiplied by cosine t plus 2 sine t as a quantity plus cosine 2t-- and that's our i component-- plus e to the 2t times 2t plus 1 quantity minus sine of 2t-- that's our j component-- plus the component t cosine t plus sine t minus cosine 2t k.