00:01
Hi there.
00:02
So in this problem we have a function t, which is defined as, let's see, v over 2u plus v.
00:19
In other words, we can make a little tree diagram as in the book.
00:23
T depends on the variables u and v.
00:28
Okay, but also, you and v depend on the variables p, q, and r.
00:37
P, q, so there's our tree diagram in the corner.
00:42
Let's also write what u and v are.
00:45
So, u equals pq squared of r.
00:52
Let's rewrite that as r to the one half, since this is calculus.
00:57
And v equals p squared of q, q to the 1ā2.
01:05
Okay.
01:10
And finally, let's remember, p is 2, q is 1, r is 4.
01:19
So let's just make a note that pqr are 2, 1, and 4.
01:32
Okay, so that'll all be useful.
01:38
So at this point, we should be ready to start.
01:44
And the first thing we are asked to find is, let's see, partial t with respect to p.
01:57
So why don't we use our tree diagram? we want to end that the p's down here, so we'll go down the left branch first, which gives us, so we go from t to u and then from u to p, plus, now the right side of this tree, we go from t to v and then from v to p.
02:27
Okay, so one thing at a time here, let's start with partial of t.
02:35
With respect to you and for that let's just look at our definition of t so partial t with respect to you let's use the quotient rule on t over here so u is our variable as we're considering it let's use the quotient rule so the quotient rule says let's see you get low so to u plus v the derivative of the numerator well the derivative of the numerator of v well the derivative of v with respect to u is zero minus the numerator which is v the derivative of the denominator the derivative of two u with respect to u is just two all right and then right all over all over the denominator squared okay so there's d t d u for d u with i'm not sure where my dell went there.
03:47
Partial of u with respect to p.
03:51
Again, for that, we can just look at our definition of u here.
03:56
And if p is our variable, the other two are constants.
04:00
So we will just get q and then square root of r.
04:09
That's our derivative of u with respect to p.
04:13
Okay.
04:14
Plus, same thing over here, partial of d with respect to to v.
04:20
Once again, we can use quotient rule.
04:23
So we get the denominator.
04:29
The derivative of the numerator now is 1 because we're going with respect to v minus the derivative of the denominator is also 1 with respect to v.
04:45
All over the denominator squared.
04:49
And finally, dvdp.
04:53
We can just look at our v over here and once again we'll get similar thing as before q1 1ā2 times r that's the derivative of v with respect to p the partial letter okay so we have our answer if we just know what to plug in can notice where this is zero and what are we going to need well we're going to need we have qp and our values from this point down here but we're also going to need u and v values at this point so let's make a note you at this point equals well p is two q is one and the square root of four will be two so two times one times two is four and v will be let's see two times now we have one to the one half power which is one times four so we have two times four which is eight okay so now we can plug everything in so we get see v was eight eight times two is sixteen so we get negative 16 on top two you plus v two times four is eight plus eight is sixteen so we get sixteen squared there times q is 1, r is 4, okay, we're just plugging things in.
06:36
Now on the other big fraction here, 2u plus v.
06:40
2 times 4 is 8 plus 8 is 16 minus v.
06:49
Our denominator will still be 16 squared, times x squared of q is 1, and r is 4.
06:59
So we're getting somewhere.
07:04
Let's see, what do we have? negative 16 over 16 squared...