00:01
So in this problem, we're rowing across a river that's three kilometers wide from a point a to a point b.
00:13
And we have that the river is across three kilometers.
00:23
And the point right across from a is point c.
00:26
And we want to know should we row directly to point b and or should we go to point c and then run to point b or maybe row to some point d and then run the rest of the way.
00:44
So we're also told that from c to b is one kilometer.
00:53
So let's go ahead and set this up here.
00:55
So let's let the distance from d to b be x.
00:59
And so that means from c to d would be 1 minus x.
01:03
And then to find the length of this hypotenuse right here, we would use a pythagorean theorem.
01:09
So our hypotenuse ad, let's call that c squared, is equal to the square root of a squared.
01:15
So 3 squared plus b squared.
01:18
1 minus x squared.
01:22
So that's the square read of 9 plus if i square 1 minus x squared, i get 1 minus 2x plus x squared.
01:34
So that's all going to simplify to x squared minus 2x plus 9.
01:46
Okay, so that's that length there.
01:48
And then what we'll do is we'll use this distance equals rate times time.
01:57
So if we want to minimize the time, we're going to use t equals distance over rate.
02:05
And we're told the rate through the water.
02:07
So rowing is six kilometers per hour.
02:20
And running is eight kilometers per hour.
02:32
All right.
02:32
So let's figure out how to go about doing this.
02:35
So we want to minimize the time.
02:38
And our time with just with respect to x is going to be again this distance divided by time.
02:48
So we want the rowing distance.
02:49
That's going to be this distance right here.
02:52
So we have the square root of x squared minus 2x plus 10.
03:01
I made a mistake up here.
03:03
This 9 plus 1 should be a 10 up here, not a 9.
03:17
And that's going to be over the rate of 6 plus the distance that they have to run is just this length x here.
03:26
So that's x over 8.
03:30
I'm going to write that as one -sixth times x squared minus two x plus 10 to the one -half power plus one -eighth x right and now of course we know the minimum happens when we take a derivative so let's go ahead and find t prime of t sorry t prime of x and again we're taking the derivative of this form over here.
04:07
So we're using the general power rule with the chain rule.
04:13
So the one half is going to come down and multiply by the one sixth.
04:15
So we're going to have a one -twelfth.
04:18
When we do the chain rule, we preserve the inside x squared minus two x plus 10.
04:26
That one -half goes down one to negative one -half.
04:29
Then we do the derivative of the inside chain rule.
04:36
So that's going to give me a two -x minus two.
04:41
So that takes the derivative.
04:42
Of that first term there and then the derivative of one eighth x is simply one eighth so let's simplify this a little bit here i'm going to go ahead and notice that there's a 12 right here that i can divide a two out of that will go down to six and each of these twos i can divide a two out of and then each go down to one so i'm going to get in a numerator i'll just get this term right here x minus one over i'll still have the six down there six and then that factor there with the negative exponent will come down to the bottom.
05:26
So x squared minus 2x plus 10 to the one half power here will be in the denominator.
05:35
And now have the plus 1 eighth.
05:38
So there's my derivative simplified.
05:40
And we know the minimum happens where the derivative is equal to zero.
05:44
The derivative is equal to zero...