00:01
In this question, an isosceles right triangle with legs of length s has an area of a equals one half s squared.
00:09
So i have an isosceles right triangle with sides of length s.
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And the area of this triangle is one half s squared.
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At the instant when s equals the squared of 32 centimeters, the area of the triangle is increasing at a rate of 12 square centimeters per second.
00:36
So what are they giving me? they are giving me the rate of change of the area with respect to time.
00:47
That's 12 square centimeters per second.
00:53
And i want to know at what rate is the length of the hypotenuse of the triangle increasing in centimeters per second at that increase.
01:03
So here is the hypotenuse of the triangle, and i want to know what is my d h d t? well, let's see.
01:16
I'm going to start with that a equals one half s squared, and i'm going to differentiate both sides with respect to t.
01:26
When i do, on the left i get d a d t.
01:34
On the right, the derivative of one half s squared with respect to t, one half times, times 2 is 1s times the s d t.
01:48
Now, my d -a -d -t, again, is 12, right? that's 12 when my s is the square root of 32.
02:05
So i am getting 12 equals the square root of 32 times dsdt.
02:13
I can divide both sides by the square root of 302.
02:18
And say 12 over the squared of 32, that's equal to my dsdt.
02:29
Now, i'm not done.
02:32
What's the issue? well, i'm looking for dhdt, right? but how do dhdt and dsdt relate to each other? you may remember from your high school geometry course that when you have an isosceles right triangle, your hypotenuse is the side length of the legs times root 2...