December 04, 2006

Scribe Post: Day 54

Terribly sorry for having this up late and on the wrong day of all things =S But I'm going to cut to the chase here, on Friday we had a pre-test for "Differentiating Rules" here are the questions a solutions:

(1) To make things basic and easy all you need to do is put this equation into your calculator (since it's already a derivative) and substitute a very very small value like 0.000001 for your h and solve the equation. You should come up with an answer relatively close to 5.55 which is (C). I'm not completely sure if this applies for all questions like this, but I'm hoping it does. =S

(2) Differentiate f(x) = x - k / x + k by using the "Quotient Rule". This should give you the derivative:

f'(x) = (x + k)(1) - (x - k)(1) / (x + k)²

Then solve for f'(0):

f'(0) = (0 + k)(1) - (0 - k)(1) / (0 + k)²
f'(0) = k - (-k) / (0 + k)²
f'(0) = 2k / (0 + k)²
f'(0) = 2 / k

(3) Differentiate g(x) = f(f(x)) by using the "Chain Rule". This should give you the derivative:

g'(x) = f'(f(x)) f'(x)

You're trying to estimate g'(1) so your equation will look like this:

g'(1) = f'(f(1)) f'(1)

You already have your value for f(1) = 2. However you must find the derivative value for f'(2) and f'(1) in order to solve the equation. You do this by applying the "Symmetric Difference Quotient to the two points x = 1 and x = 2 on the table provided. Doing this you get the values f'(1) = 0.6 and f'(2) = 2. Now you can solve the equation:

g'(1) = f'(f(1)) f'(1)
g'(1) = f'(2) f'(1)
g'(1) = (2)(0.6)
g'(1) = 1.2

(4) You will be using the "Pythagorean Theorem" to solve this question. First you must find the value for side

4² + 3² = z²
16 + 9 = z²
5 = z

Then differentiate the Pythagorean Theorem which comes out to be this:

2x dx/dt + 2y dy/dt = 2z dz/dt (NOTE: you can cancel out the 2s)
x dx/dt + y dy/dt = z dz/dt

Using this you find the value for dy/dt:

(4) 3 dy/dt + 3 dy/dt = 5(1)
15 dy/dt = 5
dy/dt = 5/15 = 1/3

Now that we have dy/dt we can find dx/dt:

dx/dt = 3 (1/3)
dx/dt = 1

(5) Differentiate the function h(x) = f[g(x)] by using the Chain Rule this gives you:

h'(x) = f'(g(x)) x g'(x)

We are asked to find the horizontal tangent lines to the graph h, this will be where the derivative is equal to 0. So first we find out where g'(x) = 0, by looking at the graph that would be at x = -3, 0, 2. Then find out where f'(x) = 0, by looking at the graph that would be at x = -2, 1. Then since you have f'(g(x)) in the equation of h' you must find out where that would equal 0. This equals 0 where g(x) = -2, 1. Those values would be x = -4, -2. Add up all the values of x you found (making sure not to count a value twice) and you come up with 6 horizontal tangent lines.

Open Response: You will be using the equation for volume to solve this question. First differentiate the equation this gives you:

dV/dt = 4pir² dr/dt

Find the volume at 4 seconds by looking at the graph, V(4) = 5pi. Then using that value we can find r which is ³√15/4. Then we find the value of dV/dt from the graph. You draw a tangent line to the graph and just estimate by using the values on the graph. (NOTE: this value varies depending on how well and where you draw your tangent line) Mr. K got a value of 4pi/3 for dV/dt. Then we just plug all these values into the equation to find dr/dt which will be the approximate rate of the radius of the balloon changing after 4 seconds:

4pi/3 = 4pi(15/4)^1/3 dr/dt
1/3(15/4)^-1/3 = dr/dt
0.21455 ≈ dr/dt

And that is all folks, I'm hoping this isn't up tooo late so that it can be if you wish as studying material for the test which is TOMORROW. You all know Christian is the scribe for today, and he's probably waiting for me to put this up right now, sorry =S. Good luck to everyone on their test tomorrow, live long and prosper!

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