Friday, November 03, 2006
The Answer to George Bush's Eye-plank Growth Problem
On Tuesday, Abe Linkum presented a problem for readers to solve that calculates the growth rate of the plank in George Bush's eye as the religious right closes in on Ground Zero. Since nobody came up with the correct answer, no one receives the autographed picture of Abe.
Here is the answer key along with the original schematic --
Using the Pythagorean Theorem and differentiating with respect to time:
w^2 + p^2 = b^2 = 400
2w(dw/dt) + 2p(dp/dt) = 0
w(dw/dt) + p(dp/dt) = 0
Now, solve for 'dp/dt' given that dw/dt = -30 ft/s, w = 1 and p = sq.root of 399 by the Pythagorean Theorem:
1(-30) + (sq.root of 399)(dp/dt) = 0
So, dp/dt = 30/(sq.root of 399) or approximately 1.5 ft/s. That means that the plank in George Bush's eye is growing at this rate when the religious right is 1 ft. away from George Bush's vision.
Here is the answer key along with the original schematic --
Using the Pythagorean Theorem and differentiating with respect to time:
w^2 + p^2 = b^2 = 400
2w(dw/dt) + 2p(dp/dt) = 0
w(dw/dt) + p(dp/dt) = 0
Now, solve for 'dp/dt' given that dw/dt = -30 ft/s, w = 1 and p = sq.root of 399 by the Pythagorean Theorem:
1(-30) + (sq.root of 399)(dp/dt) = 0
So, dp/dt = 30/(sq.root of 399) or approximately 1.5 ft/s. That means that the plank in George Bush's eye is growing at this rate when the religious right is 1 ft. away from George Bush's vision.
Labels: Bush, religious right, Ted Haggard
