If you was the scientist to discover the K.E. formula starting from the fact that K.E. is proportional to the velocity to a power by the mass.
Illustrate your steps to lead to the above formula.
Saturday, March 1, 2008
Your Strategy
All students will act as the teacher. Each of you will comment on the strategy
that you will use to teach someone about differentiation of polynomials, logs and ln.
All the common errors you have observed can now be avoided, how?
What is your strategy in your teaching to avoid these errors?
that you will use to teach someone about differentiation of polynomials, logs and ln.
All the common errors you have observed can now be avoided, how?
What is your strategy in your teaching to avoid these errors?
Tuesday, February 26, 2008
Are these solutions correct?
Solution 1
y = log base 3 (8x^7 + x)
u = 8 x^7 + x
dy/dx = 56 x^6 + 1
= (56 x^6 + 1) / (8 x^7 + x) log base 3 e
Solution 2
P = ln 6s^3 + s
u = 6s^3
dy/dx = 18 s^2
= (18s^2)/(6s^3) + 1
Solution 2
P = ln 6s^3 + s
u = 6s^3 + s
dy/dx = 18 s^2 + 1
= (18s^2 + 1)/(6s^3 + s)
Solution 3
y = 15 x^3 + 7/(2x^3) - x
y = 15x3 + 7 *2 x^-3 - x
dy/dx = 45 x^2 + 42 x^-4 - 1
Solution 3
y = 15 x^3 + 7/(2x^3) - x
dy/dx = 45 x^2 + 7/6 x^-2 - 1
y = log base 3 (8x^7 + x)
u = 8 x^7 + x
dy/dx = 56 x^6 + 1
= (56 x^6 + 1) / (8 x^7 + x) log base 3 e
Solution 2
P = ln 6s^3 + s
u = 6s^3
dy/dx = 18 s^2
= (18s^2)/(6s^3) + 1
Solution 2
P = ln 6s^3 + s
u = 6s^3 + s
dy/dx = 18 s^2 + 1
= (18s^2 + 1)/(6s^3 + s)
Solution 3
y = 15 x^3 + 7/(2x^3) - x
y = 15x3 + 7 *2 x^-3 - x
dy/dx = 45 x^2 + 42 x^-4 - 1
Solution 3
y = 15 x^3 + 7/(2x^3) - x
dy/dx = 45 x^2 + 7/6 x^-2 - 1
Saturday, February 23, 2008
Log checking
Problem 1
log (2x^2 + 6x) - 6 = log (2x)
What can I do now?
Problem 2
Am I going correct?
log (x + 6) = 2 - log (4x)
log (x + 6) = log 100 - log (4x)
log (2x^2 + 6x) - 6 = log (2x)
What can I do now?
Problem 2
Am I going correct?
log (x + 6) = 2 - log (4x)
log (x + 6) = log 100 - log (4x)
Thursday, February 14, 2008
Application of simple differentiation
Question 1
Water is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.
Question 2
The velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.
Example 3
The electric field E at a distance r from a point charge is E = k/r^2, where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
Example 4
The distance s (in m) traveled by a subway train after the brakes are applied is given by s = 20t - 2t^2. How far does it travel, after the brakes are applied, in coming to a stop?
Water is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.
Question 2
The velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.
Example 3
The electric field E at a distance r from a point charge is E = k/r^2, where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
Example 4
The distance s (in m) traveled by a subway train after the brakes are applied is given by s = 20t - 2t^2. How far does it travel, after the brakes are applied, in coming to a stop?
Tuesday, February 12, 2008
Examining the accuracy of logs
Look at the following and made some comments to assist. Remember the log base is in [] brackets
Example 1
log [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)
Example 2
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log (2x + 4) = 2
log (2x + 8) = 2
log 8
2x + 4 = 1.8
2x = 1.8 - 4
Example 3
log [8] (x + 2) = 2 - log [8] 2
log [8] x + log [8] 2 = 2 - log [8] 2
log [8] x + 2 log [8] 2 = 2
Example 4
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) = 1.67
log [8] 1.67 = x + 2
Example 5
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 2(x + 2) = 2
log [8] (2x + 4) = 2
log [8] 2x = 2 - 4
Example 6
log [8] (x + 2) = 2 - log [8] 2
log (x + 2) / log 8 = 2 - log 2 / log 8
log (x + 2) / 0.903 = 2 - 0.33
log (x + 2) 0.903 * 1.667
log x + 2 = 1.5
log x = 1.5 - 2
log x = 0.5
x = 0.32
Example 1
log [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)
Example 2
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log (2x + 4) = 2
log (2x + 8) = 2
log 8
2x + 4 = 1.8
2x = 1.8 - 4
Example 3
log [8] (x + 2) = 2 - log [8] 2
log [8] x + log [8] 2 = 2 - log [8] 2
log [8] x + 2 log [8] 2 = 2
Example 4
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) = 1.67
log [8] 1.67 = x + 2
Example 5
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 2(x + 2) = 2
log [8] (2x + 4) = 2
log [8] 2x = 2 - 4
Example 6
log [8] (x + 2) = 2 - log [8] 2
log (x + 2) / log 8 = 2 - log 2 / log 8
log (x + 2) / 0.903 = 2 - 0.33
log (x + 2) 0.903 * 1.667
log x + 2 = 1.5
log x = 1.5 - 2
log x = 0.5
x = 0.32
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