- Date for hour exam 2 could be as early as next Friday (Nov. 7)
- Exam will cover material on derivatives since first exam (e.g., chain rule, implicit differentiation, related rates, extreme values, Mean Value Theorem, optimization, etc.)
- Problem set question 2
- i = 2 cost + 2 sint
- Where are (positive) peaks?
- di/dt = -2 sint + 2 cost
- Set di/dt = 0 and solve for t
- 0 = -2 sint + 2 cost
- 2 sint = 2 cost
- tan t = 1
- t = π/4, 5π/4, ...
- Peak current is 2 cos(π/4) + 2 sin(π/4)
- Look at this with R
> i = function(x){ 2*sin(x) + 2*cos(x) } > curve( i(x), -4*pi, 4*pi ) > curve( tan(x), add=TRUE, col="blue" ) > i( pi/4 ) [1] 2.828427
- Problem set question 3
- c value(s) per mean value theorem for f(x) = √(x-1)
- c value(s) per mean value theorem for f(x) = √(x-1)
- Section 4.3
- Use mean value theorem over intervals to find local extrema by looking at whether f′ is positive or negative
- Positive derivative means function is increasing
- Negative means function decreasing
- Increasing before c and decreasing after means maximum
- Decreasing before c and increasing after means minimum
- Examples: figure out where each of the following functions is
increasing, decreasing, (locally) maximum or minimum
- f(x) = 2x3 - x2
- f′(x) = 6x2 - 2x
- = 2x (3x - 1)
- f′(x) = 0 at 1/3, 0
- f′(x) > 0 if x > 1/3, x < 0
- f′(x) < 0 if 0 < x < 1/3
- So f must look something like...
- f(x) = 2x3 - x2
- Concavity
- Read section 4.4