Problem of the Day

1. Problem of the Day If f (x) = sin(e-x), then f '(x) = A) -cos(e-x) B) cos(e-x) + e-x C) cos(e-x) - e-x D) e-xcos(e-x) E) -e-xcos(e-x)

2. Problem of the Day If f (x) = sin(e-x), then f '(x) = A) -cos(e-x) B) cos(e-x) + e-x C) cos(e-x) - e-x D) e-xcos(e-x) E) -e-xcos(e-x)

3. Euler's Method is a numerical approach to approximating the particular solution of the differential equation y' Leonhard Euler Born April 15, 1707 in Basel, Switzerland

4. Euler's Method is a numerical approach to approximating the particular solution of the differential equation y'

5. The graph of the solution passes through the point (x , y ) and has a slope of y'(x , y ) 0 0 0 0 From this starting point, you proceed in the direction indicated by the slope. Using a small step h, move along the tangent line until you arrive at the point (x , y ) where 1 1 y = y + hy'(x , y ) x = x + h 1 0 0 0 1 0

7. step length of 1.5

8. step length of 1.5 step length of .75

9. step length of .75 step length of .25

10. http://www-math.mit.edu/daimp/EulerMethod.html

11. h = 0.1 x = 0 x = 0.1 x = 0.2 x = 0.3 . . . y = 1 y' = x - y 0 1 2 3 Use Euler's Method to approximate the particular solution of the differential equation y' = x - y and point (0, 1). Use a step of h = 0.1 y = y + hy'(x , y ) 1 0 0 0

12. y = y + hy'(x , y ) x = 0 0 x = 0.1 1 h = 0.1 y = 1 y' = x - y 1 0 0 0 y = y + hy'(x , y ) = 1 + 0.1(0 - 1) = 0.9 0 0 0 1 y = y + hy'(x ,y ) = 0.9 + 0.1(0.1 - 0.9) = 0.82 1 2 1 1 x = 0.2 2 y = y + hy'(x ,y ) = 0.82 + 0.1(0.2 - 0.82) = 0.758 2 2 2 3

13. Hot coffee in a 70-degree room cools at a rate proportional to the difference between the coffee temperature and room temperature. y' (t) = k(y - 70) At a certain time, a thermometer showed a coffee temperature of 190 degrees, dropping at a rate of 12 degrees per minute.

14. Hot coffee in a 70-degree room cools at a rate proportional to the difference between the coffee temperature and room temperature. y' (t) = k(y - 70) At a certain time, a thermometer showed a coffee temperature of 190 degrees, dropping at a rate of 12 degrees per minute. y(0) = 190 degrees y'(0) = -12 degrees

15. y' (t) = k(y - 70) y(0) = 190 degrees y'(0) = -12 degrees Find the particular solution (find k)

16. y' (t) = k(y - 70) y(0) = 190 degrees y'(0) = -12 degrees Find the particular solution (find k) -12 = k (190 - 70) -12 = 120k -0.1 = k y' = -0.1 (y - 70)

17. How hot will the coffee be after 10 minutes? y' = -0.1 (y - 70) Start (0, 190) use h = 1

18. How hot will the coffee be after 10 minutes? y' = -0.1 (y - 70) Start (0, 190) use h = 1 x = 0 y1= y0 + h y'(0, 190) = 190 + 1(-12) = 178 x = 1 y2= y1 + h y'(1, 178) = 178 + 1(-10.8) = 167.2 If we continue in this manner we get -

19. How hot will the coffee be after 10 minutes? y' = -0.1 (y - 70) Start (0, 190) use h = 1

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