Now that we have calculated the first two approximate points on the graph, we can extend the procedure to calculate successive points.

14) In the following table we have sumarized the initial values and the values for the first two points. You should continue the pattern and fill in the rest of the table (on your worksheet ) for points 3, 4 and 5. Check to see if you agree with the last entry in the table. Plot and connect the points (tn,yn ) on your worksheet to see an approximation for the function Y(t).

Remember that Y’(t) = 2t+1

Point number

tn

yn

Y’(tn)

t

yn+1

yn+1

0

0

1

1

.5

.5

1.5

1

.5

1.5

2

.5

1

2.5

2

1

2.5

3

.5

 

 

3

1.5

 

 

 

 

 

4

2

 

 

 

 

 

5

2.5

11.5

link to a completed table and graph

15) Now consider that you know the results for the 20th as shown below. What will be the values for the 21st

 

20
10
106
21
.5
10.5
116.5
21

 

16) Now suppose that we have the kth point , write the expressions for:

yk+1 and yk+1

 

k

tk

yk

Y’(tk)

t

yk+1 =

yk+1 =

 

17) Given the rate of change function Y’(t) and an initial point (t0,y0 ) explain, in your own words, how the following formulas can be used to calculate coordinates for successive points.

tk+1 = tk+t

yk+1 = yk+Y’(tk) t

18) Using t =.25, approximate function Y(t) over the interval from t=0 to t=3 and plot the results on the same graph as you used in question 14.

19) What is the effect of reducing the size of t ?

20) Find a function Y(t) whose derivative is Y’(t)=2t+1 and which passes through (0,1)

 

21) Plot the graph of the function Y(t) on the same graph as the previous graphs.

 

22) In light of your answer to question 21, what is the effect of making t smaller.

 

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