Part A.
a.) When I left for American University this fall my brother bought me a sunflower as a going away gift. It began to grow on Day 5. Below is the inches the sunflower grew over time.
b.) I want to solve for the rate of change on exactly day 10.
c.)
d.)
e.) Rate of Change: (Y2-Y1)/(X2-X1)
(10, 1.5) & (5, .5): (10-5)/(1.5-.5)= 5
(10, 1.5) & (15, 2): (15-10)/(2-1.5)= 10
(10, 1.5) & (20, 3): (20-10)/(3-1.5)= 6.67
f.)
g.) Points on tangent line: (3,0) & (25,4)
Slope: (25-3)/(4-0)= 5.5
h.) I used the slope of secant and tangent lines in correspondence with the point of data I recorded on day 10 to find the most exact instantaneous rate of change. This method is effective because by finding the average rates of change between the data points closest to the point I am solving for makes the window smaller and smaller in which the instantaneous rate of change lies. We see this process above in the calculations (Y2-Y1)/(X2-X1) and the solutions to the rates of change in the tangent and secant lines.
Overall, really nice job! I think it is a little difficult to see the tangent line but from your calculations it is obvious it is there.
ReplyDeletetaylor,
ReplyDeletei like the beginning when it's a horizontal line because the seed is still in the ground! tee hee! your rate of change calculations are reversed, though. :/ you put the x's on the top and the y's on the bottom. your explanations are generally correct, though, given the calculations that you used. just remember to use the correct values for calculating slope!!!
professor little