During the flood of 1997 in Fargo, North Dakota, the Red River of the North rose
ID: 2838361 • Letter: D
Question
During the flood of 1997 in Fargo, North Dakota, the Red River of the North rose menacingly durning the month of April. Supposed that the official river level at noon on day tis given by the function R(t). We know that R(April 9) = 35.5 feet and R'(April 9)=2
a) What are the units of R'(t)
b) Construct a linear function to estimate the water level of the Red River for dates near April 9
c) Use this model from part (b) to estimate R(April 7) and R(April 11)
d) The Red River crested on April 17 at 39.79 feet . What is R'(April 17)?
e) If the Red River level on April 22 was 39 feet and R'(April 22) was -0.3, find a linear model and then use it to estime R(April 26)
I dont even know where to start on this.... please help its graded.
Step by step would be helpful
Explanation / Answer
a)
The units of R'(t) are feet/days in a month (length/time) as the term R'(t) denotes the rate of change of river level with respect to time.
b)
the linear function that estimates the water level is the equation of the line passing through the point (9,35.5) and having the slope 2.
that means we can use the formula y-y1 = m (x-x1).
R(t) - 35.5 = 2 * ( t - 9 )
R(t) = 2 * t + 17.5.
c)
Put t = 7 in R(t) = 2 * t + 17.5 to estimate R(April 7).
R(April 7) = 2 * 7 + 17.5 = 31.5 feet.
Put t = 11 in R(t) = 2 * t + 17.5 to estimate R(April 11).
R(April 11) = 2 * 11 + 17.5 = 39.5 feet.
d)
If R(April 17) = 39.79 feet then R'(April 17) is the slope of the line joining the points (0,0) and (17,39.79).
Hence R'(April 17) = (39.79-0)/(17-0)
= 2.34058823529.
e)
the linear function that estimates the water level is the equation of the line passing through the point (22,39) and having the slope -0.3.
that means we can use the formula y-y1 = m (x-x1).
R(t) - 39 = -0.3 * ( t - 22)
R(t) = -0.3 * t + 45.6
Put t = 26 in R(t) = -0.3 * t + 45.6 to estimate R(April 7).
R(April 26) = -0.3 * 26 + 45.6 = 37.8 feet.
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