Water temperature affects the growth rate of brook trout. The table shows the amount of weight gained by brook trout after 24 days in various water temperatures.
Temperature (°C) 15.5 17.7 20.0 22.4 24.4
Weight gained (g) 37.0 30.7 20.0 9.0 −9.9
If W(x) be the weight gain at temperature x, construct a table of estimated values for W′. (Use a one-sided difference quotient with an adjacent point for the first and last values, and the average of two difference quotients with adjacent points for all other values. Round your answers to two decimal places.)
x = 15.5, 17.7, 20.0, 22.4, 24.4
To find the estimated values for W′, we need to calculate the difference quotient using adjacent points.
For x = 15.5, we can use the one-sided difference quotient with the adjacent point x = 17.7.
W′(15.5) = (W(17.7) - W(15.5)) / (17.7 - 15.5)
= (30.7 - 37.0) / (17.7 - 15.5)
= -6.3 / 2.2
≈ -2.86
For x = 17.7, we can use an average of two difference quotients: one with the adjacent point x = 15.5 and one with the adjacent point x = 20.0.
W′(17.7) ≈ [(W(17.7) - W(15.5)) / (17.7 - 15.5) + (W(20.0) - W(17.7)) / (20.0 - 17.7)] / 2
≈ [(-6.3 / 2.2) + (20.0 - 30.7) / (20.0 - 17.7)] / 2
≈ (-2.86 + (-10.7)) / (2.3) / 2
≈ -13.56 / 2.3 / 2
≈ -2.94
For x = 20.0, we can use an average of two difference quotients: one with the adjacent point x = 17.7 and one with the adjacent point x = 22.4.
W′(20.0) ≈ [(W(20.0) - W(17.7)) / (20.0 - 17.7) + (W(22.4) - W(20.0)) / (22.4 - 20.0)] / 2
≈ [(20.0 - 30.7) / (20.0 - 17.7) + (9.0 - 20.0) / (22.4 - 20.0)] / 2
≈ (-10.7 / 2.3) + (-11.0 / 2.4)] / 2
≈ -4.65 + (-4.58)] / 2
≈ -9.23 / 2
≈ -4.62
For x = 22.4, we can use an average of two difference quotients: one with the adjacent point x = 20.0 and one with the adjacent point x = 24.4.
W′(22.4) ≈ [(W(22.4) - W(20.0)) / (22.4 - 20.0) + (W(24.4) - W(22.4)) / (24.4 - 22.4)] / 2
≈ [(9.0 - 20.0) / (22.4 - 20.0) + (-9.9 - 9.0) / (24.4 - 22.4)] / 2
≈ (-11.0 / 2.4) + (-18.9 / 2) / 2
≈ -4.58 + (-9.45) / 2
≈ -14.03 / 2
≈ -7.02
For x = 24.4, we can use the one-sided difference quotient with the adjacent point x = 22.4.
W′(24.4) = (W(24.4) - W(22.4)) / (24.4 - 22.4)
= (-9.9 - 9.0) / (24.4 - 22.4)
= -18.9 / 2
= -9.45
Therefore, the table of estimated values for W′ is as follows:
x | W′
---------------------
15.5 | -2.86
17.7 | -2.94
20.0 | -4.62
22.4 | -7.02
24.4 | -9.45
To construct a table of estimated values for W′ (W prime), we need to calculate the first derivative at each temperature point.
Let's start by calculating the first derivative at the temperature x = 15.5. Since this is the first value, we will use a one-sided difference quotient with the adjacent point x = 17.7.
W′(15.5) = (W(17.7) - W(15.5))/(17.7 - 15.5)
W(17.7) = 30.7 (weight gained at 17.7 degrees)
W(15.5) = 37.0 (weight gained at 15.5 degrees)
W′(15.5) = (30.7 - 37.0)/(17.7 - 15.5)
= -6.3/2.2
≈ -2.864 (rounded to two decimal places)
Next, let's calculate the first derivative at the temperature x = 17.7 using the average of two difference quotients with adjacent points (x = 15.5 and x = 20.0).
W′(17.7) = (W(20.0) - W(15.5))/(20.0 - 15.5)
W(20.0) = 20.0 (weight gained at 20.0 degrees)
W′(17.7) = (20.0 - 37.0)/(20.0 - 15.5)
= -17.0/4.5
≈ -3.778 (rounded to two decimal places)
Next, let's calculate the first derivative at the temperature x = 20.0 using the average of two difference quotients with adjacent points (x = 17.7 and x = 22.4).
W′(20.0) = (W(22.4) - W(17.7))/(22.4 - 17.7)
W(22.4) = 9.0 (weight gained at 22.4 degrees)
W′(20.0) = (9.0 - 30.7)/(22.4 - 17.7)
= -21.7/4.7
≈ -4.617 (rounded to two decimal places)
Next, let's calculate the first derivative at the temperature x = 22.4 using the average of two difference quotients with adjacent points (x = 20.0 and x = 24.4).
W′(22.4) = (W(24.4) - W(20.0))/(24.4 - 20.0)
W(24.4) = -9.9 (weight gained at 24.4 degrees)
W′(22.4) = (-9.9 - 20.0)/(24.4 - 20.0)
= -29.9/4.4
≈ -6.795 (rounded to two decimal places)
Finally, we will use a one-sided difference quotient with the adjacent point x = 22.4 to calculate the first derivative at the temperature x = 24.4 (the last value).
W′(24.4) = (W(24.4) - W(22.4))/(24.4 - 22.4)
W′(24.4) = (-9.9 - (-9.9))/(24.4 - 22.4)
= 0/2.0
= 0
Now we can construct the table of estimated values for W′:
Temperature (°C) 15.5 17.7 20.0 22.4 24.4
W′ -2.86 -3.78 -4.62 -6.80 0.00
These estimated values represent the rate at which the weight of the brook trout changes with respect to the water temperature.
To calculate the estimated values for the weight gain derivative (W′), we need to find the first-order difference quotient for the given data points using a one-sided difference quotient with an adjacent point for the first and last values, and the average of two difference quotients with adjacent points for all other values.
Let's calculate the estimated values for W′ using the given data:
For x = 15.5:
W′(15.5) ≈ (W(17.7) - W(15.5)) / (17.7 - 15.5)
= (30.7 - 37.0) / (17.7 - 15.5)
= -6.3 / 2.2
≈ -2.86
For x = 17.7:
W′(17.7) ≈ (W(20.0) - W(17.7)) / (20.0 - 17.7)
= (20.0 - 30.7) / (20.0 - 17.7)
= -10.7 / 2.3
≈ -4.65
For x = 20.0:
W′(20.0) ≈ [(W(20.0) - W(17.7)) / (20.0 - 17.7) + (W(22.4) - W(20.0)) / (22.4 - 20.0)] / 2
= [(-10.7 / 2.3) + (9.0 / 2.4)] / 2
≈ (-4.65 + 3.75) / 2
≈ -0.45
For x = 22.4:
W′(22.4) ≈ [(W(22.4) - W(20.0)) / (22.4 - 20.0) + (W(24.4) - W(22.4)) / (24.4 - 22.4)] / 2
= [(9.0 / 2.4) + (-9.9 / 2.0)] / 2
≈ (3.75 - 4.95) / 2
≈ -0.60
For x = 24.4:
W′(24.4) ≈ (W(24.4) - W(22.4)) / (24.4 - 22.4)
= (-9.9 / 2.0) / 2.0
≈ -2.48 / 2.0
≈ -1.24
Therefore, the estimated values for W′ are approximately:
W′(15.5) ≈ -2.86
W′(17.7) ≈ -4.65
W′(20.0) ≈ -0.45
W′(22.4) ≈ -0.60
W′(24.4) ≈ -1.24