Question

43. A hot-air balloon is launched from an elevation of 5400 ft above sea level. As it rises, the vertical velocity is computed using a device (called a variometer) that measures the change in atmospheric pressure. The vertical velocities at selected times are shown in the table (with units of ft/min).

c. A polynomial that fits the data reasonably well is:
g(t) = 3.49t³ - 43.21t² + 142.43t - 1.75
Estimate the elevation of the balloon after five minutes using this polynomial.

Accepted Answer

Step 1: Understand the problem. We are tasked with estimating the elevation of the hot-air balloon after 5 minutes using the given polynomial g(t) = 3.49t³ - 43.21t² + 142.43t - 1.75. The initial elevation of the balloon is 5400 ft above sea level. Step 2: Recognize that the polynomial g(t) represents the vertical velocity of the balloon as a function of time t (in minutes). To find the elevation, we need to integrate g(t) with respect to t, as the elevation is the accumulation of vertical velocity over time. Step 3: Set up the integral for the elevation. The elevation E(t) at time t is given by the integral of g(t) from 0 to t, plus the initial elevation. Mathematically, this is expressed as: E(t) = 5400 + ∫[0 to t] g(t) dt. Step 4: Compute the indefinite integral of g(t). The integral of g(t) = 3.49t³ - 43.21t² + 142.43t - 1.75 is: ∫g(t) dt = (3.49/4)t⁴ - (43.21/3)t³ + (142.43/2)t² - 1.75t + C, where C is the constant of integration. Step 5: Evaluate the definite integral from 0 to 5. Substitute t = 5 into the integrated polynomial and subtract the value of the polynomial at t = 0. Add the result to the initial elevation of 5400 ft to find the estimated elevation after 5 minutes.

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Key Concepts

Polynomial Functions

Integration

Initial Conditions