#17 step by step please The atmospheric temperature T (in degrees Celsius) at al
ID: 2879917 • Letter: #
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#17 step by step please
The atmospheric temperature T (in degrees Celsius) at altitude h meters above a certain point on Earth can be approximated by T = 15 - 0.0065h for h lessthanorequalto 12,000 m. What are the average and instantaneous rates of change of T with respect to h? Why are they the same? Sketch the graph of T for h lessthanorequalto 12,000. In Exercises 11-18, estimate the instantaneous rate of change at the point indicated. P(x) = 3x^2 - 5; x = 2 f(t) = 12t - 7; t = -4 y(x) = 1/x + 2; x = 2 y(t) = Squareroot 3t + 1; t = 1 f(x) = 3^x; x = 0 f(x) = 3^x; x = 3 f(x) = sin x; x = pi/6 f(x) = tan x; x = pi/4 The height (in meters) at time t (in seconds) of a small oscillating at the end of a spring is h(t) = 8 cos (12 pi t). (a) Calculate the mass's average velocity over the time interval [0, 0, 1] and [3, 3, 5]. (b) Estimate its instantaneous velocity at t = 3.Explanation / Answer
Given f(x) = sinx
Instantaneous rate of change of f(x) can be found by derivating it
d/dx(sinx) = cosx. (It is a standard integral)
At x = pi/6,
f'(x) = cos(pi/6)
='sqrt(3)/2
= 1.732/2
=0.866
Therefore the instantaneous rate of change of f(x) at x= pi/6 is 0.866
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