A bicyclist is competing in a tournament. He is currently in a very hilly part of the tournament's course. For his current hill, his height in metres over time in minutes is modelled by the function;
f(x)=-0.00075x^2-0.5825x+17.269
a) Estimate the bicyclists instantaneous rate of change after 100 minutes. Round your answer to two decimal places.
b) Is the bicyclist travelling uphill or downhill at this time? Explain how you know.
hey - do you mean height OVER (meaning divided by) time or height as a function of time?
If the latter, we need to take the derivative.
I think its as a function of time
ok
f'(x) = v(x) = .0015 x - .5825
at x = 100
v(x) = .15 -.5825 = -.4325 = -.43
dh/dt is negative, going downhill :)
To estimate the bicyclist's instantaneous rate of change after 100 minutes, we need to find the derivative of the given function f(x) and then evaluate it at x = 100.
a) To find the derivative of f(x), we differentiate each term with respect to x:
f'(x) = d/dx(-0.00075x^2) + d/dx(-0.5825x) + d/dx(17.269)
The derivative of a constant term is always 0, so the third term disappears:
f'(x) = -0.0015x - 0.5825
Now, we can evaluate f'(x) at x = 100:
f'(100) = -0.0015(100) - 0.5825
= -0.15 - 0.5825
= -0.7325
Therefore, the estimated instantaneous rate of change after 100 minutes is approximately -0.73 (rounded to two decimal places).
b) To determine whether the bicyclist is traveling uphill or downhill at this time, we need to analyze the sign of the derivative.
Since the derivative f'(100) is negative (-0.73), this indicates that the function f(x) is decreasing at x = 100. In other words, the bicyclist is traveling downhill at this time.
Explanation: The derivative of a function describes its instantaneous rate of change at a given point. If the derivative is positive, the function is increasing, indicating the bicyclist is going uphill. If the derivative is negative, the function is decreasing, indicating the bicyclist is going downhill.