A big clock has a minute hand 3 ft long and a second hand 4 ft long. let x be the distance between their tips. find max value of (dx)/(dt) *d means derivative
To find the maximum value of (dx)/(dt), we need to find the derivative of x with respect to time (t), and then determine the maximum value of that derivative.
First, let's establish some notation:
- Let M represent the position of the minute hand.
- Let S represent the position of the second hand.
- Let O represent the center of the clock.
Now, let's consider the triangle formed by the minute hand, the second hand, and the distance between their tips (x) when the hands are at some position.
Using the Law of Cosines, we can relate x, M, and S:
x^2 = M^2 + S^2 - 2MS * cos(theta)
Where theta is the angle between the two hands.
Now, let's differentiate both sides of the equation with respect to time (t):
2x(dx/dt) = 2M(M') + 2S(S') - 2(ds/dt) * M * S * sin(theta) * (dtheta/dt)
Since we are interested in finding the maximum value of (dx/dt), we need to set d(theta)/dt = 0. This is because the maximum (dx/dt) occurs when the hands are aligned or at a maximum or minimum angle.
Therefore, sin(theta) is maximized and we can substitute sin(theta) = 1:
2x(dx/dt) = 2M(M') + 2S(S') - 2(ds/dt) * M * S * (dtheta/dt)
Now, we need to relate (dtheta/dt) to the rates at which the hands move.
From basic trigonometry, we can determine that:
(dtheta/dt) = (dM/dt - dS/dt) / x
Notice that (dM/dt - dS/dt) represents the difference in speeds between the minute and second hands, as positive values of (dM/dt) indicate that the minute hand is moving counterclockwise, and positive values of (dS/dt) indicate that the second hand is moving clockwise.
Plugging in this expression for (dtheta/dt) back into our equation:
2x(dx/dt) = 2M(M') + 2S(S') - 2(ds/dt) * M * S * (dM/dt - dS/dt) / x
By simplifying the equation and canceling out the common factors of 2, x, S, and M:
(dx/dt) = (M(M') + S(S') - (ds/dt) * M * (dM/dt - dS/dt)) / x
Now, we can determine the maximum value of (dx/dt) by maximizing the numerator and minimizing x.
We know the following:
- M is constant at 3 ft
- S is constant at 4 ft
- (ds/dt) is constant since the second hand moves at a constant speed (we can choose it as 1 for simplicity)
Thus, we need to maximize the product of M * (dM/dt - dS/dt). We can achieve this by maximizing (dM/dt - dS/dt), which happens when the difference in speeds is maximum.
Since the minute hand moves slower than the second hand, we can assume that the minute hand moves at a constant rate, and the second hand moves at a variable rate.
Therefore, to maximize (dM/dt - dS/dt), we need the second hand to move as fast as possible, which is typically its maximum speed. We can assume (dS/dt) = 1 ft/s.
Plugging these values into the equation:
(dx/dt) = (3(0) + 4(1) - (1) * 3 * (0 - 1)) / x
(dx/dt) = (0 + 4 + 3) / x
(dx/dt) = 7 / x
Now, we can see that (dx/dt) is inversely proportional to x. Therefore, to maximize (dx/dt), we need to minimize x.
The minimum value of x occurs when the hands are completely aligned, and x is equal to the difference in the lengths of the hands: |3 - 4| = 1 ft.
Therefore, the maximum value of (dx/dt) occurs when x = 1 ft, and in this case, (dx/dt) = 7 / 1 = 7 ft/s.
Hence, the maximum value of (dx/dt) is 7 ft/s.