A certain clock has a minute hand that is 4 inches long and an hour hand is 3 inches long. How fast is the distance between the tips of the hands changing at 9:00?
To determine the rate at which the distance is changing between the tips of the clock hands, we can rely on the concept of related rates in calculus.
Let's first define the variables we will be dealing with:
- Let d represent the distance between the tips of the clock hands.
- Let x represent the horizontal distance from the center of the clock to the tip of the hour hand.
- Let y represent the vertical distance from the center of the clock to the tip of the minute hand.
Since the hour hand is 3 inches long and the minute hand is 4 inches long, we can conclude that \(x^2 + y^2 = 3^2\) and \((x+d)^2 + y^2 = 4^2\), utilizing the Pythagorean theorem.
We need to determine \(\frac{{dx}}{{dt}}\), the rate at which x is changing, to determine the rate at which the distance between the tips of the hands is changing.
To achieve this, we can find \(\frac{{dy}}{{dt}}\) (the rate at which y is changing) and \(\frac{{dd}}{{dt}}\) (the rate at which d is changing) and substitute them into the equation \(\frac{{dd}}{{dt}} = \sqrt{\left(\frac{{dx}}{{dt}}\right)^2 + \left(\frac{{dy}}{{dt}}\right)^2}\).
Now, at 9:00 on an analog clock, the minute hand is pointing straight up, while the hour hand is pointing directly at the number 9.
At this point, the minute hand is perpendicular to the x-axis, which means \(y = 4\) (the length of the minute hand). And since it is 9:00, the hour hand is pointing to the left, so \(x = -3\) (negative since it points to the left).
Using the equations \(x^2 + y^2 = 3^2\) and \(y = 4\), we can solve for \(x\) and find that \(x = -\sqrt{7}\).
Differentiating with respect to time, we obtain \(\frac{{d}}{{dt}}(y = 4)\), which gives us \(\frac{{dy}}{{dt}} = 0\) since y is not changing.
Similarly, differentiating \(x^2 + y^2 = 3^2\) with respect to time, we get \(2x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}} = 0\). Plugging in \(x = -\sqrt{7}\) and \(\frac{{dy}}{{dt}} = 0\), we can solve for \(\frac{{dx}}{{dt}}\) and find that \(\frac{{dx}}{{dt}} = \frac{{\sqrt{7}}}{2}\).
Now, substituting \(\frac{{dx}}{{dt}} = \frac{{\sqrt{7}}}{2}\) and \(\frac{{dy}}{{dt}} = 0\) into the equation \(\frac{{dd}}{{dt}} = \sqrt{\left(\frac{{dx}}{{dt}}\right)^2 + \left(\frac{{dy}}{{dt}}\right)^2}\), we get \(\frac{{dd}}{{dt}} = \frac{{\sqrt{7}}}{2}\).
Therefore, the distance between the tips of the hands is changing at a rate of \(\frac{{\sqrt{7}}}{2}\) inches per hour at 9:00.