17. A person flying a kite holds the string 5 feet above ground level and the string is let out at a rate of $2 \mathrm{ft} / \mathrm{sec}$ as the kite moves horizontally at an altitude of 105 feet (see figure). Assuming there is no sag in the string, find the rate at which the kite is moving when the string is 125 feet long. (Write a sentence explaining the meaning of your answer including units.)

Step 1 ：First, we need to calculate the value of \(x\) when \(z = 125\) feet. We can do this by rearranging the Pythagorean theorem to solve for \(x\): \(x = \sqrt{z^2 - y^2}\).

Step 2 ：Then, we can differentiate the Pythagorean theorem with respect to time \(t\) to get \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}\).

Step 3 ：Since \(y\) is constant (the height of the kite doesn't change), \(\frac{dy}{dt} = 0\).

Step 4 ：So, the equation simplifies to \(2x\frac{dx}{dt} = 2z\frac{dz}{dt}\).

Step 5 ：We can solve this equation for \(\frac{dx}{dt}\) to find the rate at which the kite is moving horizontally: \(\frac{dx}{dt} = \frac{z}{x}\frac{dz}{dt}\).

Step 6 ：Finally, we can substitute the known values into this equation to find the answer. Given that \(y = 105\), \(z = 125\), and \(\frac{dz}{dt} = 2\), we find that \(x = \sqrt{z^2 - y^2} = 67.82329983125268\).

Step 7 ：Substituting these values into the equation for \(\frac{dx}{dt}\), we find that \(\frac{dx}{dt} = \frac{z}{x}\frac{dz}{dt} = 3.6860489038724285\).

Step 8 ：Final Answer: The rate at which the kite is moving horizontally when the string is 125 feet long is approximately \(\boxed{3.69 \, \text{feet/second}}\). This means that at that moment, the kite is moving away from the person flying it at a speed of 3.69 feet per second.