Verified Answer :
- Use the equation of the circle \(x^2 + y^2 = 289\) to relate the coordinates \(x\) and \(y\).
- Apply implicit differentiation with respect to time \(t\) to find the relationship between the rates of change: \(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\).
- Calculate the possible \(y\)-coordinates by substituting \(x=15\) into the circle equation, resulting in \(y = \pm 8\).
- Solve for \(\frac{dx}{dt}\) by substituting the known values of \(x\), \(y\), and \(\frac{dy}{dt}\), yielding \(\frac{dx}{dt} = \pm 1.6\).
- Final Answer: \(\boxed{\pm 1.6}\)
Explanation
- Analysis of the Problem and Data
The particle moves along the circular path defined by the equation \(x^2 + y^2 = 289\). This represents a circle centered at the origin \((0,0)\) with a radius \(r = \sqrt{289} = 17\). We are given the position \(x = 15\) at a specific instant and the rate of change of the vertical position \(\frac{dy}{dt} = -3\). We need to determine the rate of change of the horizontal position \(\frac{dx}{dt}\) at that exact moment.
- Deriving the Relationship Between Rates of Change
Since the particle is constrained to the circle, its coordinates must satisfy \(x^2 + y^2 = 289\) for all time \(t\). Differentiating both sides of this equation with respect to \(t\) using the chain rule gives:
\[\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(289)\]
\[2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\]
Dividing the entire equation by 2 simplifies the relationship between the rates to:
\[x \frac{dx}{dt} + y \frac{dy}{dt} = 0\]
- Determining the Particle's Vertical Position
At the instant when \(x = 15\), we find the corresponding \(y\) values using the circle equation:
\[15^2 + y^2 = 289\]
\[225 + y^2 = 289\]
\[y^2 = 289 - 225 = 64\]
\[y = \pm 8\]
This means the particle is at either \((15, 8)\) or \((15, -8)\) at that instant.
- Calculating the Rate of Change \(\frac{dx}{dt}\)
Now we substitute the known values into the differentiated equation \(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\):
Case 1: When \(y = 8\)
\[15 \left(\frac{dx}{dt}\right) + 8(-3) = 0\]
\[15 \left(\frac{dx}{dt}\right) = 24\]
\[\frac{dx}{dt} = \frac{24}{15} = \frac{8}{5} = 1.6\]
Case 2: When \(y = -8\)
\[15 \left(\frac{dx}{dt}\right) + (-8)(-3) = 0\]
\[15 \left(\frac{dx}{dt}\right) = -24\]
\[\frac{dx}{dt} = -\frac{24}{15} = -\frac{8}{5} = -1.6\]
The rate of change \(\frac{dx}{dt}\) is \(\pm 1.6\) depending on whether the particle is in the upper or lower half of the circle.
- Final Answer
At the instant when \(x=15\) and \(\frac{dy}{dt}=-3\), the rate of change of the horizontal position \(\frac{dx}{dt}\) is \(\boxed{\pm 1.6}\).
Examples
Imagine you are tracking a satellite or a robot arm moving along a circular track. By understanding how the horizontal velocity depends on the vertical position and vertical velocity, you can predict the movement of the object at any point on its path. This application of related rates is fundamental in fields ranging from robotics and aerospace engineering to game physics and animation, allowing for accurate control and prediction of mechanical systems.