Problem 40 - Calculus - AP Edition

Question

A point \(P\) is moving along the curve whose equation is \(y=\sqrt{x} .\) Suppose that \(x\) is increasing at the rate of 4 units/s when \(x=3\) (a) How fast is the distance between \(P\) and the point (2,0) changing at this instant? (b) How fast is the angle of inclination of the line segment from \(P\) to (2,0) changing at this instant?

Step-by-Step Solution

Answer

(a) \( \frac{dR}{dt} = 3 \) units/s, (b) \( \frac{d\theta}{dt} = -\frac{5\sqrt{3}}{6} \) radians/s.

Step 1: Understand the Problem

We need to find how two things change as a point \( P \) on the curve \( y = \sqrt{x} \) changes its position over time. (a) The rate of change of the distance between point \( P \) and point \( (2, 0) \), and (b) the rate of change of the angle between line segment \( P \) to \( (2, 0) \). We know the rate of change of \( x \) at \( x = 3 \) is 4 units/s.

Step 2: Calculate y at x = 3

Since the point \( P \) is on the curve \( y = \sqrt{x} \), substitute \( x = 3 \) into the equation to find \( y \). Hence, \( y = \sqrt{3} \). Thus the point \( P \) has coordinates \( (3, \sqrt{3}) \).

Step 3: Determine the distance R between P and (2,0)

Use the distance formula \( R = \sqrt{(x - 2)^2 + (y - 0)^2} \). Substituting the coordinates of \( P \): \( R = \sqrt{(3 - 2)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \).

Step 4: Differentiate R with respect to time t

To find how the distance is changing with respect to time, differentiate both sides of the equation \( R = \sqrt{(x - 2)^2 + y^2} \) with respect to time \( t \):\[ \frac{dR}{dt} = \frac{1}{2}((x - 2)^2 + y^2)^{-1/2}(2(x-2)\frac{dx}{dt} + 2y\frac{dy}{dt}) \] Given \( x = 3, (x-2) = 1, \frac{dx}{dt} = 4, y = \sqrt{3}, \) and \( \frac{dy}{dt} = \frac{1}{2\sqrt{x}} \frac{dx}{dt} \) at \( x = 3 \):\( \frac{dy}{dt} = \frac{1}{2\sqrt{3}} \times 4 = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \).Thus,\[ \frac{dR}{dt} = \frac{1}{2}(4)^{-1/2}(2(1)(4) + 2\sqrt{3}\frac{2\sqrt{3}}{3}) \]After simplification:\[ \frac{dR}{dt} = \frac{1}{4}(8 + 4) = 3 \text{ units/s} \]

Step 5: Determine the angle of inclination θ

The point \( P \) forms an angle \( \theta \) with the horizontal axis as seen from the point \( (2, 0) \). We use tangent:\[ \tan \theta = \frac{y}{x-2} \]Substituting \( y = \sqrt{3} \) and \( x - 2 = 1 \),\[ \tan \theta = \sqrt{3} \] which gives \( \theta = \frac{\pi}{3} \).

Step 6: Differentiate θ with respect to time t

Find \( \frac{d\theta}{dt} \). Differentiating \( \tan \theta = \frac{y}{x - 2} \) with respect to time \( t \) gives:\[ \sec^2\theta \frac{d\theta}{dt} = \frac{(x-2)\frac{dy}{dt} - y\frac{dx}{dt}}{(x-2)^2} \]Plugging in \( x = 3, y = \sqrt{3}, \frac{dx}{dt} = 4, \frac{dy}{dt} = \frac{2\sqrt{3}}{3} \):\[ \sec^2\left(\frac{\pi}{3}\right) \frac{d\theta}{dt} = \frac{1\left(\frac{2\sqrt{3}}{3}\right) - \sqrt{3}(4)}{1} \]\[ \frac{d\theta}{dt} = \frac{1}{4}(-\frac{10\sqrt{3}}{3}) = -\frac{5\sqrt{3}}{6} \text{ radians/s} . \]

Step 7: Conclusion

The distance between point \( P \) and \( (2,0) \) is increasing at 3 units/s. The angle \( \theta \) is decreasing at a rate of \(-\frac{5\sqrt{3}}{6}\) radians/s.

Key Concepts

Differentiation

Differentiation is a key concept of calculus and is used to study how functions change. In this context, the function represents a moving point's position along a curve. It helps us understand how various elements of a problem vary over time.
At its core, differentiation involves computing a derivative, which indicates the rate of change of a function with respect to a variable. For example, consider the function describing how the point

• The function for the curve is given as \( y = \sqrt{x} \).
• Differentiating this with respect to \( x \), we have \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).

This derivative tells us how the \( y \)-coordinate changes for small changes in \( x \). Moreover, we differentiate the distance \( R \) and the angle of inclination \( \theta \) with respect to time \( t \). By using these techniques:

• We find how fast these quantities change as \( x \) changes over time.
• Differentiation is crucial to solve related rates problems where we explore relationships between varying quantities.

Instantaneous Rate of Change

The instantaneous rate of change provides an extremely valuable perspective into understanding how a dynamic system's state evolves at a specific moment. This is akin to determining the speed of a traveling car at a precise point in time.
In this exercise, we calculate the instantaneous rate of change for two measurements:

• The rate of change of the distance between the moving point \( P \) and the static point \( (2, 0) \).
• The rate of change of the angle between the line from \( P \) to \( (2, 0) \) and the horizontal axis.

To compute these rates:

• Given a constant rate of change for \( x \), \( \frac{dx}{dt} = 4 \) units/s, we derive how other components adjust by using derivatives.
• For instance, the distance \( R \) changes at \( 3 \) units/s, while the angle \( \theta \) changes at a rate of \(-\frac{5\sqrt{3}}{6}\) radians/s.

Instantaneous rates give insights into how rapid or slow changes occur for various aspects tied to a function or scenario.

Distance Formula

The distance formula is a mathematics staple, crucial for determining the straight-line distance separating two points in a plane. Its utility isn't just limited to geometry; it's fundamental in related rates problems, where you need to know how this distance changes as time progresses.
The formula itself is expressed as:

• \( R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) defines the distance \( R \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \).

In this exercise, we are assessing how this distance from point \( P \) to \( (2, 0) \) fluctuates. The distance changes because \( P \)'s position on the curve \( y = \sqrt{x} \) shifts over time as \( x \) increases.
Substituting the coordinates of \( P \) and \( (2,0) \), the formula calculated the distance between these points as \( 2 \). By differentiating this distance with respect to time, we identify the rate of change, leading us to conclude:

• This distance grows at a consistent rate of \( 3 \) units per second at the specific instant examined.

Understanding the distance formula and its differentiation provides a clearer picture of dynamic systems in continuous motion.