Problem 45 - Calculus Single Variable

Question

A particle moves along the curve \(y=\sqrt{x}\). At what point are the rates of change of the particle's coordinates equal?

Step-by-Step Solution

Answer

The rates of change are equal at the point \(\left(\frac{1}{4}, \frac{1}{2}\right)\).

Step 1: Understand the problem

You're asked to find the point on the curve \(y = \sqrt{x}\) where the rates of change of \(x\) and \(y\) are equal. This means we need to find \(x\) and \(y\) such that \(\frac{dy}{dx} = 1\).

Step 2: Differentiate the curve

Find the derivative of \(y\) with respect to \(x\). The function is \(y = \sqrt{x}\), which can be rewritten as \(y = x^{1/2}\). Use the power rule: \(\frac{dy}{dx} = \frac{1}{2}x^{-1/2}\).

Step 3: Set the derivative equal to 1

We set \(\frac{dy}{dx} = 1\) and solve for \(x\): \[ \frac{1}{2}x^{-1/2} = 1 \] Multiply both sides by \(2\) to get: \[ x^{-1/2} = 2 \].

Step 4: Solve for \(x\)

Rewrite \(x^{-1/2} = 2\) as \(x^{1/2} = \frac{1}{2}\). Square both sides to solve for \(x\):\[ x = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \].

Step 5: Find the corresponding \(y\) value

Substitute \(x = \frac{1}{4}\) back into the curve equation \(y = \sqrt{x}\): \[ y = \sqrt{\frac{1}{4}} = \frac{1}{2} \].

Step 6: Conclusion

The rates of change of the particle's coordinates are equal at the point \(\left( \frac{1}{4}, \frac{1}{2} \right)\).

Key Concepts

Understanding Derivatives

In calculus, a derivative represents how a function changes as its input changes; it is essentially a way to encode the rate of change in mathematical terms. For a given function, the derivative provides the slope of the tangent line to the function at any given point. This is crucial in analyzing how quantities evolve.
For example, in the exercise, the function is described by

• The curve equation: \(y = \sqrt{x}\)

To find how fast \(y\) changes with respect to \(x\), we find the derivative of \(y\) with respect to \(x\). Using the power rule, where differentiating \(x^{n}\) gives us \(nx^{n-1}\), we find the derivative:

• \(\frac{dy}{dx} = \frac{1}{2}x^{-1/2}\)

This result gives us a formula for calculating the rate of change of \(y\) at any point \(x\) along the curve.

Curve Sketching Insights

Curve sketching is a technique in calculus used to understand the shape and features of a graph without plotting many points. By using derivatives, you can easily find critical points such as maxima, minima, and points of inflection.
For the function \(y = \sqrt{x}\), understanding its derivative helps sketch the curve. The positive value of the derivative \(\frac{dy}{dx}\) indicates that the function is increasing.

• The curve starts from the origin (0,0), increasing gradually with \(x\).
• The derivative \(\frac{dy}{dx} = \frac{1}{2}x^{-1/2}\) suggests the slope becomes less steep as \(x\) increases.

This key insight tells us that as \(x\) moves further from zero, changes in \(y\) become less pronounced, leading to a curve that flattens out as \(x\) grows larger.

Rate of Change in Focus

Rate of change is a crucial concept in calculus, reflecting how one quantity changes in relation to another. In the context of the exercise, we're interested in finding where the rate of change of \(x\) with respect to \(y\) equals 1, implying both coordinates change at the same rate.

• We solve \(\frac{dy}{dx} = 1\) to find the point where the slopes are equal.
• Steps of solution: by equating \(\frac{1}{2}x^{-1/2} = 1\), solving gives \(x = \frac{1}{4}\).
• Substituting back: \(y = \sqrt{x} = \frac{1}{2}\).

Therefore, at point \(\left( \frac{1}{4}, \frac{1}{2} \right)\), the particle has equal rates of horizontal and vertical change, revealing an aspect of symmetry in the particle’s movement along the curve.