Problem 39
Question
A point \(P\) is moving along the line whose equation is \(y=2 x .\) How fast is the distance between \(P\) and the point (3,0) changing at the instant when \(P\) is at (3,6) if \(x\) is decreasing at the rate of 2 units/s at that instant?
Step-by-Step Solution
Verified Answer
The distance is decreasing at 4 units/s.
1Step 1: Identify Variables and Relationships
Let the coordinates of point \(P\) be \((x, y)\). Since \(P\) lies on the line \(y = 2x\), the coordinates are \((x, 2x)\). We are to find how the distance \(d\) from \(P\) to the point \((3,0)\) changes over time as \(x\) changes.
2Step 2: Express the Distance Function
The distance \(d\) between \(P(x, 2x)\) and \((3,0)\) is given by the distance formula:\[d = \sqrt{(x - 3)^2 + (2x - 0)^2}\]Simplifying gives:\[d = \sqrt{(x - 3)^2 + 4x^2} = \sqrt{x^2 - 6x + 9 + 4x^2} = \sqrt{5x^2 - 6x + 9}\]
3Step 3: Differentiate the Distance Function
We need to find \(\frac{dd}{dt}\), the rate of change of the distance. Apply the chain rule:\[\frac{dd}{dt} = \frac{d}{dx} \left(\sqrt{5x^2 - 6x + 9}\right) \cdot \frac{dx}{dt}\]First find \(\frac{d}{dx}\):\[\frac{d}{dx} (\sqrt{5x^2 - 6x + 9}) = \frac{1}{2\sqrt{5x^2 - 6x + 9}} \cdot (10x - 6)\]
4Step 4: Evaluate the Derivative at the Given Point
Substitute \(x = 3\) and \(\frac{dx}{dt} = -2\) (since \(x\) is decreasing at 2 units/s):\[\frac{dd}{dt} = \left(\frac{1}{2\sqrt{5(3)^2 - 6(3) + 9}} \cdot (10(3) - 6)\right) \cdot (-2)\]Calculate each part:- First compute inside the square root: \(5(9) - 18 + 9 = 45 - 18 + 9 = 36\), so \(\sqrt{36} = 6\)- The numerator is \(10 \times 3 - 6 = 30 - 6 = 24\)- So the derivative is \(\frac{24}{2 \times 6} \cdot (-2) = \frac{24}{12} \times (-2) = 2 \times (-2) = -4\)
Key Concepts
Distance FormulaChain RuleDerivatives
Distance Formula
The distance formula is an essential tool in mathematics, used to calculate the length of a line segment between two points in a coordinate system. When applied, the formula provides the Euclidean distance, representing the shortest path between the points.
For two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) is computed by:
In our problem, the point \(P(x, 2x)\) represents a moving point on the line \(y = 2x\), and its distance from a fixed point (3,0) is explored. By employing the distance formula, students can understand how changes in \(x\) affect overall distance.
For two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) is computed by:
- Subtracting the x-coordinates: \(x_2 - x_1\)
- Subtracting the y-coordinates: \(y_2 - y_1\)
- Plugging these differences into the formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
In our problem, the point \(P(x, 2x)\) represents a moving point on the line \(y = 2x\), and its distance from a fixed point (3,0) is explored. By employing the distance formula, students can understand how changes in \(x\) affect overall distance.
Chain Rule
In calculus, the chain rule is a powerful method for finding the derivative of a composite function. It allows us to differentiate when one function is nested inside another.
A function \(y = f(g(x))\) involves the chain rule if it can be expressed in terms of two functions where one is dependent on the other.
The chain rule is expressed as:
In the original exercise, the chain rule was crucial in calculating \(\frac{dd}{dt}\), where \(d\) is a function of \(x\), and \(x\) changes over time. By applying this rule, students enhance their problem-solving techniques, allowing for precise calculations of rates of change.
A function \(y = f(g(x))\) involves the chain rule if it can be expressed in terms of two functions where one is dependent on the other.
The chain rule is expressed as:
- Let \(u = g(x)\)
- Then \(y = f(u)\)
- So \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
In the original exercise, the chain rule was crucial in calculating \(\frac{dd}{dt}\), where \(d\) is a function of \(x\), and \(x\) changes over time. By applying this rule, students enhance their problem-solving techniques, allowing for precise calculations of rates of change.
Derivatives
Derivatives represent the rate of change of a function with respect to one of its variables. They are a fundamental concept in calculus, indicating how a function's value changes in response to variations in its input.
The derivative of a function \(f(x)\) with respect to \(x\) is often denoted as \(f'(x)\) or \(\frac{df}{dx}\). The derivative captures the slope of the function's graph at any given point.
Key aspects include:
The derivative of a function \(f(x)\) with respect to \(x\) is often denoted as \(f'(x)\) or \(\frac{df}{dx}\). The derivative captures the slope of the function's graph at any given point.
Key aspects include:
- Calculating the instantaneous rate of change
- Finding the slope of the tangent line at a certain point
- Applying the rules of differentiation such as product rule, quotient rule, and chain rule
Other exercises in this chapter
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