Question

A point describes a circle of radius \(200 \mathrm{ft}\). at the rate of \(20 \mathrm{ft}\). a seeond. How fast is its projection on a fixed diameter travelling when the distance of the point from the diameter is 100 ft. ?

Short answer

Rate: \(\frac{10}{\pi} \mathrm{ft/s}\).

Step-by-step solution

Visualize the problem and identify variables

Visualize the circle with its center at point O. Let P be the moving point on the circle, and D be its projection onto the fixed diameter. Let A be the point where the fixed diameter intercepts the circle, such that OA is the fixed diameter. Now consider the right triangle OPD with the right angle at point D. Let \(x\) be the distance OD, \(y\) be the distance PD, and \(r = 200\) ft be the radius of the circle. Given that the distance of the point from the diameter is 100 ft, we have \(y=100\) ft. The rate at which P is moving around the circle is given as 20 ft/s. We can denote this as \(\frac{\mathrm{d} \theta}{\mathrm{d} t}\), where \(\theta\) is the angle AOP. Since the circumference of the circle is \(2\pi r\), we can find the rate at which \(\theta\) is changing: \(\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{20}{2\pi(200)} = \frac{1}{20\pi}\) rad/s. Our goal is to find \(\frac{\mathrm{d}x}{\mathrm{d}t}\), the rate at which point D is traveling along the diameter OA.

Relate \(x\) and \(y\) using the Pythagorean theorem

In the right triangle OPD, we can use the Pythagorean theorem to relate \(x\) and \(y\): \(x^2 + y^2 = r^2\) Since \(y=100\) and \(r=200\), we can substitute these values into the equation: \(x^2 + 100^2 = 200^2\)

Differentiate the equation

Now differentiate the equation with respect to time \(t\): \(\frac{\mathrm{d}(x^2)}{\mathrm{d}t} + \frac{\mathrm{d}(100^2)}{\mathrm{d}t} = \frac{\mathrm{d}(200^2)}{\mathrm{d}t}\) Since 100 and 200 are constants: \(2x \cdot \frac{\mathrm{d}x}{\mathrm{d}t} = 0\)

Use angle information to relate \(x\) and \(\frac{\mathrm{d}x}{\mathrm{d}t}\)

In triangle AOP, the cosine of angle \(\theta\) can be expressed as: \(\cos(\theta) = \frac{x}{r}\) Using the chain rule, differentiate both sides with respect to time \(t\): \(-\sin(\theta) \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t} = -\frac{1}{r} \cdot \frac{\mathrm{d}x}{\mathrm{d}t}\) Solve for \(\frac{\mathrm{d}x}{\mathrm{d}t}\): \(\frac{\mathrm{d}x}{\mathrm{d}t} = r\cdot\sin(\theta) \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t}\) When y = 100, use triangle OPD to find the sine of the angle: \(\sin(\theta)=\frac{y}{r} = \frac{100}{200} =\frac{1}{2}\) Now we can plug in the values for \(\sin(\theta)\) and \(\frac{\mathrm{d}\theta}{\mathrm{d}t}\): \(\frac{\mathrm{d}x}{\mathrm{d}t} = 200\cdot\frac{1}{2} \cdot \frac{1}{20\pi}=\frac{10}{\pi} \mathrm{ft/s}\)

Conclude the solution

The projection of the point onto the fixed diameter of the circle is traveling at \(\frac{10}{\pi} \mathrm{ft/s}\) when the distance of the point from the diameter is 100 ft.

Key concepts

Circle Geometry

Circle geometry involves exploring properties and relationships that exist within a circle. In this exercise, we're dealing with a point tracing along the circumference of a circle with a fixed radius of 200 ft. The circle's radius is crucial since it acts as the circle's fingerprint, defining all distances from the center to any point on the circle's edge.
Understanding the movement of a point in a circle gives insight into various geometric principles.

• Radius: The distance from the center of the circle to any point on it, consistently 200 ft in this problem.
• Circumference: Given by the formula \(2\pi r\), which helps in calculating how quickly the angle \(\theta\) changes as the point moves.
• Projection on Diameter: The exercise involves projecting the point's movement onto a fixed line or diameter—another core aspect based on the circle's geometry.

Manipulating these properties allows determining how fast this projection travels along the diameter.

Trigonometry

Trigonometry connects the angles in a triangle to the lengths of its sides. In this exercise, we focus on the angle \(\theta\) between the center of the circle (point O), the moving point (point P), and the projection of P onto the diameter (point D).

• Angle and Arc Relation: The angle \(\theta\) correlates with the point's position as it circles around.
• Sine and Cosine Ratios: In triangle OPD, the sine of angle \(\theta\) is given by \(\sin( heta) = \frac{OP}{OD}\), useful for finding movement rates in the circle's context.

Using trigonometric functions, we bridge the gap between the rotation's angular speed and the linear speed along the circle's diameter.

Differentiation

Differentiation is a powerful tool used to calculate the rate of change. Here, we utilize it to find how fast the projection of the moving point across the circle's diameter travels.
The exercise involves determining \(\frac{dx}{dt}\) when the point's distance from the diameter is fixed at 100 ft, which ties the problem to a familiar real-world context.

• Angular Speed: Deriving \(\frac{d\theta}{dt}\) helps track angular motion, given as \(\frac{1}{20\pi}\) radians per second.
• Relating Rates: Using the chain rule, differentials \(\frac{dx}{dt}\) and \(\frac{d\theta}{dt}\) are linked together, reflecting how changes in angle induce changes in travel rates along the diameter.

Through differentiation, we transition from knowing the angular speed to learning about the linear speed using calculated derivatives.

Pythagorean Theorem

The Pythagorean Theorem is a cornerstone of geometry, establishing relationships between the sides of a right triangle. This exercise applies the theorem to triangle OPD, made up of:

• The hypotenuse OD, which equals the circle’s radius \(200\) ft.
• The vertical distance PD, involving the fixed 100 ft from the problem.
• The horizontal distance along the diameter, represented as side OD or \(x\).

According to the theorem:x^2 + 100^2 = 200^2. Differentiating this equation with respect to time harmonizes with other rate changes; thus, helping solve for \(\frac{dx}{dt}\). In linking these geometric principles, the Pythagorean Theorem relates directly to calculating rates in this dynamic setting.