Question

The luminous intensity \(I\) candelas of a lamp at varying voltage \(V\) is given by: \(I=5 \times 10^{-4} \mathrm{~V}^{2}\). Determine the voltage at which the light is increasing at a rate of \(0.4\) candelas per volt.

Short answer

The voltage is 400 volts.

Step-by-step solution

Understand the Given Formula

We are given the luminous intensity equation as a function of voltage: \(I = 5 \times 10^{-4} V^2\). This equation models how the intensity \(I\) changes with the voltage \(V\).

Differentiate the Intensity Function

To find how quickly the intensity increases with voltage, we need to differentiate \(I = 5 \times 10^{-4} V^2\) with respect to \(V\). This gives us the rate of change of intensity with respect to voltage. Let's differentiate:\(\frac{dI}{dV} = \frac{d}{dV}(5 \times 10^{-4} V^2) = 2 \times 5 \times 10^{-4} V = 10^{-3} V.\)

Set the Derivative Equal to the Given Rate

We are asked to find the voltage where the rate of change of intensity is \(0.4\) candelas per volt. Set the derivative \(\frac{dI}{dV} = 10^{-3} V\) equal to \(0.4\): \(10^{-3} V = 0.4.\)

Solve for the Voltage

Solve the equation \(10^{-3} V = 0.4\) for \(V\). Divide both sides by \(10^{-3}\):\(V = \frac{0.4}{10^{-3}} = 400.\)Thus, the voltage at which the luminous intensity is increasing at a rate of \(0.4\) candelas per volt is \(400\) volts.

Key concepts

Differentiation

Differentiation is a mathematical process that helps us understand how a function changes. Specifically, it gives us a way to find the rate at which one quantity changes with respect to another. In this exercise, we need to differentiate the function of luminous intensity with respect to voltage. The function is given by:

• \(I = 5 \times 10^{-4} V^2\)

To find the rate at which intensity changes as voltage changes, we take the derivative of this function with respect to \(V\). Differentiating means calculating the slope of the tangent line to the curve at any point. It is like finding out how steep or flat the curve is at a specific moment.
When we differentiate \(I = 5 \times 10^{-4} V^2\), we get:

• \(\frac{dI}{dV} = 10^{-3} V\)

This derivative shows the rate of change of intensity per unit of voltage. Understanding how to perform differentiation is crucial for determining how processes affect one another, especially when they change continuously.

Voltage

Voltage is an essential concept in understanding how electrical components operate. In simple terms, voltage can be thought of as the "push" that makes electric charges move through a conductor. It is measured in volts and is the driving force behind the operation of any electrical device.
In the problem we are solving, voltage plays a key role because it affects the luminous intensity of the lamp. The relationship is described by the equation:

• \(I = 5 \times 10^{-4} V^2\)

This shows that the luminous intensity is proportional to the square of the voltage. This means even a small change in voltage can lead to a significant change in intensity. That's why knowing the exact voltage at which certain changes happen is important, as it directly impacts the performance of the lamp. The problem specifically asks for the voltage when the rate of intensity change is given, which is calculated using differentiation.

Rate of Change

The rate of change is a measure of how much a quantity changes as another quantity changes. In this exercise, we are concerned with how quickly luminous intensity increases as voltage changes. To determine this, we use the derivative of the intensity function:

• \(\frac{dI}{dV} = 10^{-3} V\)

This represents the rate of change of luminous intensity with respect to voltage.
The given problem asks us to find the voltage at which this rate of change is 0.4 candelas per volt. By setting the derivative equal to the desired rate (0.4 candelas/volt), we solve for the voltage, finding that:

• \(V = 400\) volts

Understanding rates of change is vital in many real-world applications, as it helps predict how systems will react under various conditions. It's particularly useful in fields that require precise control and optimization, such as engineering and physics.