91Ó°ÊÓ

Implicit differentiation is a technique used in calculus when we have an equation that defines a relationship between two variables, like \( x \) and \( y \), but is not explicitly solved for one variable in terms of the other. It allows us to find the derivative of a dependent variable with respect to an independent variable even if the equation isn't in the standard form \( y=f(x) \). Instead, we differentiate each term separately, applying the chain rule as necessary.
For the equation \( x^2 + y^2 = 25 \), we treat each variable as dependent on time \( t \). This means we apply implicit differentiation to find how changes in \( x \) and \( y \) with respect to time \( t \) relate to each other. By differentiating \( x^2 \) and \( y^2 \) with respect to \( t \), we apply the chain rule:

• Differentiate \( x^2 \) to get \( 2x \frac{dx}{dt} \).
• Differentiate \( y^2 \) to get \( 2y \frac{dy}{dt} \).
• Set the sum equal to zero because the derivative of a constant (25) is zero.

This is how we deal with problems where variables are related but not given as functions of one another directly.

Differentiation with respect to time is essentially about finding the rate at which a variable changes over time. When dealing with physical problems or any scenario where quantities vary over time, the rate of change is usually with respect to time.
In this specific problem, both \( x \) and \( y \) are functions of time \( t \), meaning they change as time progresses. The rate of change of \( x \) with respect to time is given as \( \frac{dx}{dt} = -2 \).
This negative value tells us that \( x \) is decreasing over time.

• We differentiate each expression in the original equation \( x^2 + y^2 = 25 \) with respect to \( t \).
• Using implicit differentiation allows us to calculate \( \frac{dy}{dt} \), which is the rate of change of \( y \) over time.
• Once we differentiate, we can substitute known values, like \( x = 3 \) and \( y = -4 \), to solve for \( \frac{dy}{dt} \).

Understanding how these rates interact is key to solving problems involving steady and changing quantities.

Solving differential equations involves finding a function that satisfies the given equation. In problems like these, we often have an equation with derivatives, and our goal is to find the unknown rate of change, such as \( \frac{dy}{dt} \). This involves algebraic manipulation after we've set up our differentiated equation.
Through the differentiation process, we've derived the equation: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] To find the value of \( \frac{dy}{dt} \), we did the following steps:

• Plug in the known values into the differentiated equation. For instance, when \( x = 3 \) and \( y = -4 \).
• Simplify the equation to isolate \( \frac{dy}{dt} \). This requires some basic algebraic techniques, like subtraction and division.
• For this example, it led to the solution: \( \frac{dy}{dt} = -\frac{3}{2} \), indicating how fast \( y \) changes as \( x \) and \( y \) move along the derived geometric path, reflecting their continuous relationship at that specific instance.

These processes illuminate not only the individual derivatives but also quantify the associated rates of change in real-time scenarios.