Math Problem Statement

when finding an wuation that describes the relationship of rates of change of x and y, for (x^2)+3y=4, why do we multiply 2x by dx/dt

Solution

When differentiating an equation like x2+3y=4x^2 + 3y = 4 to find the relationship between the rates of change of xx and yy, we use implicit differentiation with respect to time tt. Here's why 2x2x is multiplied by dx/dtdx/dt:

1. Implicit Differentiation Explanation

  • The equation x2+3y=4x^2 + 3y = 4 involves two variables, xx and yy, that may be functions of time tt. Therefore, we differentiate both sides of the equation with respect to tt, treating xx and yy as functions of tt.

  • For the term x2x^2, we apply the chain rule: ddt(x2)=2xdxdt.\frac{d}{dt}(x^2) = 2x \cdot \frac{dx}{dt}. The reasoning is that x2x^2 depends on xx, and xx depends on tt. So, when we differentiate x2x^2, we not only compute the derivative 2x2x but also multiply by the rate of change of xx with respect to tt, which is dxdt\frac{dx}{dt}.

2. Why Multiply by dxdt\frac{dx}{dt}?

  • If xx were a constant, its derivative would be 00, and there would be no dx/dtdx/dt.
  • However, since xx changes with tt, the chain rule ensures that any derivative involving xx must account for how xx itself varies over time.

Step-by-Step Differentiation of x2+3y=4x^2 + 3y = 4:

  1. Differentiate x2x^2: ddt(x2)=2xdxdt.\frac{d}{dt}(x^2) = 2x \cdot \frac{dx}{dt}.

  2. Differentiate 3y3y (using the chain rule): ddt(3y)=3dydt.\frac{d}{dt}(3y) = 3 \cdot \frac{dy}{dt}.

  3. Differentiate the constant 44: ddt(4)=0.\frac{d}{dt}(4) = 0.

3. Combine Results:

2xdxdt+3dydt=0.2x \cdot \frac{dx}{dt} + 3 \cdot \frac{dy}{dt} = 0.

This equation relates dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}, the rates of change of xx and yy with respect to time.

Key Idea:

The multiplication by dxdt\frac{dx}{dt} comes from the chain rule, which accounts for the dependency of xx on tt.


Would you like to see further examples or clarification? Here are five related questions you might consider exploring:

  1. How does the chain rule apply to more complex functions of multiple variables?
  2. What is the physical significance of rates of change (dx/dtdx/dt and dy/dtdy/dt) in real-world contexts?
  3. How would this differentiation look if we had x3+y2=10x^3 + y^2 = 10?
  4. Can implicit differentiation help solve for specific dx/dtdx/dt or dy/dtdy/dt values?
  5. What happens if xx or yy is a constant during differentiation?

Tip: Always think of implicit differentiation as "tracking dependencies" between variables as they change over time!

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Rates of Change
Chain Rule

Formulas

d/dt(x^2) = 2x * dx/dt
d/dt(3y) = 3 * dy/dt
2x * dx/dt + 3 * dy/dt = 0

Theorems

Chain Rule of Differentiation

Suitable Grade Level

Grades 10-12