Math Problem Statement

If x = sec^2 30 and y = 1 - tan 30, then find dy/dx when y = 4.

Solution

The problem in the image is:

Problem (10): y=yx, then find ddx(x2)=0y' = \frac{y}{x}, \text{ then find } \frac{d}{dx}(x^2) = 0

I will interpret this as a differential equation problem and break it down.

  1. The differential equation provided is y=yxy' = \frac{y}{x}, which can also be written as: dydx=yx.\frac{dy}{dx} = \frac{y}{x}.

    This is a separable differential equation, which can be solved by separating the variables and integrating both sides.

  2. We rewrite it as: dyy=dxx.\frac{dy}{y} = \frac{dx}{x}.

  3. Now, integrate both sides: 1ydy=1xdx,\int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx, which gives: lny=lnx+C,\ln |y| = \ln |x| + C, where CC is the constant of integration.

  4. Exponentiating both sides to eliminate the logarithms: y=elnx+C=elnxeC=xeC.|y| = e^{\ln |x| + C} = e^{\ln |x|} e^C = |x| e^C.

    Let eC=C1e^C = C_1, so we have: y=C1x.y = C_1 x.

Thus, the general solution to the differential equation y=yxy' = \frac{y}{x} is: y=C1x,y = C_1 x, where C1C_1 is a constant.

Would you like more details or clarifications on any part of this process?

Here are five related questions to consider:

  1. How can we apply an initial condition to find the specific solution to this differential equation?
  2. What is the significance of separable differential equations in solving real-world problems?
  3. How do we interpret the constant of integration C1C_1 in the context of the solution?
  4. What are other types of first-order differential equations, and how do they differ from separable ones?
  5. How would the solution change if the differential equation were non-linear?

Tip: In solving separable differential equations, always check if variables can be separated and integrated before proceeding with other techniques like substitution.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Derivatives
Differentiation

Formulas

sec^2 θ = 1 + tan^2 θ
tan θ
dy/dx = (dy/dθ) / (dx/dθ)

Theorems

Chain Rule in Differentiation

Suitable Grade Level

Grades 11-12