Math Problem Statement

A particle is moving along the parabola y^2 = 4(x + 1). As it passes through the point (3, 16), its y-coordinate is increasing at the rate of 3 units per second. How fast is the x-coordinate changing at this instant?

Solution

We are given the equation of a parabola y2=4(x+1)y^2 = 4(x + 1), and we need to find how fast the xx-coordinate is changing at the given point when the yy-coordinate is changing at a rate of 3 units per second.

Step 1: Implicit differentiation

Start by differentiating the given equation with respect to time tt. The equation is:

y2=4(x+1)y^2 = 4(x + 1)

Differentiating both sides with respect to tt, using the chain rule:

2ydydt=4dxdt2y \frac{dy}{dt} = 4 \frac{dx}{dt}

Step 2: Solve for dxdt\frac{dx}{dt}

Rearranging the equation to solve for dxdt\frac{dx}{dt}:

dxdt=ydydt2\frac{dx}{dt} = \frac{y \frac{dy}{dt}}{2}

Step 3: Substitute known values

The particle passes through the point (3,16)(3, 16), so x=3x = 3 and y=16y = 16. We are also given that dydt=3\frac{dy}{dt} = 3 units per second.

Substitute these values into the equation:

dxdt=16×32=482=24\frac{dx}{dt} = \frac{16 \times 3}{2} = \frac{48}{2} = 24

Final Answer:

The rate at which the xx-coordinate is changing is dxdt=24\frac{dx}{dt} = 24 units per second.

Would you like more details or have any questions?

Here are 5 questions related to this problem for further practice:

  1. How would the rate of change of xx vary if dydt\frac{dy}{dt} were different?
  2. What is the geometric interpretation of the relationship between dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}?
  3. How does implicit differentiation apply to curves that are not functions?
  4. How can this method be applied to curves in 3D, such as surfaces?
  5. What happens if the point of interest on the curve is at the vertex of the parabola?

Tip: Always check that the point you are analyzing satisfies the given equation before performing differentiation.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Related Rates
Parabolas

Formulas

2y(dy/dt) = 4(dx/dt)
dx/dt = (y(dy/dt)) / 2

Theorems

Chain Rule of Differentiation

Suitable Grade Level

Grades 10-12