Math Problem Statement

Koordinat dari suatu partikel yang bergerak di bidang-xy terdirensialkan dengan dx/dt = -1 m/s dan dy/dt = 2 m/s. Seberapa cepat jarak partikel dengan titik asal bertambah ketika partikel tersebut berada di titik (5,12)?

Solution

The problem asks how fast the distance of a particle from the origin is increasing. Let’s summarize and solve it step by step.

Problem summary:

  • A particle moves in the xyxy-plane with the following rates of change:
    • dxdt=1m/s\frac{dx}{dt} = -1 \, \text{m/s}
    • dydt=2m/s\frac{dy}{dt} = 2 \, \text{m/s}
  • We need to determine the rate at which the distance from the origin is changing when the particle is at the point (5,12)(5, 12).

Solution:

  1. Distance from origin formula: r=x2+y2r = \sqrt{x^2 + y^2} where rr is the distance from the origin.

  2. Differentiate rr with respect to tt (using the chain rule): drdt=xx2+y2dxdt+yx2+y2dydt\frac{dr}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}} \frac{dy}{dt}

  3. Substitute the values at the given point (x,y)=(5,12)(x, y) = (5, 12): r=52+122=25+144=169=13r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

  4. Use the given rates dxdt=1m/s\frac{dx}{dt} = -1 \, \text{m/s} and dydt=2m/s\frac{dy}{dt} = 2 \, \text{m/s}. Now substitute everything into the equation:

    drdt=513(1)+1213(2)\frac{dr}{dt} = \frac{5}{13}(-1) + \frac{12}{13}(2)

  5. Simplify the equation: drdt=513+2413=1913\frac{dr}{dt} = -\frac{5}{13} + \frac{24}{13} = \frac{19}{13}

  6. Final answer: drdt=1913m/s1.46m/s\frac{dr}{dt} = \frac{19}{13} \, \text{m/s} \approx 1.46 \, \text{m/s}


The rate at which the particle's distance from the origin is increasing is approximately 1.46 m/s.


Would you like further clarifications, or do you have any additional questions?

Here are five follow-up questions related to this topic:

  1. How would the answer change if dxdt\frac{dx}{dt} were positive?
  2. What does the sign of dxdt\frac{dx}{dt} or dydt\frac{dy}{dt} tell us about the particle's movement?
  3. How can we find the angle of the particle's velocity with respect to the x-axis?
  4. What happens to drdt\frac{dr}{dt} if the particle were at the origin?
  5. How would the equation change if we were calculating speed instead of the rate of change of the distance?

Tip: Using the chain rule correctly is essential for related rates problems like these. Make sure to carefully identify what quantities are changing with time.

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Math Problem Analysis

Mathematical Concepts

Calculus
Related Rates
Distance Formula

Formulas

Distance formula: r = sqrt(x^2 + y^2)
Rate of change formula: dr/dt = (x/sqrt(x^2 + y^2)) * dx/dt + (y/sqrt(x^2 + y^2)) * dy/dt

Theorems

Chain Rule for Derivatives

Suitable Grade Level

Undergraduate or Advanced High School Calculus