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Drone Flight Path Linearization (Robotics & Computer Engineering) Solution

Formulas & Definitions

Problem Overview

We linearize the quadcopter flight path y(x)=4xβˆ’x2y(x) = 4x - x^2y(x)=4xβˆ’x2 at waypoint x=3x = 3x=3 and evaluate local path estimation accuracy.


Step-by-Step Mathematical Solution

Step 1: Compute General Derivative yβ€²(x)y'(x)yβ€²(x)

Set up the difference quotient limit:

yβ€²(x)=lim⁑hβ†’0[4(x+h)βˆ’(x+h)2]βˆ’(4xβˆ’x2)hy'(x) = \lim_{h \to 0} \frac{[4(x+h) - (x+h)^2] - (4x - x^2)}{h}yβ€²(x)=hβ†’0lim​h[4(x+h)βˆ’(x+h)2]βˆ’(4xβˆ’x2)​
=lim⁑hβ†’04x+4hβˆ’(x2+2xh+h2)βˆ’4x+x2h= \lim_{h \to 0} \frac{4x + 4h - (x^2 + 2xh + h^2) - 4x + x^2}{h}=hβ†’0lim​h4x+4hβˆ’(x2+2xh+h2)βˆ’4x+x2​
=lim⁑hβ†’04hβˆ’2xhβˆ’h2h=lim⁑hβ†’0(4βˆ’2xβˆ’h)=4βˆ’2x= \lim_{h \to 0} \frac{4h - 2xh - h^2}{h} = \lim_{h \to 0} (4 - 2x - h) = 4 - 2x=hβ†’0lim​h4hβˆ’2xhβˆ’h2​=hβ†’0lim​(4βˆ’2xβˆ’h)=4βˆ’2x

Step 2: Evaluate Point of Tangency and Slope at x=3x = 3x=3

  • Altitude coordinate:
    y(3)=4(3)βˆ’(3)2=12βˆ’9=3Β mβ€…β€ŠβŸΉβ€…β€Š(3,3)y(3) = 4(3) - (3)^2 = 12 - 9 = 3\text{ m} \implies (3, 3)y(3)=4(3)βˆ’(3)2=12βˆ’9=3Β m⟹(3,3)
  • Instantaneous tangent slope:
    m=yβ€²(3)=4βˆ’2(3)=4βˆ’6=βˆ’2m = y'(3) = 4 - 2(3) = 4 - 6 = -2m=yβ€²(3)=4βˆ’2(3)=4βˆ’6=βˆ’2

Step 3: Construct the Tangent Line (Linearization L(x)L(x)L(x))

Using point-slope form yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1)yβˆ’y1​=m(xβˆ’x1​):

yβˆ’3=βˆ’2(xβˆ’3)β€…β€ŠβŸΉβ€…β€Šyβˆ’3=βˆ’2x+6y - 3 = -2(x - 3) \implies y - 3 = -2x + 6yβˆ’3=βˆ’2(xβˆ’3)⟹yβˆ’3=βˆ’2x+6
L(x)=βˆ’2x+9L(x) = -2x + 9L(x)=βˆ’2x+9

Step 4: Compare Linear Approximation with Exact Nonlinear Value at x=3.1x = 3.1x=3.1

  • Linear Tangent Approximation:
    L(3.1)=βˆ’2(3.1)+9=βˆ’6.2+9=2.80Β mL(3.1) = -2(3.1) + 9 = -6.2 + 9 = 2.80\text{ m}L(3.1)=βˆ’2(3.1)+9=βˆ’6.2+9=2.80Β m
  • Exact Trajectory Value:
    y(3.1)=4(3.1)βˆ’(3.1)2=12.4βˆ’9.61=2.79Β my(3.1) = 4(3.1) - (3.1)^2 = 12.4 - 9.61 = 2.79\text{ m}y(3.1)=4(3.1)βˆ’(3.1)2=12.4βˆ’9.61=2.79Β m
  • Approximation Error:
    Error=∣2.79βˆ’2.80∣=0.01Β m(β‰ˆ0.36%Β relativeΒ error)\text{Error} = |2.79 - 2.80| = 0.01\text{ m} \quad (\approx 0.36\% \text{ relative error})Error=∣2.79βˆ’2.80∣=0.01Β m(β‰ˆ0.36%Β relativeΒ error)

Engineering Conclusion

Within a narrow window near x=3x = 3x=3, the tangent line linearization L(x)=βˆ’2x+9L(x) = -2x + 9L(x)=βˆ’2x+9 predicts the vehicle position to within 1Β centimeter1\text{ centimeter}1Β centimeter of accuracy. In robotics and embedded flight control, substituting tangent line models for nonlinear curves allows low-power flight microprocessors to execute guidance loop corrections at over 500Β Hz500\text{ Hz}500Β Hz without computational latency.