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Δs_{1} | + | Δs_{2} | + | Δs_{3} | + ... = | Δs_{1} + Δs_{2} + Δs_{3} + ... |
Δx | Δx | Δx | Δx |
∑ n |
Δs_{n} Δx |
Δx = Δs_{n} |
(Δs)^{2} | ≈ 1 + | (Δy)^{2} | |
(Δx)^{2} | (Δx)^{2} | so now I can take the square root of both sides |
Δs | ≈ | √( | 1 + | (Δy)^{2} | ) |
Δx | (Δx)^{2} |
Δs Δx |
≈ √(1 + (Δy / Δx)^{2}) |
Line length between a and b = | ∫ | b | √(1 + (dy / dx)^{2}) dx |
a |
lim | ∑ y_{i}•Δx = ∫ y dx |
Δx → 0 |
Δs Δx |
≈ √(1 + (Δy / Δx)^{2}) |
ds dx |
≈ √(1 + (dy / dx)^{2}) |
Line length between a and b = | ∫ | b | √(1 + (dy / dx)^{2}) dx |
a |
Line length = | ∫ | b | √(1 + (dy / dx)^{2}) dx |
a |
Circumference = 2 | ∫ | r | √(1 + (dy / dx)^{2}) dx |
-r |
and if b(x) = √a(x), then the gradient is b'(x) = | -1 | = | -1 |
2√(a(x)) | 2√(r^{2} - x^{2}) |
dy | = -2x | -1 | = | x |
dx | 2√(r^{2} - x^{2}) | √(r^{2} - x^{2}) |
Circumference = 2 | ∫ | r | √(1 + (dy / dx)^{2}) dx |
-r |
x^{2} |
r^{2} - x^{2} |
1 + (dy / dx)^{2} = | r^{2} |
r^{2} - x^{2} |
Circumference = 2r | ∫ | r | 1 | dx = 2r(arcsin (1) - arcsin(-1) ) |
-r | r^{2} - x^{2} |
Have you found an error or do you want to add more
information to these pages? You can contact me at the bottom of the home page. |