What is the unit tangent vector T(t)?
Type the components of a curve r(t) = (x(t), y(t), z(t)), and a value of t. The page gives the unit tangent vector T(t), its value at t = a, and r′(t).
- T(a)
- (4/sqrt(61), -6/sqrt(61), 3/sqrt(61))
The unit tangent vector at t = 1 is (4/sqrt(61), -6/sqrt(61), 3/sqrt(61)).
- T(t) =
- (6t + 2, -12t^2, 6)/sqrt(144t^4 + 36t^2 + 24t + 40)
- r′(t) =
- (6t + 2, -12t^2, 6)
- ‖r′(a)‖
- 15.62049935
T(a): (4/sqrt(61), -6/sqrt(61), 3/sqrt(61)). The unit tangent vector at t = 1 is (4/sqrt(61), -6/sqrt(61), 3/sqrt(61)).
How to calculate
Finds the unit tangent vector T(t) = r′(t)/‖r′(t)‖ of a curve r(t) in 2D or 3D, and its value at t = a.
Example with the default inputs (x(t) 3t^2 + 2t, y(t) 2 - 4t^3, z(t) (empty for a plane curve) 6t + 5, At t = a 1): The unit tangent vector at t = 1 is (4/sqrt(61), -6/sqrt(61), 3/sqrt(61)).
Method: T(t) = r′(t) / ‖r′(t)‖, each component of r′ from a computer algebra system, checked numerically.
- The variable is t; angles are in radians.
- T(a) needs r′(a) ≠ 0.
Worked examples
Each example is checked against the calculator on every build.
- x(t) 3t^2 + 2t, y(t) 2 - 4t^3, z(t) (empty for a plane curve) 6t + 5, At t = a 1 gives r′(t) = (6t + 2, -12t^2, 6), T(t) = (6t + 2, -12t^2, 6)/sqrt(144t^4 + 36t^2 + 24t + 40), T(a) (4/sqrt(61), -6/sqrt(61), 3/sqrt(61)), ‖r′(a)‖ 15.620499.Source: OpenStax, Calculus Volume 3, section 3.2 Calculus of Vector-Valued Functions, Example 3.7 (https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions), part b: u′(t) = (6t + 2, −12t², 6)
- x(t) cos(t), y(t) sin(t), At t = a pi/2 gives r′(t) = (-sin(t), cos(t)), T(t) = (-sin(t), cos(t)), T(a) (-1, 0), ‖r′(a)‖ 1.Source: OpenStax, Calculus Volume 3, section 3.2 Calculus of Vector-Valued Functions, Example 3.7 (https://openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions), part a: T(t) = −sin t i + cos t j
- x(t) t^2 - 3, y(t) 2t + 1, z(t) (empty for a plane curve) t - 2, At t = a 2 gives T(a) (4/sqrt(21), 2/sqrt(21), 1/sqrt(21)), ‖r′(a)‖ 4.582576.
How it works
For a curve r(t) = (x(t), y(t), z(t)), or (x(t), y(t)) when z(t) is empty, the page computes:
- r′(t) = (x′(t), y′(t), z′(t)), each derivative from a computer algebra system (nerdamer, open source) and checked against a numeric difference quotient at 18 points.
- ‖r′(t)‖² = x′² + y′² + z′², expanded and also simplified; the shorter form is shown (cos(t)² + sin(t)² becomes 1).
- T(t) = r′(t)/√(‖r′(t)‖²). When ‖r′(t)‖² is 1, T(t) is r′(t).
- ‖r′(a)‖, the speed at t = a: the square root of the sum of the squares of r′(a), computed after scaling by the largest component so no square overflows.
- T(a) = r′(a)/‖r′(a)‖. Each component is also simplified exactly from the typed a; the exact form shows when every component equals the decimal value to 10⁻⁹. Otherwise each component shows to 10 significant figures.
a may be a number or a constant such as pi/3. If r′(a) is the zero vector, or a component of r′ is not defined at a, there is no answer.
What you can type
- Each component is a function of t. pi (or π) is π and e is Euler’s number.
- Operations: + − * / and ^; numbers and t side by side multiply (3t^2).
- Functions: sqrt, ln, exp, abs, sin, cos, tan and their inverses. Angles are in radians.
Worked examples by hand
u(t) = (3t² + 2t, 2 − 4t³, 6t + 5) (OpenStax Calculus Volume 3, Example 3.7b). u′(t) = (6t + 2, −12t², 6). ‖u′(t)‖² = (6t + 2)² + 144t⁴ + 36 = 144t⁴ + 36t² + 24t + 40, so T(t) = (6t + 2, −12t², 6)/√(144t⁴ + 36t² + 24t + 40). OpenStax writes the same vector with 2 taken out of the top and the bottom: (3t + 1, −6t², 3)/√(36t⁴ + 9t² + 6t + 10). At t = 1: u′(1) = (8, −12, 6), ‖u′(1)‖ = √244 = 2√61, so T(1) = (4/√61, −6/√61, 3/√61).
r(t) = (cos t, sin t) (Example 3.7a). r′(t) = (−sin t, cos t), and sin²t + cos²t = 1, so T(t) = (−sin t, cos t). At t = π/2, T = (−1, 0).
r(t) = (t² − 3, 2t + 1, t − 2) at t = 2. r′(t) = (2t, 2, 1), so r′(2) = (4, 2, 1) and ‖r′(2)‖ = √(16 + 4 + 1) = √21. T(2) = (4/√21, 2/√21, 1/√21).
Other questions people ask
What is the unit tangent vector?
For a smooth curve r(t), the unit tangent vector is T(t) = r′(t)/‖r′(t)‖. It points the way the curve moves as t grows, and its length is 1. It exists wherever r′(t) is not the zero vector.
How do I find the unit tangent vector?
Differentiate each component to get r′(t). Find its length ‖r′(t)‖ = √(x′² + y′² + z′²). Divide each component of r′(t) by that length. For r(t) = (cos t, sin t), r′(t) = (−sin t, cos t) has length 1, so T(t) = (−sin t, cos t).
Why is T not defined at some points?
Where r′(t) = 0, the length is 0 and the division is not possible. The curve may have a sharp point there: r(t) = (t³, t²) has r′(0) = (0, 0) and a cusp at the origin. The page says so instead of giving a vector.
Does the unit tangent vector depend on the parametrization?
Its direction depends only on the direction of travel. A new parameter that runs the same way (t = 2s) gives the same T at each point; one that runs the other way (t = −s) gives −T.
How does T relate to curvature?
The curvature is κ = ‖T′(t)‖/‖r′(t)‖: how fast the unit tangent turns per unit of arc length. The principal unit normal is N(t) = T′(t)/‖T′(t)‖.
How do I enter a plane curve?
Leave z(t) empty. The page then works with r(t) = (x(t), y(t)), and the answers have two components.
How is the answer checked?
Each derivative comes from a computer algebra system and must match a numeric difference quotient. The exact value of T(a) is shown only when each of its components equals the decimal value to 10⁻⁹; otherwise the page shows decimals to 10 significant figures.