Worked examples · Sheet 2

Tangent lines to parametric curves

Theory used: Tangent vector and tangent line Regular curve

Exercise 1

Write the equation of the tangent line to the curve

γ (t) = (t, t^{2}), t \in [- 2, 2],

at the point $γ (1)$ .

γ (1) = (1, 1), γ^{'} (t) = (1, 2 t), γ^{'} (1) = (1, 2)

r (s) = (1 + s, 1 + 2 s), s \in R

y - 1 = 2 (x - 1) ⟹ y = 2 x - 1

The point of tangency is $γ (1) = (1, 1)$ and the direction of the line is given by the tangent vector $γ^{'} (1) = (1, 2)$ , which is not zero, hence the tangent line is defined. The parametric equation is written by adding to the point the direction vector multiplied by the parameter $s$ . For the Cartesian form one observes that the direction $(1, 2)$ corresponds to the slope $2$ , therefore the line through $(1, 1)$ with slope $2$ has equation $y = 2 x - 1$ . The trace of the curve is the parabola $y = x^{2}$ and the result coincides with the tangent to the parabola computed with the derivative of $x^{2}$ at $x = 1$ .

Exercise 2

Write the equation of the tangent line to the curve

γ (t) = (e^{t}, e^{- t}), t \in [- 1, 1],

at the point $γ (0)$ .

γ (0) = (1, 1), γ^{'} (t) = (e^{t}, - e^{- t}), γ^{'} (0) = (1, - 1)

r (s) = (1 + s, 1 - s), s \in R

y - 1 = - (x - 1) ⟹ y = 2 - x

The trace of

γ

, an arc of the hyperbola

x y = 1

, and the tangent line

y = 2 - x

at the point

(1, 1)

The components are differentiated with the rules for the exponential: the derivative of $e^{- t}$ is $- e^{- t}$ by the chain rule. At the point $t = 0$ the tangent vector is $(1, - 1)$ , not zero, hence the tangent line exists and passes through $γ (0) = (1, 1)$ with that direction. The slope associated with the direction $(1, - 1)$ is $- 1$ , therefore the Cartesian form is $y = 2 - x$ . The trace of the curve lies on the hyperbola $x y = 1$ , since $e^{t} \cdot e^{- t} = 1$ , and the line found is the tangent to the hyperbola at the point $(1, 1)$ .

Exercise 3

Write the parametric equation of the tangent line to the helix

γ (t) = (cos t, sin t, t), t \in [0, 2 π],

at the point $γ (\frac{π}{2})$ .

γ (\frac{π}{2}) = (0, 1, \frac{π}{2})

γ^{'} (t) = (- sin t, cos t, 1), γ^{'} (\frac{π}{2}) = (- 1, 0, 1)

r (s) = (- s, 1, \frac{π}{2} + s), s \in R

The point of tangency is obtained by evaluating the parametrization at $t = \frac{π}{2}$ , where $cos \frac{π}{2} = 0$ and $sin \frac{π}{2} = 1$ . The tangent vector at the same instant is $(- 1, 0, 1)$ , not zero because the third component of $γ^{'}$ is always $1$ . In space the tangent line does not admit a single Cartesian equation, hence the answer stays in parametric form: each value of $s$ corresponds to the point obtained by starting from $(0, 1, \frac{π}{2})$ and moving by $s$ times the vector $(- 1, 0, 1)$ . The second coordinate of the line is constant, since the $y$ component of the tangent vector at that point is zero.