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Curvature Calculator

Calculate curvature step by step

The calculator will find the curvature of the given explicit, parametric, or vector-valued function at the given point, with steps shown.

Related calculators: Unit Binormal Vector Calculator, Torsion Calculator

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If you have an explicit function y=f(x)y = f{\left(x \right)}, enter it as xx, f(x)f{\left(x \right)}, 00. For example, the curvature of y=x2y = x^{2} can be found here.
Leave empty if you don't need the curvature at a specific point.

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Your Input

Find the curvature of r⃗(t)=⟨t,3t+1,t2−5⟩\mathbf{\vec{r}\left(t\right)} = \left\langle t, 3 t + 1, t^{2} - 5\right\rangle.

Solution

Find the derivative of r⃗(t)\mathbf{\vec{r}\left(t\right)}: r⃗′(t)=⟨1,3,2t⟩\mathbf{\vec{r}^{\prime}\left(t\right)} = \left\langle 1, 3, 2 t\right\rangle (for steps, see derivative calculator).

Find the magnitude of r⃗′(t)\mathbf{\vec{r}^{\prime}\left(t\right)}: ∣r⃗′(t)∣=4t2+10\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\right\rvert} = \sqrt{4 t^{2} + 10} (for steps, see magnitude calculator).

Find the derivative of r⃗′(t)\mathbf{\vec{r}^{\prime}\left(t\right)}: r⃗′′(t)=⟨0,0,2⟩\mathbf{\vec{r}^{\prime\prime}\left(t\right)} = \left\langle 0, 0, 2\right\rangle (for steps, see derivative calculator).

Find the cross product: r⃗′(t)×r⃗′′(t)=⟨6,−2,0⟩\mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)} = \left\langle 6, -2, 0\right\rangle (for steps, see cross product calculator).

Find the magnitude of r⃗′(t)×r⃗′′(t)\mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)}: ∣r⃗′(t)×r⃗′′(t)∣=210\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)}\right\rvert} = 2 \sqrt{10} (for steps, see magnitude calculator).

Finally, the curvature is κ(t)=∣r⃗′(t)×r⃗′′(t)∣∣r⃗′(t)∣3=5(2t2+5)32.\kappa\left(t\right) = \frac{\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)}\right\rvert}}{\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\right\rvert}^{3}} = \frac{\sqrt{5}}{\left(2 t^{2} + 5\right)^{\frac{3}{2}}}.

Answer

The curvature is κ(t)=5(2t2+5)32\kappa\left(t\right) = \frac{\sqrt{5}}{\left(2 t^{2} + 5\right)^{\frac{3}{2}}}A.