Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 3 / 3A (Vector Calculus)

For a regular differentiable curve $\mathbf{x}(t)$, its speed is $ds/dt=|{\mathbf{x}}^{\prime }(t)|>0$, where $s$ is arc length. The oriented unit tangent vector is $\hat{\mathbf{T}}={\mathbf{x}}^{\prime }/|{\mathbf{x}}^{\prime }|$. The curvature of a space curve is $\kappa =|d\hat{\mathbf{T}}/ds|$. Equivalently, for a twice differentiable regular curve,For the helix, differentiation gives ${\mathbf{x}}^{\prime }=(-a\sin t,a\cos t,b)$ and speed $v=\sqrt{a^2+b^2}$. ConsequentlyIndeed $d\hat{\mathbf{T}}/dt=(-a\cos t,-a\sin t,0)/v$ has magnitude $|a|/v$, and division by $ds/dt=v$ gives the result. This requires $(a,b)\ne (0,0)$; the excluded constant curve has no well-defined unit tangent vector. For $a=0,b\ne 0$ the curve is a straight line with zero curvature of a space curve.

Write the surface as the level set $F(x,y,z)=x^2{y}^3-y+1-z=0$. The gradient is normal to a regular level surface because differentiating $F$ along any surface curve gives $\nabla F\cdot \mathbf{v}=0$ for its tangent vector. At the specified point,Thus a normal vector and a unit normal are $(2,2,-1)$ and $(2,2,-1)/3$, respectively. The tangent-plane displacement is perpendicular to that normal, givingReversing the normal's sign gives the same plane.