Ugur's copy of Find points of intersection, tangents, and angles between parametric curves

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Created this as a copy of Find points of intersection, tangents, and angles between parametric curves.

Name	Status	Author	Last Modified
Find points of intersection, tangents, and angles between parametric curves	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Ugur's copy of Find points of intersection, tangents, and angles between parametric curves	Ready to use	Ugur Efem	07/06/2023 16:48

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Question statement

The pair of curves

$\begin{align}\mathcal{C}_1&:t\mapsto\pmatrix{\simplify{{a1}*t},\simplify{{b1}*t},\simplify{{c1}*t}},-\infty\leqslant t\leqslant\infty\\\mathcal{C}_2&:\tau\mapsto\pmatrix{\simplify{{d1}*tau},\simplify{{e1}*tau^2},\simplify{{f1}*tau^3}},-\infty\leqslant \tau\leqslant\infty\end{align}$

intersect at two distinct points $\boldsymbol{p}$ and $\boldsymbol{q}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

f1	integer	-3
phi	number	177.75
lenu	number	1184.3107700262
dotuw	number	-2094072
tau	rational	-14
e1	integer	5
dotuv	number	-8
a1	integer	-4
u	vector	vector(-4,140,1176)
t	integer	7
w	vector	vector(2,-140,-1764)
v	vector	vector(2,0,0)
lenw	number	1769.5479648769
lenv	number	2
d1	integer	2
q	vector	vector(-28,980,8232)
theta	number	90.19
c1	number	1176
b1	number	140

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Generated value: `integer`

-3

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Enter the least value of $t$ , and the corresponding value of $\tau$ , defining the first intersection point. Hence enter the values of the intersection point $\boldsymbol{p}$ for these values of $t$ and $\tau$ .

$t=$ Gap 0.; $\tau=$ Gap 1..

$\boldsymbol{p}=($ Gap 2. $,$ Gap 3. $,$ Gap 4. $)$ .

Advice

Note that in this advice, the full calculator display is used in the calculation of each step; any rounding is purely for display clarity.

The two curves $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect where

$\begin{align}\simplify{{a1}*t}&=\simplify{{d1}*tau}\tag{1},\\\simplify{{b1}t}&=\simplify{{e1}*tau^2},\tag{2}\\\simplify{{c1}*t}&=\simplify{{f1}*tau^3}.\tag{3}\end{align}$

From equation (1)

$\tau=\frac{\var{a1}}{\var{d1}}t=\simplify{{a1}/{d1}t},\tag{4}$

which we substitute into equation (2) to determine that

$\var{b1}t=\var{e1}\times\left(\simplify{{a1}/{d1}t}\right)^2=\simplify{{e1*a1^2}/{d1^2}t^2}.$

Then either $t=0$ or $t=\simplify{{b1*d1^2}/{e1*a1^2}}$ .

Substitute these two expressions into equation (4), then either $\tau=0$ (when $t=0$ ), or $\tau=\simplify{{b1*d1}/{e1*a1}}$ (when $t=\var{t}$ ).

(As a check, substitute these pairs of values into equation (3), to show that equality holds.)

To determine the intersection points $\boldsymbol{p}$ and $\boldsymbol{q}$ , substitute the values of $t$ and $\tau$ into either expression for the curves $\mathcal{C}_1$ and $\mathcal{C}_2$ .

The point $\boldsymbol{p}$ is given by the least value of $t$ , which is $t=0$ (and correspondingly $\tau=0$ ). The point $\boldsymbol{p}$ is therefore $\boldsymbol{p}=\pmatrix{0,0,0}$ .

The point $\boldsymbol{q}$ is given by the greatest value of $t$ , which is $t=\var{t}$ (and correspondingly $\tau=\var{tau}$ ). The point $\boldsymbol{q}$ is therefore $\boldsymbol{q}=\pmatrix{\var{a1}\times\var{t},\var{b1}\times\var{t},\var{c1}\times\var{t}}=\pmatrix{\var{q[0]},\var{q[1]},\var{q[2]}}$ .

In general, the tangent vector $\boldsymbol{u}$ , of a curve $t\rightarrow\pmatrix{x(t),y(t),z(t)}$ , is given by $\boldsymbol{u}\equiv\pmatrix{\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t},\frac{\mathrm{d}z}{\mathrm{d}t}}$ .

The tangent vector of the curve $\mathcal{C}_1$ is therefore given by $\boldsymbol{u}=\pmatrix{\var{u[0]},\var{u[1]},\var{u[2]}}$ , which is constant, and independent of $t$ .

The tangent vector of $\mathcal{C}_2$ is given by $\pmatrix{\var{d1},\var{2*e1}\tau,\var{3*f1}\tau^2}$ , so the tangent vector at the point $\boldsymbol{p}$ (where $\tau=0$ ) is given by $\boldsymbol{v}=\pmatrix{\var{v[0]},\var{v[1]},\var{v[2]}}$ .

In a similar way, the tangent vector of $\mathcal{C}_2$ at the point $\boldsymbol{q}$ (where $\tau=\var{tau}$ ) is given by $\boldsymbol{w}=\pmatrix{\var{w[0]},\var{w[1]},\var{w[2]}}$ .

The angle $\theta$ between any two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ can be calculated using

$\cos(\theta)=\frac{\boldsymbol{a\cdot b}}{\lvert\boldsymbol{a}\rvert\lvert\boldsymbol{b}\rvert},$

where $\lvert\boldsymbol{x}\rvert=\sqrt{x_1^2+x_2^2+x_3^2}$ is the length of the vector $\boldsymbol{x}$ .

The angle $\theta$ between the tangent vectors at the point $\boldsymbol{p}$ is the angle between the vectors $\boldsymbol{u}$ and $\boldsymbol{v}$ , so

$\cos(\theta)=\frac{(\var{u[0]}\times\var{v[0]})+(\var{u[1]}\times\var{v[1]})+(\var{u[2]}\times\var{v[2]})}{\sqrt{(\var{u[0]})^2+(\var{u[1]})^2+(\var{u[2]})^2}\sqrt{(\var{v[0]})^2+(\var{v[1]})^2+(\var{v[2]})^2}}=\frac{\var{dotuv}}{\var{precround(lenu,4)}\times\var{precround(lenv,4)}}=\var{precround(dotuv/(lenu*lenv),4)}\;\text{to 4d.p.}$

Then $\theta=\arccos(\var{precround(dotuv/(lenu*lenv),4)})=\var{theta}^\circ$ to 2d.p.

In an identical way, the angle $\phi$ between the tangent vectors at the point $\boldsymbol{q}$ is the angle between the vectors $\boldsymbol{u}$ and $\boldsymbol{w}$ , so $\phi=\var{phi}^\circ$ to 2d.p.

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This question is used in the following exam:

Vector Calculus by Ugur Efem in Ugur's workspace.

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Ugur Efem 1 year, 9 months ago

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Generated value: `integer`