Find points of intersection, tangents, and angles between parametric curves

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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

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

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

Give any introductory information the student needs.

			Name	Type	Generated Value

f1	integer	-4
phi	number	167.23
lenu	number	37.148351242
dotuw	number	-1842
tau	rational	-2
e1	integer	4
dotuv	number	-50
a1	integer	10
u	vector	vector(10,16,32)
t	integer	1
w	vector	vector(-5,-16,-48)
v	vector	vector(-5,0,0)
lenw	number	50.8428952755
lenv	number	5
d1	integer	-5
q	vector	vector(10,16,32)
theta	number	105.62
c1	number	32
b1	number	16

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Generated value: `integer`

-4

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Ask the student a question, and give any hints about how they should answer this part.

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.

Give a worked solution to the whole question.

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Find points of intersection, tangents, and angles between parametric curves

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Bill Foster 6 years, 6 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Find points of intersection, tangents, and angles between parametric curves

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Taxonomy: mathcentre

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Contributors

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Bill Foster 6 years, 6 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`