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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 $\\tau$.

$t=$ [[0]]; $\\tau=$ [[1]].

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

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Enter the greatest value of $t$, and the corresponding value of $\\tau$, defining the second intersection point. Hence enter the values of the intersection point $\\boldsymbol{q}$ for these values of $t$ and $\\tau$.

$t=$ [[0]]; $\\tau=$ [[1]].

$\\boldsymbol{q}=($[[2]]$,$[[3]]$,$[[4]]$)$.

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Find the tangent vector $\\boldsymbol{u}$ of the curve $\\mathcal{C}_1$.

$\\boldsymbol{u}=($[[0]]$,$[[1]]$,$[[2]]$)$.

Find the tangent vector $\\boldsymbol{v}$ of the curve $\\mathcal{C}_2$ at the point $\\boldsymbol{p}$.

$\\boldsymbol{v}=($[[0]]$,$[[1]]$,$[[2]]$)$.

Find the tangent vector $\\boldsymbol{w}$ of the curve $\\mathcal{C}_2$ at the point $\\boldsymbol{q}$.

$\\boldsymbol{w}=($[[0]]$,$[[1]]$,$[[2]]$)$.

Calculate the angle $\\theta$ (in degrees) between the tangent vectors of each curve, at the point $\\boldsymbol{p}$.

$\\theta=$ [[0]]$^\\circ$. (Enter your answer to 2d.p.)

Calculate the angle $\\phi$ (in degrees) between the tangent vectors of each curve, at the point $\\boldsymbol{q}$.

$\\phi=$ [[0]]$^\\circ$. (Enter your answer to 2d.p.)

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 $\\boldsymbol{q}$.

", "tags": ["checked2015", "intersection of curves", "parametric curves", "tangent vectors"], "rulesets": {}, "extensions": [], "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Intersection points, tangent vectors, angles between pairs of curves, given in parametric form.

"}, "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.

", "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}