Ugur's copy of Parametric representations of a curve - find tangent and length

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Created this as a copy of Parametric representations of a curve - find tangent and length.

Name	Status	Author	Last Modified
Parametric representations of a curve - find tangent and length	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Ugur's copy of Parametric representations of a curve - find tangent and length	Ready to use	Ugur Efem	24/05/2024 11:29

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Question statement

The space curve $\mathcal C$ is parametrically given by $\boldsymbol{r} = \pmatrix{\var{a}\cos(\simplify{{b}t}),\var{-a}\sin(\simplify{{b}t})}$ , where $t_1\pi\leqslant t\leqslant t_2\pi$ .

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			Name	Type	Generated Value

a	integer	9
b	integer	9
t2	integer	4

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Parts

Part a) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Part b) Gap-fill
1. Gap Mathematical expression
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Part c) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Find the tangent vector $\boldsymbol{u}$ to the curve.

$\boldsymbol{u}=($ Gap 0. $,$ Gap 1. $)$ . (Do not enter decimals in your answers.)

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a) The tangent vector to the curve is given by $\boldsymbol{u} = \frac{\mathrm{d}r}{\mathrm{d}t} \equiv\pmatrix{\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}} = \pmatrix{\simplify{{-a*b}sin({b}*t)}, \simplify{{-a*b}*cos({b}*t)}}$ (note that, since the curve is not parametrised by arc length, $\frac{\mathrm{d}r}{\mathrm{d}t}$ does not give the unit tangent!).

b) The length $s$ of the curve in the range $t_1\pi\leqslant t\leqslant t_2\pi$ is given by

$s = \int_{t_1\pi}^{t_2\pi}{u\mathrm{d}t},$

where $u^2=\lvert\boldsymbol{u}\rvert^2=\boldsymbol{u\cdot u}$ .

In this question, therefore, $\boldsymbol{u}=\pmatrix{\frac{\mathrm{d}}{\mathrm{d}t}\left\{\var{a}\cos(\simplify{{b}*t})\right\},\frac{\mathrm{d}}{\mathrm{d}t}\left\{\var{-a}\sin(\simplify{{b}*t})\right\}}=\pmatrix{\var{-a*b}\sin(\simplify{{b}*t}),\var{-a*b}\cos(\simplify{{b}*t})}$ , and so $u^2=\var{(a*b)^2}$ .

Then

$s=\int_{t_1\pi}^{t_2\pi}{u\mathrm{d}t}=\var{a*b}\int_{t_1\pi}^{t_2\pi}{\mathrm{d}t}=\var{a*b}(t_2-t_1)\pi.$

Finally, substitute $t_1=-\simplify{1/{t2}}$ and $t_2=\simplify{1/{t2}}$ into this expression for $s$ , to find the length of the curve over the given range of $t$ .

Hence $s=\simplify{{2*a*b}/{t2}}\pi$ .

c) An alternative parametric representation, using $s$ as the curve parameter is given by

$s=\int_{t_1\pi}^{t}{u\mathrm{d}\tau}=\var{a*b}\int_{t_1\pi}^{t}{\mathrm{d}\tau}=\var{a*b}(t-t_1\pi).$

Now rearrange this expression for $t(s)$ , so

$t(s)=\frac{s}{\var{a*b}}+t_1\pi,$

and substitute into the original representation of the curve $\mathcal C\equiv\pmatrix{\var{a}\cos(\simplify{{b}t}),\var{-a}\sin(\simplify{{b}t})}$ with $t_1\pi\leqslant t\leqslant t_2\pi$ . Hence

$\mathcal C \equiv\pmatrix{\var{a}\cos\left(\simplify{s/{a}}+\var{b}t_1\pi\right),\var{-a}\sin\left(\simplify{s/{a}}+\var{b}t_1\pi\right)},$

with $0\leqslant s\leqslant\var{a*b}(t_2-t_1)\pi$ .

Finally, substitute $t_1=-\simplify{1/{t2}}$ and $t_2=\simplify{1/{t2}}$ into the above expressions, to find the specific parametric representation corresponding to the given range of t:

$\mathcal C \equiv\pmatrix{\var{a}\cos\left(\simplify{s/{a}}-\simplify{{b*pi}/{t2}}\right),\var{-a}\sin\left(\simplify{s/{a}}-\simplify{{b*pi}/{t2}}\right)},$

with $0\leqslant s\leqslant \simplify{{2*a*b}/{t2}}\pi$ .

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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's copy of Parametric representations of a curve - find tangent and length

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Ugur's copy of Parametric representations of a curve - find tangent and length

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Taxonomy: mathcentre

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Taxonomy: Context

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Ugur Efem 1 year, 10 months ago

Ugur Efem 1 year, 10 months ago

Ugur Efem 1 year, 10 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`