Ugur's copy of Parametric curve - find tangent and coordinates at given point,

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Created this as a copy of Parametric curve - find tangent and coordinates at given point, .

Name	Status	Author	Last Modified
Parametric curve - find tangent and coordinates at given point,	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Ugur's copy of Parametric curve - find tangent and coordinates at given point,	Ready to use	Ugur Efem	07/06/2023 16:49

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Question statement

You are given the following curve, $t\mapsto\pmatrix{\simplify{{a1}*cos({b1}t)},\simplify{{c1}*sin({d1}t)}}$ , defined with respect to the parameter $t$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

f1	number	-8
dxdte1	number	-48.901
dxdtf1	number	-41.4857517115
dydte1	number	-7.661
dydtf1	number	6.6737868841
a1	number	9
b1	number	6
y	number	0.576
x	number	3.818
c1	number	2
e1	number	-4
speed	number	42.019
d1	number	4

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Name

Data type

Value

xxxxxxxxxx
 
random(1..9)*sign(random(1,-1))

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `number`

-8

→ Used by:

dxdtf1
dydtf1

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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Part c) Gap-fill
1. Gap Number entry
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Gap-fill

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

Find the coordinates $\pmatrix{x,y}$ of the point corresponding to $t=\var{e1}$ .

$\pmatrix{x,y}=($ Gap 0. $,$ Gap 1. $)$ . (Enter your answers to 3d.p.)

Advice

To find the coordinates of the point corresponding to $t=\var{e1}$ , substitute $t=\var{e1}$ into the expression for the curve, i.e.

$\pmatrix{x,y}=\pmatrix{\simplify{{a1}*cos({b1*e1})},\simplify{{c1}*cos({d1*e1})}}=\pmatrix{\var{x},\var{y}}.$

Differentiate each component of the vector in part a) to find the tangent vector $\boldsymbol{u}$ , i.e.

$\boldsymbol{u}=\pmatrix{\frac{\mathrm{d}}{\mathrm{d}t}\left(\simplify{{a1}*cos({b1}*t)}\right),\frac{\mathrm{d}}{\mathrm{d}t}\left(\simplify{{c1}*sin({d1}*t)}\right)}=\pmatrix{\simplify{{-a1*b1}*sin({b1}*t)},\simplify{{c1*d1}*cos({d1}*t)}}.$

The tangent vector at $t=\var{e1}$ is found by substituting $t=\var{e1}$ into the tangent vector $\boldsymbol{u}$ , i.e.

$\boldsymbol{u}\vert_{t=\var{e1}}=\pmatrix{\simplify{{-a1*b1}*sin({b1*e1})},\simplify{{c1*d1}*cos({d1*e1})}}=\pmatrix{\var{dxdte1},\var{dydte1}}.$

The velocity $u$ is given by $u=\lvert\boldsymbol{u}\rvert=\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}$ . We must calculate the speed at $t=\var{f1}$ , however, therefore

$u\vert_{t=\var{f1}}=\sqrt{\left(\simplify{{-a1*b1}*sin({b1*f1})}\right)^2+\left(\simplify{{c1*d1}*cos({d1*f1})}\right)^2}=\sqrt{\var{dxdtf1}^2+\var{dydtf1}^2}=\var{speed} \; \text{to 3d.p.}$

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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's copy of Parametric curve - find tangent and coordinates at given point,

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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

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Ugur's copy of Parametric curve - find tangent and coordinates at given point,

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

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: number

Parts

Gap-fill

Generated value: `number`