Equation of tangent to Parametric curve

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Simon Thomas

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Simon Thomas 5 years, 2 months ago

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Simon Thomas 5 years, 2 months ago

Created this as a copy of Parametric Differentiation.

Name	Status	Author	Last Modified
Parametric Differentiation	draft	Simon Thomas	12/02/2020 10:50
Equation of tangent to Parametric curve	draft	Simon Thomas	12/02/2020 10:51
Parametric 2nd derivative	draft	Simon Thomas	12/02/2020 10:52

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Question statement

A curve is defined by the parametric equations $x=\var{a}t+\frac{\var{b}}{t^2}, y = \var{a}t-\frac{\var{b}}{t^2}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
b	integer	2
t	integer	5
m	number	1.0325203252
c	number	-0.487804878
xdot	number	1.968
ydot	number	2.032
x	number	10.08
y	number	9.92

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Generated value: `integer`

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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Ask the student a question, and give any hints about how they should answer this part.

(i) At the point $P$ on the curve, $t=\var{t}$ . Find the co-ordinates of $P$ .

$P$ = (Gap 0.,Gap 1.)

(ii) Find the equation of the tangenet to the curve at point $P$ , in the form $y = mx+c$ .

$m$ = Gap 2. (give your answer as a fraction)

$c$ = Gap 3.

Advice

(i) To find $P$ we substitute $t=\var{t}$ into $x=\var{a}t+\frac{\var{b}}{t^2}, y = \var{a}t-\frac{\var{b}}{t^2}$

Hence $x=\var{a}\times\var{t}+\frac{\var{b}}{\var{t}^2} = \var{a*t} + \var{b/t^2}=\var{x}$

and $y=\var{a}\times\var{t}-\frac{\var{b}}{\var{t}^2} = \var{a*t} - \var{b/t^2}=\var{y}$

Hence $P = (\var{x},\var{y})$

(ii) To find the equation of the tangent, we need to find the gradient $m$ , which is simply the value of $dy/dx$ at the point $P$ . Since we are working with parametric equations, it is easiest to calculate $dy/dx=\frac{dy}{dt} / \frac{dx}{dt}$ , and evaluate these when $t = \var{t}$ , corresponding to the point $P$

Now $\frac{dy}{dt} = \var{a}-\frac{\var{b}}{t^3} = \var{a}-\frac{\var{b}}{\var{t}^3} =\var{ydot}$ at the point $P$

and $\frac{dx}{dt} = \var{a}+\frac{\var{b}}{t^3} = \var{a}+\frac{\var{b}}{\var{t}^3}= \var{xdot}$ at the point $P$

Hence $m=dy/dx = \frac{\var{ydot}}{\var{xdot}}=\simplify[all,fractionNumbers]{{m}}$

Now we can find $c$ by substituting $m=\simplify[all,fractionNumbers]{{m}}, x=\var{x}$ and $y=\var{y}$ into $y = mx+c$

Hence $c=y-mx=\var{y}-\simplify[all,fractionNumbers]{{m}}\times \var{x} = \simplify[all,fractionNumbers]{{c}}$

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This question is used in the following exam:

Parametric Differentiation by Simon Thomas in Maths support.

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Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

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Equation of tangent to Parametric curve

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`