Implicit Differentiation

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Kevin Bohan 5 years, 11 months ago

Created this as a copy of SFY0004 Implicit 2.

Name	Status	Author	Last Modified
SFY0004 Implicit 2	Ready to use	Bill Foster	01/06/2016 09:47
CATORCE DOS Derivacion implicita	draft	Marlon Arcila	08/09/2021 00:47
Implicit 2	draft	Leticija Dubickaite	07/06/2018 11:23
Implicit 2	draft	Leticija Dubickaite	05/07/2018 16:52
Nick's copy of Implicit 2	draft	Nick Walker	22/06/2018 00:48
Praneetha's copy of Nick's copy of Implicit 2	draft	Praneetha Singh	28/06/2018 05:50
Implicit Differentiation	Ready to use	Gareth Woods	25/01/2022 12:32
Aoife's copy of Implicit Differentiation	draft	Aoife O'Brien	21/09/2018 12:44
Simon's copy of SFY0004 Implicit 2	draft	Simon Thomas	11/03/2019 14:58
Implicit Differentiation	draft	Kevin Bohan	05/06/2019 14:44
Implicit Differentiation	draft	Kevin Bohan	05/06/2019 14:50

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Question statement

Given the following relation between $x$ and $y$

$\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\var{c}$
answer the following questions.

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			Name	Type	Generated Value

a	integer	-4
c	integer	7
b	integer	6
d	integer	2

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Description

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Generated value: `integer`

-4

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Parts

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1. Gap Mathematical expression
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1. Step Information only

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Using implicit differentiation find $\displaystyle \frac{dy}{dx}$ in terms of $x$ and $y$ .

Input your answer here:

$\displaystyle \frac{dy}{dx}=$ Unnamed gap

Input all numbers as integers not as decimals.

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Advice

a)

On differentiating both sides of the equation implicitly we get

$2x + \simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0$
Collecting terms in

$\displaystyle\frac{dy}{dx}$ and rearranging the equation we get

$( \simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\frac{dy}{dx} = \simplify[all,!collectNumbers]{{ -a} -2x-{d}y}$ and hence on further rearranging:

$\frac{dy}{dx} = \simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}$

b)

On putting $x=0$ in the relation we get:

$\simplify{y^2+{b}y={c}} \Rightarrow \simplify{y^2+{b}y-{c}=0 }\Rightarrow (y+\var{c})(y-1)=0$

Hence $a=-\var{c}$ and $b=1$ .

c)

First we find the tangent at the point $(0,-\var{c})$ .

We find using the formula we found for $\frac{dy}{dx}$ in part a) that the gradient at $(0,-\var{c})$ is:

$\frac{dy}{dx}=\frac{\simplify[all,!collectnumbers]{{-a}+{d*c}}}{\var{b}-\var{2*c}}=\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}$

As the tangent goes through the point $(0,\var{-c})$ i.e. at $x=0,\;\;y=-\var{c}$ we see that the equation of the tangent is:

$y=\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}x-\var{c}$

Next we find that the gradient at $(0,1)$ is:

$\frac{dy}{dx}=\frac{\simplify[all,!collectnumbers]{{-a}-{d}}}{\var{b}+2}=\simplify[all,fractionNumbers]{-{a+d}/{c+1}}$

As the tangent goes through the point $(0,1)$ i.e. at $x=0,\;\;y=1$ we see that the equation of the tangent is:

$y=\simplify[all,fractionNumbers]{-{a+d}/{c+1}}x+1$

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Kevin Bohan 5 years, 11 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

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Implicit Differentiation

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Kevin Bohan 5 years, 11 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

Generated value: `integer`