// Numbas version: exam_results_page_options {"name": "Maths Support: Implicit differentiation", "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": false, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "duration": 0.0, "metadata": {"notes": "", "description": "

Implicit differentiation including finding tangents

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On differentiating both sides of the equation implicitly we get
\$2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) + {a} + {b} *Diff(y,x,1)} = 0\$
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\$(\\var{b} + 2y) \\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x}\$ and hence on further rearranging:
\$\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x) / ({b} + (2 * y))}\$

", "rulesets": {}, "parts": [{"prompt": "\n

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

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$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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Given the following relation between $x$ and $y$
\$\\simplify[all,!collectNumbers]{x^2+y^2+{a}x+{b}y}=\\var{c}\$

", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "-random(1..9)", "name": "a"}, "c": {"definition": "random(1..9)", "name": "c"}, "b": {"definition": "random(1..9)", "name": "b"}}, "metadata": {"notes": "\n \t\t \t\t \t\t

20/06/2012:

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Improved display using \\displaystyle where appropriate.

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Changed marks to 2.

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3/07/2012:

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Implicit differentiation.

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Given $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

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\n \t\t \t\t \n \t\t \n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SFY0004 Implicit 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "d"], "tags": ["Calculus", "Differentiation", "calculus", "derivative", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "equation of tangent", "first derivative using implicit differentiation", "gradient", "implicit differentiation", "implicit relation", "tangent at a point"], "preamble": {"css": "", "js": ""}, "advice": "

a)

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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:

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

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b)

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

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

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

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c)

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

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

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

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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:

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

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

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

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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:

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

", "rulesets": {"std": ["all", "fractionNumbers"]}, "parts": [{"stepsPenalty": 0, "prompt": "\n

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

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$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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Input all numbers as integers not as decimals.

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If you want more help click on Show steps - you will not lose any marks.

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Input all numbers as integers or as fractions, not as decimals.

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Note that we regard $y$ as a function of $x$.

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Hence we have (using the chain rule):

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$\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$

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And , using the product rule:

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$\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.

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Now differentiate both sides of the relation with respect to $x$.

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Find the two points $(0,a),\\;\\;(0,b),\\;\\;a \\lt b$ which lie on the curve given by the relation.

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$a=\\;$[[0]]

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$b=\\;$[[1]]

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(remember that $a \\lt b$).

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Hence find the equations of the tangents at the points $(0,a)$ and $(0,b)$.

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Input all numbers as integers or as fractions, not as decimals.

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Equation of tangent at $(0,a)$:

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Find the gradient of the tangent at $(0,a)$.

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Hence the equation of the tangent at $(0,a)$ is:

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$y = \\;$[[1]]

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Equation of tangent at $(0,b)$:

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Find the gradient of the tangent at $(0,b)$.

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Hence the equation of the tangent at $(0,b)$ is:

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$y = \\;$[[3]]

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Input as an integer or as a fraction, not as a decimal.

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Input all numbers as integers or as fractions, not as decimals.

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Given the following relation between $x$ and $y$
\$\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\$

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30/04/2013

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Created new question for SFY0004 out of 1041 CBA2_5.

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Implicit differentiation.

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Given $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

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Also find two points on the curve where $x=0$ and find the equation of the tangent at those points.

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