Finding the rate of change of x in the equation

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Homework Help Overview

The discussion revolves around finding the rate of change of \( x \) in the context of a given equation involving \( x \) and \( y \). The original poster presents two different approaches to differentiate the equation, leading to conflicting results.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to differentiate the rearranged equation and the original equation directly, leading to different values for \( \frac{dx}{dt} \). Some participants question the validity of the input values used in the differentiation process.

Discussion Status

Participants are exploring the implications of the original poster's calculations and questioning the consistency of the input values. There is a recognition of potential flaws in the problem setup, with some participants expressing concern about the validity of the given point in relation to the equation.

Contextual Notes

There is mention of the problem being part of an ISC mathematics exam, raising concerns about the quality of exam questions and their clarity. The discussion reflects on the challenges faced under exam pressure and the potential for flawed questions in educational assessments.

tellmesomething
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Homework Statement
##\frac{1}{y} +\frac{1}{x} = \frac{1}{12}##. Find the rate of change of x with respect to time atx=5,y=1, if dy/dt=-1
Relevant Equations
Ok
I rearranged the equation and got
$$12x+12y=xy$$
Differentiating this I get
$$12\frac{dx}{dt} + 12\frac{dy}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$$
Substituting the values I get
$$12\frac{dx}{dt} -12= -5+\frac{dx}{dt}$$
SO
$$\frac{dx}{dt}=\frac{7}{11}$$
But if I directly differentiate the given equation
##\frac{1}{y} +\frac{1}{x} = \frac{1}{12}##
I get
$$\frac{-1}{x²} \frac{dx}{dt}-\frac{1}{y²} \frac{dy}{dt}=0$$
And here if I substitute the values I get
dx/dt=25

Im not sure where I went wrong in the first method, I'm guessing the differentiation of d(xy)/dt is a little more complicated. Can someone help me out please?
 
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x=5 and y=1 would imply 1/x + 1/y > 1 > 1/12 … your given point is not on the given curve …

Inconsistent input leads to inconsistent output.
 
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Orodruin said:
x=5 and y=1 would imply 1/x + 1/y > 1 > 1/12 … your given point is not on the given curve …

Inconsistent input leads to inconsistent output.
Okay so if dy/dx doesn't exist then that would mean dy/dt also doesn't? At x=5 and y=1
 
tellmesomething said:
Okay so if dy/dx doesn't exist then that would mean dy/dt also doesn't? At x=5 and y=1
Is this a problem you made up yourself?
 
PeroK said:
Is this a problem you made up yourself?
No it came in isc year 2024
Screenshot_2025-03-02-13-42-19-055_com.whatsapp.jpg
 
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What is ISC?
 
PeroK said:
What is ISC?
It's an Indian education board. They certify you with 12th grade diploma if you pass their exam. This is from their mathematics paper last year, and tomorrow I'm taking this same exam, hopefully I won't get absurd questions like these :-( cause under pressure i will not be able to identify what's wrong even.
 
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tellmesomething said:
It's an Indian education board. They certify you with 12th grade diploma if you pass their exam. This is from their mathematics paper last year, and tomorrow I'm taking this same exam, hopefully I won't get absurd questions like these :-( cause under pressure i will not be able to identify what's wrong even.
Sadly, it seems, there are a lot of flawed questions in exam papers these days. You can see how someone who doesn't really understand what they are doing could construct a problem like this. It never occurs to them that the first equation in ##x## and ##y## must be satisfied by the point they specify.
 
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PeroK said:
Sadly, it seems, there are a lot of flawed questions in exam papers these days. You can see how someone who doesn't really understand what they are doing could construct a problem like this. It never occurs to them that the first equation in ##x## and ##y## must be satisfied by the point they specify.
It is what happens when someone takes last year’s exam question and change the parameters randomly to make a new one …
 

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