Solve the problem that involves implicit differentiation

In summary: I think the problem is simple...find when the tangent to the curve is parallel to ##y=x##. This is so when their gradients are the same, that is, when ##\dfrac{dy}{dx}=\dfrac{-x^2-y}{x-y}=\dfrac{dy}{dx} (x=1)=\dfrac{-1-1}{1-1}=\dfrac{-2}{0}## does not exist...this is not dependent on any point of the curve...hope that is clear. It provided too much...I think the problem is simple...find when the tangent to the curve is parallel to ##y=x##. This is so when their gradients are the same, that is
  • #1
chwala
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Homework Statement
see attached
Relevant Equations
differentiation
1655822204671.png
My take;
##6x^2+6y+6x\dfrac{dy}{dx}-6y\dfrac{dy}{dx}=0##
##\dfrac{dy}{dx}=\dfrac{-6x^2-6y}{6x-6y}##
##\dfrac{dy}{dx}=\dfrac{-x^2-y}{x-y}##

Now considering the line ##y=x##, for the curve to be parallel to this line then it means that their gradients are the same at the point##(1,1)##
##\dfrac{dy}{dx} (x=1)=\dfrac{-1-1}{1-1}=\dfrac{-2}{0}## does not exist...
i hope this is the correct approach unless i am missing something.

* I do not have the solution.
 
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  • #2
You need to show that the tangent is not parallel to ## y = x ## anywhere, not just at ## (1, 1) ## (edit: ##(1, 1)## is not even on the curve).
 
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  • #3
pbuk said:
You need to show that the tangent is not parallel to ## y = x ## anywhere, not just at ## (1, 1) ## (edit: ##(1, 1)## is not even on the curve).
True, i meant that the straight line has a gradient value of ##1##. At this value , the corresponding gradient ( where ##x=1##) at the curve is non existent.
 
  • #4
chwala said:
Now considering the line ##y=x##, for the curve to be parallel to this line then it means that their gradients are the same at the point##(1,1)##
##\dfrac{dy}{dx} (x=1)=\dfrac{-1-1}{1-1}=\dfrac{-2}{0}## does not exist...
The above is incorrect (as already pointed out). What this should say is that ##\frac{dy}{dx}## is equal to 1, not that this derivative should be evaluated at x = 1.
chwala said:
* I do not have the solution.
This shouldn't matter all that much. A problem that states "Show that such and such is true" already is giving you the answer. The most important part, assuming you reach the desired conclusion, is that you have not arithmetic or algebraic errors.
 
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  • #5
Mark44 said:
The above is incorrect (as already pointed out). What this should say is that ##\frac{dy}{dx}## is equal to 1, not that this derivative should be evaluated at x = 1.

This shouldn't matter all that much. A problem that states "Show that such and such is true" already is giving you the answer. The most important part, assuming you reach the desired conclusion, is that you have not arithmetic or algebraic errors.
My bad, I will amend that...true I should be using the comparison of gradient at a given point on curve and the gradient of the tangent...
 
  • #6
chwala said:
##\dfrac{dy}{dx} (x=1)##
A more commonly used notation would be ##\left.\dfrac{dy}{dx}\right|_{x=1}##
 
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  • #7
chwala said:
My bad, I will amend that...true I should be using the comparison of gradient at a given point on curve and the gradient of the tangent...
What is the difference between the "gradient at a given point on curve" and the "gradient of the tangent"? How are you getting on with your revised answer?
 
  • #8
chwala said:
I should be using the comparison of gradient at a given point on curve and the gradient of the tangent...

pbuk said:
What is the difference between the "gradient at a given point on curve" and the "gradient of the tangent"?
In the US, we use the phrase "slope of the tangent line" rather than "gradient," and reserve gradient for functions of two or more variables.
 
  • #9
Mark44 said:
In the US, we use the phrase "slope of the tangent line" rather than "gradient," and reserve gradient for functions of two or more variables.
In the US, what is the difference between the "slope at a given point on curve" and the "slope of the tangent"? :wink:
 
  • #10
pbuk said:
What is the difference between the "gradient at a given point on curve" and the "gradient of the tangent"? How are you getting on with your revised answer?
will respond later in the day @pbuk ...
 
  • #11
pbuk said:
What is the difference between the "gradient at a given point on curve" and the "gradient of the tangent"? How are you getting on with your revised answer?
They are the same...have same value...Consider a tangent at a point of the curve say ##(x,y)## then it follows that gradient of a curve at this point ##(x,y)## is the same as the gradient of the tangent line. I hope that's clear...this is basic/straightforward to me...maybe its the language that i am using that is confusing...
 
  • #12
chwala said:
it follows that gradient of a curve at this point ##(x,y)## is the same as the gradient of the tangent line. I hope that's clear...this is basic/straightforward to me...
Yes, this is basic for me too, so what did you mean by...
chwala said:
I should be using the comparison of gradient at a given point on curve and the gradient of the tangent...
... what is the point in "comparing" something with itself?

Anyway this is not really relevant, the point is that as far as anyone can tell, you still have not determined what it is about the expression
chwala said:
##\dfrac{dy}{dx}=\dfrac{-x^2-y}{x-y}##
that tells you that the tangent to the curve is nowhere parallel to ## y = x ##.
 
  • #13
pbuk said:
Yes, this is basic for me too, so what did you mean by...

... what is the point in "comparing" something with itself?

Anyway this is not really relevant, the point is that as far as anyone can tell, you still have not determined what it is about the expression

that tells you that the tangent to the curve is nowhere parallel to ## y = x ##.
Agreed, let me look at the problem later in the day...appreciated.
I will solve for ##f(x)=0## using fact ##y=x## then use discriminant to establish that...now that you had asked of the revision progress on the question. Talk later mate.
 
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  • #14
Ehm sorry why my post got deleted here? Was I revealing too much or my post was completely off/out so you deleted it as some sort of spam?
 
  • #15
chwala said:
Agreed, let me look at the problem later in the day...appreciated.
I will solve for ##f(x)=0## using fact ##y=x##
What makes you think that the point on the curve where ## y = x ## is relevant (if it even exists)?
 
  • #16
Delta2 said:
Ehm sorry why my post got deleted here? Was I revealing too much
It provided too much help.
 
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  • #17
Mark44 said:
It provided too much help.
Ok fine then, but I didn't receive any message telling me why the post got deleted, that's why I asked.
 
  • #18
Delta2 said:
Ok fine then, but I didn't receive any message telling me why the post got deleted, that's why I asked.
I thought I had included a message, but apparently I didn't do that.
 
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  • #19
pbuk said:
What makes you think that the point on the curve where ## y = x ## is relevant (if it even exists)?
I will respond as to why that is relevant...been slightly unwell bear with me...
 
  • #20
pbuk said:
What makes you think that the point on the curve where y=x is relevant (if it even exists)?

chwala said:
I will respond as to why that is relevant
Hint: The point on the curve (if it exists) where y = x is not relevant.
chwala said:
been slightly unwell bear with me...
Roger that... Get well soon.
 
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  • #21
chwala said:
I will respond as to why that is relevant...been slightly unwell bear with me...
Stay away from bears, well or unwell ones!
 
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  • #22
chwala said:
been slightly unwell bear with me...
I'm sorry to hear that, I hope it clears up soon 💐
 
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  • #23
pbuk said:
I'm sorry to hear that, I hope it clears up soon 💐
Yes am good...thanks guys. I shall get back to business in a few hours...then we can smack at each other :biggrin: :cool:
 
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  • #24
chwala said:
Homework Statement:: see attached
Relevant Equations:: differentiation

View attachment 303129My take;
##6x^2+6y+6x\dfrac{dy}{dx}-6y\dfrac{dy}{dx}=0##
##\dfrac{dy}{dx}=\dfrac{-6x^2-6y}{6x-6y}##
##\dfrac{dy}{dx}=\dfrac{-x^2-y}{x-y}##

Now considering the line ##y=x##, for the curve to be parallel to this line then it means that their gradients are the same at the point##(1,1)##
##\dfrac{dy}{dx} (x=1)=\dfrac{-1-1}{1-1}=\dfrac{-2}{0}## does not exist...
i hope this is the correct approach unless i am missing something.

* I do not have the solution.

##\dfrac{dy}{dx}=\dfrac{-x^2-y}{x-y}##

##\dfrac{-x^2-y}{x-y}=1##

##-x^2-y=x-y##
##-y+y=x+x^2##
##x(x+1)=0##
##x=0⇒ 3y^2=-2## but ##y^2≥0## for solution to exist, so no solution.
##x=-1⇒3y^2+6y+4=0## using discriminant, ##-12<0## hence no real roots.
 
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  • #25
chwala said:
I will respond as to why that is relevant...been slightly unwell bear with me...
You were right! my assertion was not relevant.
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly defined in terms of one variable. This means that the function may have multiple variables and cannot be easily solved for one variable in terms of the others.

2. When should I use implicit differentiation?

Implicit differentiation is often used when a function is difficult or impossible to solve for one variable, or when the function contains both dependent and independent variables. It can also be used when the function is defined implicitly, such as in an implicit equation.

3. How do I perform implicit differentiation?

To perform implicit differentiation, you will need to use the chain rule, product rule, and quotient rule to take the derivative of each term in the function. Then, you can solve for the derivative of the dependent variable in terms of the independent variable.

4. What are some common mistakes when using implicit differentiation?

One common mistake when using implicit differentiation is forgetting to apply the chain rule correctly. Another mistake is not isolating the derivative of the dependent variable and leaving it in terms of both the dependent and independent variables.

5. Can implicit differentiation be used to find higher order derivatives?

Yes, implicit differentiation can be used to find higher order derivatives by repeating the process of taking the derivative of each term in the function. However, the calculations can become more complex as the order of the derivative increases.

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