Solve the problem that involves implicit differentiation

[chwala]

Homework Statement

see attached

Relevant Equations

differentiation

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.

 

[pbuk]

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).

 

[chwala]

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.

 

[Mark44]

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.

* 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.

 

[chwala]

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...

 

[Mark44]

$\dfrac{dy}{dx} (x=1)$

A more commonly used notation would be $\left.\dfrac{dy}{dx}\right|_{x=1}$

 

[pbuk]

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?

 

[Mark44]

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"?

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.

 

[pbuk]

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"?

 

[chwala]

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 ...

 

[chwala]

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...

 

[pbuk]

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...

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

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

that tells you that the tangent to the curve is nowhere parallel to $y = x$.

 

[chwala]

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.

 

[Delta2]

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?

 

[pbuk]

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)?

 

[Mark44]

Ehm sorry why my post got deleted here? Was I revealing too much

It provided too much help.

 

[Delta2]

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.

 

[Mark44]

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.

 

[chwala]

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...

 

[Mark44]

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

Hint: The point on the curve (if it exists) where y = x is not relevant.

been slightly unwell bear with me...

Roger that... Get well soon.

 

[WWGD]

I will respond as to why that is relevant...been slightly unwell bear with me...

Stay away from bears, well or unwell ones!

 

[pbuk]

been slightly unwell bear with me...

I'm sorry to hear that, I hope it clears up soon

 

[chwala]

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

 

[chwala]

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.

 

[chwala]

I will respond as to why that is relevant...been slightly unwell bear with me...

You were right! my assertion was not relevant.