Find the equation of the tangent and normal to.

[Nemo1]

Hi Community,

I have this question.

View attachment 5059I start by creating the derivative,

\(\displaystyle \d{}{x}(x^2-xy+y^2) = \d{}{x}(3) \)

and solve for \(\displaystyle \d{}{x}(3) \) = 0 (Derivative of a constant is alway 0)

I understand how to solve the derivative of

\(\displaystyle \d{}{x}(x^2-xy+y^2)\)

and get \(\displaystyle 2x-y\)

In the worked example it says:

\(\displaystyle 2x-y-x\d{}{x}+2y\d{}{x}=0 \implies 2x-y=(x-2y)\d{y}{x}\)

Which becomes,

\(\displaystyle \d{y}{x}=\frac{2x-y}{x-2y}\)

I am lost at the previous step as I don't know where \(\displaystyle -x\d{y}{x}\) and the \(\displaystyle +2y\d{y}{x}\) come from?

I am wondering if I should be using the sum/differential rule \(\displaystyle (f \pm g)^{\prime}= f^{\prime} \pm g^{\prime}\)

I would like to fully understand the process for finding the derivative and how to plug it into solve for the tangent and normal to lines.

Many thanks for your time in advance.

Nemo

 

[kaliprasad]

Hi Community,

I have this question.

I start by creating the derivative,

\(\displaystyle \d{}{x}(x^2-xy+y^2) = \d{}{x}(3) \)

and solve for \(\displaystyle \d{}{x}(3) \) = 0 (Derivative of a constant is alway 0)

I understand how to solve the derivative of

\(\displaystyle \d{}{x}(x^2-xy+y^2)\)

and get \(\displaystyle 2x-y\)

In the worked example it says:

\(\displaystyle 2x-y-x\d{}{x}+2y\d{}{x}=0 \implies 2x-y=(x-2y)\d{y}{x}\)

Which becomes,

\(\displaystyle \d{y}{x}=\frac{2x-y}{x-2y}\)

I am lost at the previous step as I don't know where \(\displaystyle -x\d{y}{x}\) and the \(\displaystyle +2y\d{y}{x}\) come from?

I am wondering if I should be using the sum/differential rule \(\displaystyle (f \pm g)^{\prime}= f^{\prime} \pm g^{\prime}\)

I would like to fully understand the process for finding the derivative and how to plug it into solve for the tangent and normal to lines.

Many thanks for your time in advance.

Nemo

The reason is that you have taken y to be a constant but it is a function of x and so the product rule shall apply and the worked example is correct

 

[MarkFL]

Hello Nemo! :)

In your following step:

...I understand how to solve the derivative of

\(\displaystyle \d{}{x}(x^2-xy+y^2)\)

and get \(\displaystyle 2x-y\)...

you are treating $y$ as a constant, when in fact it is a function of $x$, so you need to differentiate term by term, and use the product, power and chain rules:

\(\displaystyle \d{}{x}\left(x^2-xy+y^2\right)=\d{}{x}\left(x^2\right)-\d{}{x}\left(xy\right)+\d{}{x}\left(y^2\right)\)

Let's look at each term individually:

\(\displaystyle \d{}{x}\left(x^2\right)=2x\)

\(\displaystyle \d{}{x}\left(xy\right)=x\d{y}{x}+y\)

\(\displaystyle \d{}{x}\left(y^2\right)=2y\d{y}{x}\)

And so we have:

\(\displaystyle \d{}{x}\left(x^2-xy+y^2\right)=2x-x\d{y}{x}-y+2y\d{y}{x}=0\)

Now we solve for \(\displaystyle \d{y}{x}\) by arranging all terms with this as a factor on the left, and everything else on the right:

\(\displaystyle -x\d{y}{x}+2y\d{y}{x}=y-2x\)

Factor:

\(\displaystyle \d{y}{x}(2y-x)=y-2x\)

And finally divide through by $2y-x$ to get:

\(\displaystyle \d{y}{x}=\frac{y-2x}{2y-x}\)

So, what must the slope of the tangent and normal lines be at the given point on the curve?

 

[Nemo1]

Light bulb moment! thanks Mark & kaliprasad,

So following on from \(\displaystyle \d{y}{x}=\frac{y-2x}{2y-x}\) in Marks reply.

Looks very familiar to the \(\displaystyle mx=\frac{y2-y1}{x2-x1}\)

I could then write:

\(\displaystyle mx=\frac{2-2(1)}{2(2)-1}\)

to get the slope which when calculated equals \(\displaystyle 0\) so I know my tangent line is horizontal at point \(\displaystyle (1,2)\)

Therefore the normal to is going to be \(\displaystyle 90\) degrees which is \(\displaystyle x=1\)

As per my graph below.

View attachment 5068

Please let me know if I am in the ball park and if there is a clearer way to prove it.

Cheers

Nemo.

 

[MarkFL]

What I would do is use the expression we obtained for \(\displaystyle \d{y}{x}\) and the given point to determine the slope of the tangent line:

\(\displaystyle \left.\d{y}{x}\right|_{(x,y)=(1,2)}=\frac{2-2(1)}{2(2)-1}=0\)

So, we know the tangent line is horizontal, and since it must pass through the point $(1,2)$, this line is:

\(\displaystyle y=2\)

Since a line normal to a horizontal line must be vertical, and it must pass through the given point, it will be:

\(\displaystyle x=1\)

If the tangent line had not been horizontal, then we we take the value of the slope, along with the given point and used the point-slope formula to determine the equation of the tangent line. Then to find the normal line, we would use the fact that the product of the slope of two normal lines is -1, and then again used the point-slope formula.

We kind of had it easy here because the tangent line is horizontal. :)