Finding equation of tangent and normal lines

S

Sonarks

New member
Joined
Jul 10, 2019
Messages
5
13032

I am struggling to answer the above question could someone please show me how?

Thanks
 
I would begin by implicitly differentiating the equation with respect to \(x\)...what do you get?
 
Sonarks said:
hopefully (-2x+y)/(2y-x)
Click to expand...

Let's see:

[MATH]2x-\left(x\d{y}{x}+y\right)+2y\d{y}{x}=0[/MATH]
[MATH]\d{y}{x}=\frac{y-2x}{2y-x}\quad\checkmark[/MATH]
So, what must the slope of the tangent and normal lines be at the given point?
 
pka said:
When you say struggling what exactly do you mean?
As Mark says, can you find the the value of \(\displaystyle \frac{dy}{dx}\) at \(\displaystyle (1,2)~?\)
Click to expand...

If i am honest i have not done any applications of differentiation, i have simply learnt how to differentiate so i don't even really know what the question wants me to do so no i am unsure what the (1,2) have to do with the question. Perhaps i am attempting something too difficult for me or missing something blatantly obvious.
 
Have you considered evaluating the expression with the given values?
 
Last edited:
Sonarks said:
hopefully (-2x+y)/(2y-x)
Click to expand...
You posted as I was typing. Have a good long look at this plot.
Mark told you that \(\displaystyle \frac{dy}{dx}=\frac{y-2x}{2y-x}\) is correct.
Can you evaluate \(\displaystyle \frac{dy}{dx}(1,2)~?\). Then look at that graph again. If your evaluation correct?
Post & we will help you.
 
f i am honest i have not done any applications of differentiation, i have simply learnt how to differentiate so i don't even really know what the question wants me to do so no i am unsure what the (1,2) have to do with the question. Perhaps i am attempting something too difficult for me or missing something blatantly obvious.
Click to expand...
When you learned to differentiate did you not learn what the derivative is? Typically, one of the first things a person learns aboiut the derivative, even before how to differentiate, is that the derivative is the "slope of the tangent line". Once you have determined that the derivative is (y- 2x)/(x- 2y) then you know that the derivative at (2, 4), is (4- 4)/(2- 8)= 0. The tangent line has slope 0 so is horizontal, y= 4. And the "normal" is perpendicular to the tangent line so is x= 2.
 
You must log in or register to reply here.
Top