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View the curve $ \ (y-x)^2+2=xy-3 \ $ as a contour of $ \ f(x,y) \ $

Use $ \ \nabla f(3,2) \ $ to find a vector normal to the curve at $ \ (3,2)\ $

Answer:

Let $ \ f(x,y)=(y-x)^2-xy+5=0 \ $

Then,

$ \nabla f(x,y)=\left\langle f_x,f_y \right\rangle \ = \left\langle 2(x-y)-y,2(y-x)-x \right\rangle $

Therefore,

$ \nabla f(3,2)=\left\langle0,-5 \right\rangle \ $ , which is normal t the level curves but not on the curve $ \ f(x,y) \ $

How to find find the vector normal to the curve $ \ f(x,y) \ $ using $ \ \nabla f(3,2) \ $ ?

Help me doing this.

MAS
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  • You’ve already done it. – amd May 29 '18 at 00:12
  • what is the normal vector to the curve $ \ f(x,y) \ $ ? Because $ \ \nabla f(3,2) \ $ is not normal to $ \ f(x,y) \ $ but normal to the level curve as given in the question . – MAS May 29 '18 at 00:14
  • Remember that vectors really only have a direction and length, you can move them around as you please (you don't need to have the starting point on the curve). – Michael Burr May 29 '18 at 00:20
  • “View the curve ... as a contour of $f(x,y)$.” – amd May 29 '18 at 00:23
  • the answer is not correct which show incorrect. what would be the vector normal to $ f(x,y) \ $ at $ \ (3,2) \ $ ? – MAS May 29 '18 at 00:25
  • Your phrase “the curve $f(x,y)$” is nonsensical: $f(x,y)$ is not a curve; $f(x,y)=0$ is. You’ve already found a normal to this curve. What is it that you’re really asking? – amd May 29 '18 at 00:25
  • I have put the answer in the answer box but it shows incorrect. The question says we have to use $ \ \nabla f(3,2) \ $ to find the normal vector to the curve $ \ f(x,y)=0 \ $. – MAS May 29 '18 at 00:27
  • There is an infinite number of normal vectors to the curve at that point. You’ve correctly identified one. You now need to figure out what your course material means by the normal vector. Perhaps it’s insisting on a unit vector. – amd May 29 '18 at 01:40

1 Answers1

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Here is a graph of your curve, plotted by Maple.enter image description here

It shows clearly that the curve is horizontal at $(3,2)$, so the normal is vertical, so your answer is correct and the software marking it is wrong.

The only suggestion I could make is that any vertical vector is normal to the curve at this point, that is, any vector $(0,b)$ with $b\ne0$. Try a unit vector, $(0,1)$ or $(0,-1)$ and see if that gets marked correct.

David
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