Finding the Gradient of a Difficult Curve: Tips and Hints

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To find the gradient of the curve defined by the equation x² + xy + y² = 3 at the point (-1, k), implicit differentiation is recommended. For part (b), the tangent is parallel to the x-axis, indicating that the gradient must equal zero at that point. Participants express confusion about how to approach the problem and seek hints rather than complete solutions. The discussion emphasizes the importance of understanding implicit differentiation to solve for the gradient effectively. Engaging with the problem collaboratively can help clarify the necessary steps to find the correct value of k.
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Difficult gradient problem!

Consider the curve with equation x2 + xy + y2 = 3.
(a) Find in terms of k, the gradient of the curve at the point (−1, k).
(b) Given that the tangent to the curve is parallel to the x-axis at this point, find the
value of k.
 
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don1231915 said:
Consider the curve with equation x2 + xy + y2 = 3.
(a) Find in terms of k, the gradient of the curve at the point (−1, k).
(b) Given that the tangent to the curve is parallel to the x-axis at this point, find the
value of k.

I hope you aren't done posting yet. You are going to show an attempt at solving it, right?
 


Dick said:
I hope you aren't done posting yet. You are going to show an attempt at solving it, right?

yes, I have no idea how to do (a) but for b I tried substituting 1 and solve for y. I don't know if it is right
I am really confused when it asks for the slope

PLease help if you can!

Thank you so much

P.S. You don't have to provide me with a solution, just some hint or help would be more than helpful.
 


don1231915 said:
yes, I have no idea how to do (a) but for b I tried substituting 1 and solve for y. I don't know if it is right
I am really confused when it asks for the slope

PLease help if you can!

Thank you so much

P.S. You don't have to provide me with a solution, just some hint or help would be more than helpful.

Ok, here's a hint. Try using implicit differentiation to find the gradient dy/dx. You know about that, yes?
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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