Find direction of tangent line to equation

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SUMMARY

The discussion focuses on finding the direction of the tangent line to the implicit curve defined by the equation x4 + y4 = 32 at the point (2, -2). The gradient of the curve is correctly identified as 4x3i + 4y3j, which at the specified point evaluates to 32i - 32j. However, this vector represents the direction of greatest slope, not the tangent line itself. To find the tangent line, one must determine a vector that is perpendicular to the gradient vector, which involves rotating the gradient vector by 90 degrees.

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  • Understanding of implicit differentiation
  • Knowledge of vector calculus concepts
  • Familiarity with gradient vectors
  • Ability to compute perpendicular vectors
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  • Study implicit differentiation techniques in calculus
  • Learn how to compute gradients for multivariable functions
  • Explore methods for finding perpendicular vectors in vector calculus
  • Review the concept of tangent lines in the context of curves
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Students studying calculus, particularly those focusing on vector calculus and implicit functions, as well as educators looking for examples of tangent line calculations.

ArcanaNoir
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Homework Statement



Find the direction of the line tangent to the curve x^4+y^4=32 at the point (2, -2)

Homework Equations



Anything goes, we're in vector calculus now.

The Attempt at a Solution



So, to find the tangent line, I was thinking of taking the gradient, but I'm not sure how to do that since the curve is in a squirrely format. I think they call that "implicit".

Perhaps the gradient is 4x^3i+4y^3j?

and then at the given point we would have 32i-32j

Is this actually the direction of the line tangent to the curve at the given point? Is this actually the answer? I don't really know.
 
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ArcanaNoir said:

Homework Statement



Find the direction of the line tangent to the curve x^4+y^4=32 at the point (2, -2)

Homework Equations



Anything goes, we're in vector calculus now.

The Attempt at a Solution



So, to find the tangent line, I was thinking of taking the gradient, but I'm not sure how to do that since the curve is in a squirrely format. I think they call that "implicit".

Perhaps the gradient is 4x^3i+4y^3j?

and then at the given point we would have 32i-32j

Is this actually the direction of the line tangent to the curve at the given point? Is this actually the answer? I don't really know.
No. 32\hat{i}-32\hat{j},, is the direction of greatest slope. The tangent line is perpendicular to this.
 
darn. so how do I find the perpendicular to that?
 
ArcanaNoir said:
darn. so how do I find the perpendicular to that?
Rotate the vector 90°.
 
Remember doing lines perpendicular to a line how did you figure out the slope?

Negative inverse right?
 

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