Finding equation of tangent plane & normal line to a given surface

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SUMMARY

The discussion focuses on finding the equations of the tangent plane and the normal line to the surface defined by the equation x - z = 4arctan(yz) at the point (1+π, 1, 1). The initial attempt yielded incorrect partial derivatives, specifically fx=0 and fy=0, which are not valid given the presence of both x and y in the equation. The correct approach involves defining the function f(x,y,z) = x - z - 4arctan(yz) and recalculating the partial derivatives to derive the accurate normal vector and tangent plane equation.

PREREQUISITES
  • Understanding of partial derivatives in multivariable calculus
  • Familiarity with the concept of tangent planes and normal lines
  • Knowledge of the arctangent function and its properties
  • Ability to work with implicit functions and their gradients
NEXT STEPS
  • Review the calculation of partial derivatives for multivariable functions
  • Study the derivation of tangent planes and normal lines in calculus
  • Explore the properties of the arctangent function and its derivatives
  • Practice solving similar problems involving implicit functions and their geometrical interpretations
USEFUL FOR

Students studying multivariable calculus, particularly those focusing on tangent planes and normal lines, as well as educators seeking to clarify these concepts in a classroom setting.

physicsidiot1
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Homework Statement



Find equations of the tangent plant and the normal line to x-z=4arctan(yz) at (1+∏, 1, 1)

Homework Equations





The Attempt at a Solution


I took the partials and got fx=0 fy=0 fz=(-4y)/((yz)2+1) so for the plane i got -2z-2=0 and for the normal line I got .5z-2=0. I feel like I may have messed up somewhere, probably in taking the partials...
 
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physicsidiot1 said:

Homework Statement



Find equations of the tangent plant and the normal line to x-z=4arctan(yz) at (1+∏, 1, 1)

Homework Equations





The Attempt at a Solution


I took the partials and got fx=0 fy=0 fz=(-4y)/((yz)2+1) so for the plane i got -2z-2=0 and for the normal line I got .5z-2=0. I feel like I may have messed up somewhere, probably in taking the partials...

Start by writing your function like this:

f(x,y,z) = x - z - 4arctan(yz) = 0

Then look again at your partials (they are all three wrong). In particular, for example, how do you get fx = 0 out of that, for starters? Show us what you get for your normal vector then we can talk.
 
When I look at x- z= 4 arctan(yz), I see both x and y variables in it. So the derivatives fx and fy can't be 0.

But since you never say exactly what "f" is, it is impossible to tell what you are doing.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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