Implicit Diff: Tangent Line at (-3*31/2,1)

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Homework Help Overview

The discussion revolves around using implicit differentiation to find the equation of the tangent line to the curve defined by the equation x2/3 + y2/3 = 4 at the point (-3√3, 1).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of implicit differentiation and share their attempts at deriving the derivative. Questions arise regarding the substitution of the x-value into the derivative, particularly concerning the handling of negative numbers in the context of cube roots.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the calculations involved. Some guidance has been offered regarding the simplification of expressions, and there is an acknowledgment of the validity of taking cube roots of negative numbers.

Contextual Notes

There is a mention of restrictions on using calculators, which may affect how participants approach the simplification of expressions.

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Homework Statement


Implicit Differentiation:Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x2/3 + y2/3 = 4 at (-3*31/2,1)


Homework Equations


? None?


The Attempt at a Solution


2/3 * x-1/3 + 2/3y-1/3*y' = 0
then after a few steps of switching the sides of the variables

y' = (-2x1/3)/ 2/3*y-1/3

The part that I'm just confused on is putting x into the variables, especially since we're not suppose to use calculators, any help?
 
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3*3^(1/2)=3^(3/2). Does that help you simplify the x part?
 
Dick said:
3*3^(1/2)=3^(3/2). Does that help you simplify the x part?

I think I got it, does the slope end up being -3*31/2? Otherwise I did it incorrectly :P
 
Yes, I think it does. It always helps to show your work when asking question like this.
 
Wait wait wait, okay, when I plug in x... how can I plug in a negative number into that...? -3^3/2 into x^- 1/3?
 
x^(-1/3) is one over the cube root of x. There isn't any problem with taking the cube roots of negative numbers.
 

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