The discussion revolves around finding the equation of a tangent line to the curve defined by the equation x^(1/3) + y^(1/3) = 4 at the point (-3√3, 1). Participants are working through the process of implicit differentiation to determine the slope of the tangent line.
There is an ongoing exploration of the derivative and its implications for finding the slope of the tangent line. Participants have provided some guidance and clarification regarding the original equation, but there is no explicit consensus on the correct slope yet.
There is mention of a possible typo in the original equation, which has led to confusion in the differentiation process. The specific values being plugged into the derivative have resulted in complications, contributing to the uncertainty in the discussion.
Looks like there is a typo here.b521 said:Homework Statement
Find the equation of a tangent line at the curve at point (-3√3, 1)
x^(1/3) = y^(1/3) = 4
This would be the line with slope m, that passes through (1, 1).b521 said:Homework Equations
Point-slope:
y-1=m(x-1)
It's hard to tell whether this is correct, since the original equation likely has a typo. If you typed '=' in place of '+', then you have a sign error in your derivative. Also, it would be helpful to have an equation for your derivative; e.g., y' = y^(2/3)/x^(2/3).b521 said:The Attempt at a Solution
I took the derivative of that equation and resulted in
y^(2/3)/x^(2/3)
b521 said:When I tried plugging in x and y to get slope, the equation got very messy and I couldn't get a number out of it. Can anyone help me out?
Mark44 said:What is the original equation? I can't tell if your work is correct without knowing the original equation.
Mark44 said:(-3√3)^(2/3) can also be written as ((-3√3)^2)^(1/3)