Calculus Question

Question

Find the equation of the tangent line to the curve 3xy = x³+y³, at the point (2/3,4/3).

HELP!?

Answer 1

Hi Don,

 

Need to find the slope of the tangent line at (2/3, 4/3). Since the original equation cannot be easily solved for y, use implicit differentiation. The understanding is that y is some function of x, whenever we take the derivative of y it is with respect to x, i.e. dy/dx. In the following y' is used instead of dy/dx.

 

On the left hand side use the product rule where the "first function is 3x" and the second is y, at least that is one way to think about it.

 

3xy' + 3y = 3x² + 3y²y'

 

Notice at this point we can divide all terms by 3 as part of the simplification. Transposing terms containing y' to left side and others to right side we have:

xy' - y²y' = x² - y

 

Factor out the y' from both terms on the left. Divide the right side by the coefficient of y'. Now use (2/3, 4/3) in the formula for y' to get the slope of the tangent line at that point (4/5). Now use point-slope to write the equation.

 

Here is a picture:

 

https://www.desmos.com/calculator/uhxhft0zvl