2x+2y.y'-4+7y'=0
y'(2y+7)=(4-2x)
y'=(4-2x)/(2y+7) as you found:)
(a) Find dy/dx given that x^2+y^2-4x+7y=15.
(b) Under what conditions on x and/or y is the tangent line to this curve horizontal? Vertical?
For part A, I simplifid it to
-(2x+4/2y+7)
For part B,I am confused on what to do?
2 answers
Assuming the y' formula is correct, the tangent is zero when y' = 0 (x = 2) and vertical when the denominator of y' is zero (y = -7/2). The corresponding y and x values can be solve for using the original equation.
1 answer