Posted by **jump** on Sunday, September 30, 2012 at 2:11pm.

The slope of the curve x^3y^2 + 2x - 5y + 2 = 0 at the point (1,1) is

- calculus -
**Reiny**, Sunday, September 30, 2012 at 3:34pm
using implicit derivatives :

x^3(2y)dy/dx + y^2 (3x^2) + 2 - 5 dy/dx = 0

dy/dx(2x^3 y - 5) = -3x^2 y^2 - 2

dy/dx = (-3x^2y^2 - 2)/(2x^3y - 5)

at the point (1,1)

dy/dx = (-3 - 2)/(2-5) = 5/3

now you have the slope of the tangent, and a point on that tangent.

Use the method you commonly use to find the equation of that straight line

## Answer this Question

## Related Questions

- 12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...
- calculus - 1. Given the curve a. Find an expression for the slope of the curve ...
- Calculus - A curve passes through the point (7,6) and has the property that the ...
- Calculus Help Please!!! Check - A curve passes through the point (0, 2) and has...
- calculus - Determine the equation of a curve in the xy-plane that passes through...
- calculus - The slope of a curve is at the point (x,y) is 4x-3. Find the curve if...
- Calc AB - Suppose that f(x) is an invertible function (that is, has an inverse ...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
- calculus - 8. Find the derivative of the function f(x) = cos^2(x) + tan^2(x) 9...
- Calculus - Find the equation of the curve passing through P (1,2) whose slope is...

More Related Questions