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Calculus/Slope of line tangent to curve

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Expert: Scotto - 11/4/2009


Question
The slope of the line tangent to the curve y squared + (xy +1)cubed= 0 at (2,-1) is...

Answer
Let x be a function on y, so x is f(y).
That is, y² + (xy + 1)³ = 0.
This says that (xy+1)³ = -y².
Taking the cuberoot of both sides gives xy + 1 = cr((-y²)^(1/3)), were cr is for cube-root.
Subtract 1 from both sides and then divide both sides by y.
This gives x = (cr((-y²)^(1/3)) - 1)/y = g(y)/y where g(y) = cr((-y²)^(1/3)) - 1.

The chain rule says that dx/dy = (lo d hi - hi d lo)/lo² = (yg'(y) - g(y)(1))y².

We were given g(y) = cr((-y²)^(1/3)) - 1,
so g'(y) = (1/3)/cr(-y^(1/3)) = 1/(3cr(-y^(1/3)).

Put in -1 for y and get -1/slope of the curve.

Calculus

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Scotto

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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 25 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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