Question

(a) Confirm that \((-1,1)\) is on the curve defined by $$x^{3} y^{2}=\cos (\pi y)$$ (b) Use part (a) to find the slope of the line tangent to the curve at \((-1,1) .\)

Short answer

The point (-1,1) lies on the curve, and the slope of the tangent line at this point is 3\pi.

Step-by-step solution

Substitute Point into Equation

Substitute (-1,1) into the equation \(x^{3} y^{2}=\cos (\pi y)\). This gives \((-1)^{3} (1)^{2}=\cos (\pi (1))\). Simplify this to obtain -1 = -1, which is a true statement. Therefore, the point (-1,1) is on the curve.

Derive the Equation

Next, take the derivative of both sides of the equation with respect to x, implicitly. This gives \(3x^{2}y^{2} + 2x^{3}yy' = -\pi y\sin(\pi y)y'\).

Solve for y'

Rearrange the derivative to solve for y'. This yields \(y' = \frac{3x^{2}y^{2}}{-\pi y\sin(\pi y) - 2x^{3}y}\).

Substitute Point into Derivative

Substitute (-1,1) into the derivative to find the slope of the tangent line at the point. This gives \(y' = \frac{3(-1)^{2}(1)^{2}}{-\pi (1)\sin(\pi (1)) - 2(-1)^{3}(1)} = 3\pi = 3\pi\). Thus, the slope of the tangent line at the point (-1,1) is 3\pi.