a. verify that the given point lies on the curve.\nb. determine an equation of the line tangent to the curve…

a. verify that the given point lies on the curve.\nb. determine an equation of the line tangent to the curve at the given point.\n\\(\\sin y + 2x=y^{2};\\left(\\frac{\\pi^{2}}{2},\\pi\\right)\\)\na. verify that the point is on the given curve. evaluate each side of the equation separately.\nwhen \\(x = \\frac{\\pi^{2}}{2}\\) and \\(y = \\pi\\), \\(\\sin y+2x=\\square\\) and \\(y^{2}=\\square\\).\n(type exact answers, using \\(\\pi\\) as needed.)

a. verify that the given point lies on the curve.\nb. determine an equation of the line tangent to the curve at the given point.\n\\(\\sin y + 2x=y^{2};\\left(\\frac{\\pi^{2}}{2},\\pi\\right)\\)\na. verify that the point is on the given curve. evaluate each side of the equation separately.\nwhen \\(x = \\frac{\\pi^{2}}{2}\\) and \\(y = \\pi\\), \\(\\sin y+2x=\\square\\) and \\(y^{2}=\\square\\).\n(type exact answers, using \\(\\pi\\) as needed.)

Answer

Explanation:

Step1: Evaluate left - hand side

Substitute (x = \frac{\pi^{2}}{2}) and (y=\pi) into (\sin y+2x). We know that (\sin\pi = 0), so (\sin y+2x=0 + 2\times\frac{\pi^{2}}{2}=\pi^{2}).

Step2: Evaluate right - hand side

Substitute (y = \pi) into (y^{2}). Then (y^{2}=\pi^{2}).

Step3: Implicit differentiation

Differentiate (\sin y+2x=y^{2}) with respect to (x). The derivative of (\sin y) with respect to (x) is (\cos y\frac{dy}{dx}) (by the chain - rule), the derivative of (2x) with respect to (x) is (2), and the derivative of (y^{2}) with respect to (x) is (2y\frac{dy}{dx}). So we have (\cos y\frac{dy}{dx}+2 = 2y\frac{dy}{dx}).

Step4: Solve for (\frac{dy}{dx})

Rearrange the equation (\cos y\frac{dy}{dx}+2 = 2y\frac{dy}{dx}) to get (\cos y\frac{dy}{dx}-2y\frac{dy}{dx}=-2), then (\frac{dy}{dx}(\cos y - 2y)=-2), and (\frac{dy}{dx}=\frac{2}{2y-\cos y}).

Step5: Find the slope at the given point

Substitute (y = \pi) into (\frac{dy}{dx}). We know that (\cos\pi=-1), so (\frac{dy}{dx}\big|_{y = \pi}=\frac{2}{2\pi-(-1)}=\frac{2}{2\pi + 1}).

Step6: Find the equation of the tangent line

Use the point - slope form (y - y_{0}=m(x - x_{0})), where (x_{0}=\frac{\pi^{2}}{2}), (y_{0}=\pi), and (m = \frac{2}{2\pi+1}). (y-\pi=\frac{2}{2\pi + 1}(x-\frac{\pi^{2}}{2})) (y=\frac{2}{2\pi + 1}x-\frac{\pi^{2}}{2\pi + 1}+\pi)

Answer:

a. When (x=\frac{\pi^{2}}{2}) and (y = \pi), (\sin y+2x=\pi^{2}) and (y^{2}=\pi^{2}). b. The equation of the tangent line is (y=\frac{2}{2\pi + 1}x-\frac{\pi^{2}}{2\pi + 1}+\pi)