A parametric curve is given by the equations&nbsp;𝑥 = 𝑡 − 𝑠𝑖𝑛𝑡 and 𝑦 = 1 − 𝑐𝑜𝑠𝑡&nbsp;Find dy/dx and d^2y/dx^2 at the point (𝜋,2).&nbsp;

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A parametric curve is given by the equations&amp;nbsp;𝑥 = 𝑡 − 𝑠𝑖𝑛𝑡 and 𝑦 = 1 − 𝑐𝑜𝑠𝑡&amp;nbsp;Find dy/dx and d^2y/dx^2 at the point (𝜋,2).&amp;nbsp;

Nick Camelot · Accepted Answer

To find the derivatives of the parametric curve, we need to express the equations in terms of a single parameter, usually denoted by &#039;t&#039;. Given the parametric equations:x=t−sin(t)y=1−cos(t)To find dydx, we&#039;ll use the chain rule. First, we need to find dxdt and dydt.Taking the derivative of x with respect to t, we have:dxdt=1−cos(t)Taking the derivative of y with respect to t, we have:dydt=sin(t)Next, we&#039;ll find dydx by dividing dydt by dxdt:dydx=dydtdxdtdydx=sin(t)1−cos(t)Now, we&#039;ll evaluate dydx and d2ydx2 at the point (π,2).Substituting π into the equations, we have:x=π−sin(π)=πy=1−cos(π)=1+1=2To find dydx at (π,2), we substitute π into dydx:dydx=sin(π)1−cos(π)=02=0To find d2ydx2, we&#039;ll differentiate dydx with respect to t and then divide it by dxdt.Differentiating dydx with respect to t, we have:d2ydx2=ddt(dydx)dxdtDifferentiating dydx with respect to t:d2ydx2=ddt(sin(t)1−cos(t))1−cos(t)To simplify this expression, we&#039;ll differentiate the numerator and denominator separately.Differentiating the numerator, we have:ddt(sin(t))=cos(t)Differentiating the denominator, we have:ddt(1−cos(t))=sin(t)Substituting these results back into d2ydx2, we get:d2ydx2=cos(t)sin(t)=cot(t)To find d2ydx2 at the point (π,2), we substitute π into d2ydx2:d2ydx2=cot(π)=cot(π2)The cotangent function is undefined at π2 because it corresponds to a vertical asymptote. Therefore, at the point (π,2), the value of d2ydx2 is undefined.In summary:dydx=sin(t)1−cos(t)At the point (π,2):dydx=0d2ydx2 is undefined.

A parametric curve is given by the equations 𝑥

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