Ответы на вопрос:
X=(1+(cos(t))^2)^2 y=cos(t)/(sin(t))^2 решение. найдем вначале первую производную dy/dx =(dy/dt)/(dx/dt) отдельно находим производные xt' и yt' dx/dt = 2(1+(cos(t))^2)*2cos(t)*(-sin() = -4(1+(cos(t))^2)*cos(t)*sin(t) dy/dt = (t))^3-2(cos(t))^2*sin(t))/(sin(t))^4 = (t))^2+2(cos(t))^2)/(sin(t))^3 = = -(1+(cos(t))^2)/(sin(t))^3 следовательно: dy/dx = [-(1+(cos(t))^2)/(sin(t))^3]/[-4(1+(cos(t))^2)*cos(t)*sin(t)] = =1/(4*(sin(t))^4*cos(t)) найдем yx'' (вторую производную): y’’ = [d(dy/dx)/dt]/[dx/dt] d(dy/dx)/dt = ((1/4)*(sin(t))^(-4)*(cos(t))^(-1))’ = =(1/4)*)*(sin(t))^(-5)*cos(t)*(cos(t))^(-1) + (sin(t))^(-4)*(-1)(cos(t))^(-2)*sin(t))= = (1/4)*(-4/(sin(t))^(5) – 1/[(sin(t))^(3)*(cos(t))^(2)]) = = (-1/4)*(4(cos(t))^2+(sin(t))^2)/((sin(t))^5*(cos(t))^2)= = -(3(cos(t))^2+1)/(4(sin(t))^5*(cos(t))^2) тогда y’’ = -(3(cos(t))^2+1)/(4(sin(t))^5*(cos(t))^2)/(-4(1+(cos(t))^2)*cos(t)*sin(t))= =(3(cos(t))^2+1)/(16*(sin(t))^6*(cos(t))^3*(1+(cos(t))^2)
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