Question

Вычислите dy/dx и d^2y/dx^2 , если функция y(x) задана параметрически, x=(1+cos^2t)^2y=cost/sin^2t

Answer 1

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 = (-(sin(t))^3-2(cos(t))^2*sin(t))/(sin(t))^4 = -((sin(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)*((-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)