Respuesta :
The given equations are
[tex]x(t+1)-4t \sqrt{x} =9[/tex] (1)
[tex]2y+4y^{3/2}=t^{3}+t[/tex] (2)
When t=0, obtain
[tex]x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0[/tex]
Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means [tex] \frac{dx}{dt} [/tex].
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means [tex] \frac{dy}{dt} [/tex].
Because [tex] \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} [/tex], obtain
[tex] \frac{dy}{dx} |_{t=0}\, = \frac{1/2}{3}= \frac{1}{6} [/tex]
Answer:
The slope of the curve at t=0 is 1/6.
[tex]x(t+1)-4t \sqrt{x} =9[/tex] (1)
[tex]2y+4y^{3/2}=t^{3}+t[/tex] (2)
When t=0, obtain
[tex]x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0[/tex]
Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means [tex] \frac{dx}{dt} [/tex].
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means [tex] \frac{dy}{dt} [/tex].
Because [tex] \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} [/tex], obtain
[tex] \frac{dy}{dx} |_{t=0}\, = \frac{1/2}{3}= \frac{1}{6} [/tex]
Answer:
The slope of the curve at t=0 is 1/6.
