Question

Assuming that the equations define x and y implicitly as differentiable functions x=f(t),y=g(t) find the slope of the curve x=f(t), y=g(t) at the given value of t x(t+1)-4tsqrt(x)=9, 2y+4y^3/2=t^3+t , t=0

Answer 1

The given equations are
$x(t+1)-4t \sqrt{x} =9$            (1)
$2y+4y^{3/2}=t^{3}+t$           (2)

When t=0, obtain
$x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0$

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 $\frac{dx}{dt}$.

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 $\frac{dy}{dt}$.

Because $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$, obtain
$\frac{dy}{dx} |_{t=0}\, = \frac{1/2}{3}= \frac{1}{6}$

Answer:
The slope of the curve at t=0 is 1/6.