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.
-$$2 x + 4 x ^ { 3 / 2 } = t ^ { 3 } + t , y ( t + 1 ) - 4 t \sqrt { y } = 4 , t = 0$$
A) -8
B) 2
C) 8
D) 4

Question

Quizplus · Accepted Answer

To find the slope of the curve at the given value of t, we need to find $\frac{dy}{dx}$. Since x and y are both functions of t, we can use the chain rule to find $\frac{dy}{dx}$ by finding $\frac{dy}{dt}$ and $\frac{dx}{dt}$, and then dividing the former by the latter.First, differentiate both given equations with respect to t.For $2x + 4x^{3/2} = t^3 + t$, differentiating both sides with respect to t gives:$$2\frac{dx}{dt} + 6x^{1/2}\frac{dx}{dt} = 3t^2 + 1$$At $t = 0$, $x = f(0)$ can be found by substituting $t = 0$ into the original equation for x, which gives $2x = 0$, so $x = 0$. Substituting $x = 0$ and $t = 0$ into the differentiated equation gives:$$2\frac{dx}{dt} = 1$$So, $\frac{dx}{dt} = \frac{1}{2}$.For $y(t + 1) - 4t\sqrt{y} = 4$, differentiating both sides with respect to t gives:$$y' + y - 4\sqrt{y} = 0$$At $t = 0$, we need to find $y$. Substituting $t = 0$ into the original equation for y gives $y(1) - 0 = 4$, so $y = 4$. Substituting $y = 4$ into the differentiated equation gives:$$y' + 4 - 8 = 0$$So, $y' = 4$.Therefore, the slope of the curve at $t = 0$ is:$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4}{1/2} = 8$$

Assuming That the Equations Define X and Y Implicitly as Differentiable