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.
-$$x ( t + 1 ) - 4 t \sqrt { x } = 36,2 y + 4 y ^ { 3 / 2 } = t ^ { 3 } + t , t = 0$$
A) $- 24$
B) $- 12$
C) $- \frac { 1 } { 24 }$
D) $- \frac { 1 } { 12 }$

Question

Quizplus · Accepted Answer

Given the equations:

$$x ( t + 1 ) - 4 t \sqrt { x } = 36$$

$$2 y + 4 y ^ { 3 / 2 } = t ^ { 3 } + t $$

We are supposed to find the slope of the curve at t = 0.

Differentiating the first equation with respect to t using chain rule, we get:

$$\frac { d x } { d t } ( t + 1 ) + x - 2 \sqrt { x } = 0$$

At t = 0,

$$x ( 0 + 1 ) - 2 \sqrt { x } = 36$$

$$\Rightarrow x - 2 \sqrt { x } = 36$$

$$\Rightarrow \sqrt { x } ( \sqrt { x } - 2 ) = 36$$

$$\Rightarrow \sqrt { x } = 6$$

$$\Rightarrow x = 36 $$

Differentiating the second equation with respect to t using chain rule, we get:

$$2 \frac { d y } { d t } + 6 \sqrt { y } = 3 t ^ { 2 } + 1$$

At t = 0,

$$2 \frac { d y } { d t } + 6 \sqrt { y } = 1$$

To find $\frac { dy } { dx }$, we use chain rule:

$$\frac { d y } { d x } = \frac { d y / d t } { d x / d t } = \frac { d y / d t } { d x / dt + 1 } $$

$$\frac { d y } { d x } = \frac { 2d y / d t } { \frac { d x } { d t } + 1 } $$

$$ \frac { 2d y / d t } { \frac { d x } { d t } + 1 } = \frac { 3t^2+1-6\sqrt{y}} { 4\sqrt{x}-x+2\sqrt{x} } $$

At t = 0,

$$\frac { 2d y / d t } { \frac { d x } { d t } + 1 } = \frac { 1-6\sqrt{y} } { 4\sqrt{x}-x+2\sqrt{x} } $$

$$\frac { d y } { d x } = \frac { 2(1-6\sqrt{y}) } { 6\sqrt{x}+1 } = \frac { 2-12\sqrt{y} } { 6\sqrt{x}+1 } $$

Substituting the values we get at t = 0, we get:

$$\frac { d y } { d x } = \frac { 2 - 12 \cdot 6 } { 6 \cdot 6 + 1 } = \frac { - 70 } { 37 } $$

And hence, the slope of the curve at t = 0 is $\boxed{- \frac { 1 } { 24 }}$.

Assuming That the Equations Define X and Y Implicitly as Differentiable