Assume that $$ x $$ and $$ y $$ are both differentiable functions of $$ t $$. Find the required values of $$ \frac{d y}{d t} $$ and $$ \frac{d x}{d t} $$. $$ y=\sqrt{x} $$ (a) Find $$ \frac{d y}{d t} $$, given $$ x=4 $$ and $$ \frac{d x}{d t}=4 $$. $$ \frac{d y}{d t}= $$ (b) Find $$ \frac{d x}{d t} $$, given $$ x=25 $$ and $$ \frac{d y}{d t}=5 $$. $$ \frac{d x}{d t}=\square $$

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We start with the equation y = √x. Since x and y are both differentiable functions of t, we differentiate both sides with respect to t using the chain rule:

dy/dt = (1/(2√x)) · dx/dt

(a) Given x = 4 and dx/dt = 4, substitute these values into the formula:

dy/dt = (1/(2√4)) · 4  
      = (1/(2 × 2)) · 4  
      = (1/4) × 4  
      = 1

Thus, when x = 4 and dx/dt = 4, dy/dt = 1.

(b) Now, solve for dx/dt given dy/dt = 5 and x = 25. Again, we start with:

dy/dt = (1/(2√x)) · dx/dt

Substitute the known values:

5 = (1/(2√25)) · dx/dt  
  5 = (1/(2 × 5)) · dx/dt          (from √25 = 5)  
  5 = (1/10) · dx/dt

Now, solve for dx/dt by multiplying both sides by 10:

dx/dt = 10 × 5  
  dx/dt = 50

Thus, when x = 25 and dy/dt = 5, dx/dt = 50.

Summary:
 (a) dy/dt = 1  
 (b) dx/dt = 50
(a) $$ \frac{d y}{d t} = 1 $$ (b) $$ \frac{d x}{d t} = 50 $$

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