It is given that $$ y=(x-3) \sqrt{2 x+1} $$. Find the value of $$ x $$ when the rate of decrease of $$ y $$ is twice the rate of increase of $$ x $$.

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Given 
$$
y = (x-3)\sqrt{2x+1},
$$
we are asked to find the value of $$x$$ when the rate of decrease of $$y$$ is twice the rate of increase of $$x$$.

This means that if $$x$$ is increasing at a rate of $$ \frac{dx}{dt} $$, then $$y$$ is decreasing at a rate of $$ 2\frac{dx}{dt} $$. Since $$y$$ is decreasing, its rate of change with respect to time is negative. Hence, we have
$$
\frac{dy}{dt} = -2\frac{dx}{dt}.
$$
Using the chain rule
$$
\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt},
$$
we obtain
$$
\frac{dy}{dx}\frac{dx}{dt} = -2\frac{dx}{dt}.
$$
Assuming $$\frac{dx}{dt} \neq 0$$, we can cancel $$\frac{dx}{dt}$$ to get
$$
\frac{dy}{dx} = -2.
$$

**Step 1. Compute $$\frac{dy}{dx}$$.**

The function is
$$
y = (x-3)\sqrt{2x+1}.
$$
Let
$$
u = x-3 \quad \text{and} \quad v = \sqrt{2x+1} = (2x+1)^{1/2}.
$$
Then,
$$
u' = 1,
$$
and
$$
v' = \frac{1}{2}(2x+1)^{-1/2}\cdot (2) = \frac{1}{\sqrt{2x+1}}.
$$
Using the product rule,
$$
\frac{dy}{dx} = u'v + uv' = \sqrt{2x+1} + \frac{x-3}{\sqrt{2x+1}}.
$$
Writing with a common denominator:
$$
\frac{dy}{dx} = \frac{(2x+1) + (x-3)}{\sqrt{2x+1}}.
$$
Simplify the numerator:
$$
(2x+1)+(x-3) = 3x-2.
$$
Thus,
$$
\frac{dy}{dx} = \frac{3x-2}{\sqrt{2x+1}}.
$$

**Step 2. Set the derivative equal to $$-2$$.**

We have
$$
\frac{3x-2}{\sqrt{2x+1}} = -2.
$$
Multiply both sides by $$\sqrt{2x+1}$$ (noting that $$\sqrt{2x+1}>0$$ for $$x>- \frac{1}{2}$$):
$$
3x-2 = -2\sqrt{2x+1}.
$$

**Step 3. Solve the equation.**

Square both sides to eliminate the square root:
$$
(3x-2)^2 = 4(2x+1).
$$
Expand the left side:
$$
9x^2 - 12x + 4 = 8x + 4.
$$
Subtract $$8x + 4$$ from both sides:
$$
9x^2 - 12x + 4 - 8x - 4 = 0,
$$
which simplifies to
$$
9x^2 - 20x = 0.
$$
Factor out $$x$$:
$$
x(9x-20) = 0.
$$

Thus,
$$
x=0 \quad \text{or} \quad x=\frac{20}{9}.
$$

**Step 4. Check the solutions.**

Substitute $$x=0$$ into the derivative:
$$
\frac{dy}{dx} = \frac{3(0)-2}{\sqrt{2\cdot0+1}} = \frac{-2}{1} = -2.
$$
This matches the condition.

For $$x=\frac{20}{9}$$:
$$
\frac{dy}{dx} = \frac{3\left(\frac{20}{9}\right)-2}{\sqrt{2\left(\frac{20}{9}\right)+1}} = \frac{\frac{60}{9}-2}{\sqrt{\frac{40}{9}+1}}.
$$
Simplify the numerator:
$$
\frac{60}{9}-2 = \frac{60-18}{9} = \frac{42}{9} = \frac{14}{3}.
$$
Simplify the denominator:
$$
\sqrt{\frac{40}{9}+1} = \sqrt{\frac{40+9}{9}} = \sqrt{\frac{49}{9}} = \frac{7}{3}.
$$
Thus,
$$
\frac{dy}{dx} = \frac{\frac{14}{3}}{\frac{7}{3}} = 2,
$$
which does not satisfy the condition $$ \frac{dy}{dx} = -2$$.

**Conclusion:**

The only valid solution is
$$
x=0.
$$
The value of $$ x $$ is 0.

It is given that .
Find the value of when the rate
of decrease of is twice the rate of
increase of .

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