A particle moves along the curve y=23x+1y = \frac{2}{\sqrt{3x+1}} in such a way that the yy-coordinate of the particle is decreasing at a constant rate of 3 units per second. Find the yy-coordinate of the particle at the instant when the xx-coordinate is increasing at 0.125 units per second. Show that y=xln⁡x2y = x \ln x^2 is an increasing function when x>1ex > \frac{1}{e}.

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Worked solution

8(a)

1. Use the chain rule to find the value of $\frac{dy}{dx}$ at the given instant.

By the chain rule, dydt=dydx×dxdt\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}.
Given dydt=−3\frac{dy}{dt} = -3 and dxdt=0.125\frac{dx}{dt} = 0.125,
−3=dydx×0.125-3 = \frac{dy}{dx} \times 0.125
dydx=−30.125=−24\frac{dy}{dx} = \frac{-3}{0.125} = -24

2. Differentiate the equation of the curve with respect to $x$.

y=2(3x+1)−12y = 2(3x+1)^{-\frac{1}{2}}
dydx=2(−12)(3x+1)−32×3\frac{dy}{dx} = 2 \left(-\frac{1}{2}\right)(3x+1)^{-\frac{3}{2}} \times 3
dydx=−3(3x+1)−32\frac{dy}{dx} = -3(3x+1)^{-\frac{3}{2}}

3. Equate the two expressions for $\frac{dy}{dx}$ to solve for $x$.

−3(3x+1)−32=−24-3(3x+1)^{-\frac{3}{2}} = -24
(3x+1)−32=8(3x+1)^{-\frac{3}{2}} = 8
(3x+1)32=18(3x+1)^{\frac{3}{2}} = \frac{1}{8}
3x+1=(18)233x+1 = \left(\frac{1}{8}\right)^{\frac{2}{3}}
3x+1=143x+1 = \frac{1}{4}
3x=−343x = -\frac{3}{4}
x=−14x = -\frac{1}{4}

4. Substitute the value of $x$ into the original equation to find the $y$-coordinate.

When x=−14x = -\frac{1}{4},
y=23(−14)+1y = \frac{2}{\sqrt{3\left(-\frac{1}{4}\right)+1}}
y=214y = \frac{2}{\sqrt{\frac{1}{4}}}
y=212=4y = \frac{2}{\frac{1}{2}} = 4

Answer

y=4y = 4

8(b)

1. Differentiate the function with respect to $x$.

Rewrite y=xln⁡x2y = x \ln x^2 as y=2xln⁡xy = 2x \ln x (since x>0x > 0).
Using the product rule:
dydx=2ln⁡x+2x(1x)\frac{dy}{dx} = 2 \ln x + 2x \left(\frac{1}{x}\right)
dydx=2ln⁡x+2\frac{dy}{dx} = 2 \ln x + 2

2. Show that the derivative is positive for the given domain.

For yy to be an increasing function, we need dydx>0\frac{dy}{dx} > 0.
Given x>1ex > \frac{1}{e}, we have x>e−1x > e^{-1}.
Taking the natural logarithm on both sides (since ln⁡x\ln x is an increasing function):
ln⁡x>ln⁡(e−1)\ln x > \ln(e^{-1})
ln⁡x>−1\ln x > -1
Multiplying by 2 and adding 2:
2ln⁡x>−22 \ln x > -2
2ln⁡x+2>02 \ln x + 2 > 0
Since dydx>0\frac{dy}{dx} > 0 for all x>1ex > \frac{1}{e}, y=xln⁡x2y = x \ln x^2 is an increasing function in this domain.

Answer

Shown that dydx>0\frac{dy}{dx} > 0 for x>1ex > \frac{1}{e}.

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