At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$. The surface area of the cube is increasing at a constant rate of $8$ cm$^2$s$^{-1}$. Show that (a) $rac{dx}{dr} = rac{k}{x}$, where $k$ is a constant to be found, (b) $rac{dV}{dr} = 2V^{rac{1}{3}}$. Given that $V = 8$ when $t = 0$, (c) solve the differential equation in part (b), and find the value of $t$ when $V = 16 ext{/}2$.

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At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 7

Question 2

At-time-$t$-seconds-the-length-of-the-side-of-a-cube-is-$x$-cm,-the-surface-area-of-the-cube-is-$S$-cm$^2$,-and-the-volume-of-the-cube-is-$V$-cm$^3$-Edexcel-A-Level Maths Pure-Question 2-2006-Paper 7.png

At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$. The surface a... show full transcript

Worked Solution & Example Answer:At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 7

Step 1

Show that $ \frac{dx}{dt} = \frac{k}{x} $

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Answer

From the surface area of the cube, we know that $S = 6x^2.$
Differentiating with respect to time $t$ gives: $\frac{dS}{dt} = 12x \frac{dx}{dt}.$
Since the surface area is increasing at a constant rate of $8$ cm $^2$ s $^{-1}$ , we have: $\frac{dS}{dt} = 8.$
Setting these equations equal, we get: $12x \frac{dx}{dt} = 8.$
Solving for (\frac{dx}{dt}), we find: $\frac{dx}{dt} = \frac{8}{12x} = \frac{2}{3x}.$
This implies that: $\frac{dx}{dt} = k \frac{1}{x},$
where (k = \frac{2}{3}).

Step 2

Show that $ \frac{dV}{dt} = 2V^{\frac{1}{3}} $

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Answer

Given that the volume of the cube is given by $V = x^3,$
we differentiate with respect to time: $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}.$
Substituting our expression for (\frac{dx}{dt}): $\frac{dV}{dt} = 3x^2 \left( \frac{2}{3x} \right) = 2x.$
Since we know from the relation (x = V^{\frac{1}{3}}), substituting gives: $$\frac{dV}{dt} = 2V^{\frac{1}{3}}.$

Step 3

Solve the differential equation in part (b), and find the value of $t$ when $V = 8$

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Answer

To solve the differential equation: $\frac{dV}{dt} = 2V^{\frac{1}{3}}$
we can separate variables: $\frac{dV}{V^{\frac{1}{3}}} = 2dt.$
Integrating both sides: $\int V^{\frac{-1}{3}} dV = \int 2 dt.$
The left side evaluates to: $\frac{3}{2} V^{\frac{2}{3}} = 2t + C.$
Now, applying the initial condition $V = 8$ when $t = 0$ :

ightarrow \frac{3}{2} \cdot 4 = C = 6.$$ Hence, the equation becomes: $$\frac{3}{2} V^{\frac{2}{3}} = 2t + 6.$$ To solve for $t$ when $V = 16$: $$\frac{3}{2} \cdot 16^{\frac{2}{3}} = 2t + 6.$$ Evaluating $16^{\frac{2}{3}}$ gives $16^{\frac{2}{3}} = 16^{\frac{2}{3}} = 4^2 = 16$. Thus: $$\frac{3}{2} \cdot 16 = 2t + 6 \rightarrow 24 = 2t + 6 \rightarrow 2t = 18 \rightarrow t = 9.$

At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 7

Worked Solution & Example Answer:At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 7

Show that \( \frac{dx}{dt} = \frac{k}{x} \)

Show that \( \frac{dV}{dt} = 2V^{\frac{1}{3}} \)

Solve the differential equation in part (b), and find the value of \(t\) when \(V = 8\)

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