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In an experiment testing solid rocket fuel, some fuel is burned and the waste products are collected - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 1

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In an experiment testing solid rocket fuel, some fuel is burned and the waste products are collected. Throughout the experiment, the sum of the masses of the unburne... show full transcript

Worked Solution & Example Answer:In an experiment testing solid rocket fuel, some fuel is burned and the waste products are collected - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 1

Step 1

Explain, in the context of the problem, what $\frac{dx}{dt}$ and $M$ represent.

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Answer

dxdt\frac{dx}{dt} represents the rate of change of the mass of waste products with respect to time. It indicates how quickly the waste products are accumulating as the solid rocket fuel is burned. On the other hand, MM is the total mass of unburned fuel and waste fuel combined at the start of the experiment. It represents the initial mass of the solid rocket fuel before any burning occurs.

Step 2

solve the differential equation, expressing $x$ in terms of $k$, $M$ and $t$.

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Answer

To solve the differential equation dxdt=k(M−x),\frac{dx}{dt} = k(M - x), we first separate the variables:

dxM−x=kdt.\frac{dx}{M - x} = k dt.

Integrating both sides gives:

−ln⁡∣M−x∣=kt+c,-\ln|M - x| = kt + c,

where cc is the constant of integration.

Exponentiating both sides leads to:

M−x=e−kt−c.M - x = e^{-kt - c}.

Therefore,

x=M−e−kt−c.x = M - e^{-kt - c}.

To find the constant cc, we use the initial condition x(0)=0x(0) = 0:

early 0=M−e−c⇒e−c=M⇒c=−ln⁡(M).0 = M - e^{-c} \Rightarrow e^{-c} = M \Rightarrow c = -\ln(M).

Thus,

x=M−Me−kt=M(1−e−kt).x = M - Me^{-kt} = M(1 - e^{-kt}).

Step 3

find the value of $x$ when $t = \ln 9$, expressing $x$ in terms of $M$, in its simplest form.

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Answer

Using x=M(1−e−kt)x = M(1 - e^{-kt}), we substitute t=ln⁡9t = \ln 9:

x=M(1−e−kln⁡9)=M(1−9−k)=M(1−19k).x = M(1 - e^{-k\ln 9}) = M(1 - 9^{-k}) = M(1 - \frac{1}{9^k}).

Thus, the value of xx when t=ln⁡9t = \ln 9 can be expressed as:

x=M(1−19k).x = M\left(1 - \frac{1}{9^k}\right).

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