2019 NYJC P1 Q12

2019 NYJC P1 Q12

12 marks
Prelims

In the study of force field, we are often interested in whether the work done in moving an object from point \({A}\) to point \({B}\) is independent of the path taken. If a force field is such that the work done is independent of the path taken, it is said to be a conservative field.

A force field \({\mathbf{F}}\) can be regarded as a vector \({\mathbf{F}=M\left( x,y \right)\mathbf{i}+N\left( x,y \right)\mathbf{j}}\) where \(M\) and \(N\) are functions of \(x\) and \(y\). The path that the object is moving along is denoted by \({C}\). The work done in moving the object along the curve \({C}\) from the point where \({x=a}\) to the point where \(x=b\) is given by

\(W = \int_{a}^{b} \left[ M(x, y) + N(x, y) \frac{\mathrm{d}y}{\mathrm{d}x} \right] \mathrm{d}x,\)

where \(y=\mathrm{f}\left( x \right)\) is the equation of the curve \({C}\).

  1. Sketch the curve \({C}\) with equation \({{{y}^{2}}=4\left( 1-x \right)}\), for \(x\le 1\).[2]
  2. Find an expression of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \({y}\).[1]
  3. The points \({P}\) and \({Q}\) are on \({C}\) with \({x=1}\) and \({x=-3}\) respectively and \({Q}\) is below the \(x\)-axis.

    Find the equation of the line \(PQ\).

    [2]

For the rest of the question, the force field is given by \({\mathbf{F}={{x}^{2}}\mathbf{i}+x{{y}^{2}}\mathbf{j}}\).

  1. Show that the work done in moving an object along the curve \({C}\) from \({Q}\) to \(P\) is given by the integral \(\int_{-3}^{1} \left( x^2 + 4x(1 - x)^{\frac{1}{2}} \right) \mathrm{d}x\). Hence evaluate the exact work done in moving the object along the curve \({C}\) from \({Q}\) to \(P\).

    [4]
  2. Find the work done in moving an object along the line \({PQ}\) from \({Q}\) to \(P\) to \({2}\) decimal places.[2]
  3. Determine, with reason, whether \({\mathbf{F}}\) is a conservative force field.[1]

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Answer:(ii) \({-\frac{2}{y}}\) (iii) \({y = x - 1}\) (v) \({-33.33}\) (vi) Since the work done for the two paths are different, the force field \({\mathbf{F}}\) is not conservative.

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