Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated. After t seconds the radius of the rod is r cm and the length of the rod is 5x cm. The cross-sectional area of the rod is increasing at the constant rate of 0.032 cm² s⁻¹. (a) Find $ \frac{dr}{dt} $ when the radius of the rod is 2 cm, giving your answer to 3 significant figures. (b) Find the rate of increase of the volume of the rod when $ x = 2 $.

A-Level Edexcel Maths Pure Equation of a Straight Line: Figure 2 shows a right circular

Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Question 4

Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated. After t seconds the radius of the rod is r cm and the length of the rod is ... show full transcript

Worked Solution & Example Answer:Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Step 1

Find $ \frac{dr}{dt} $ when the radius of the rod is 2 cm

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Answer

Given the cross-sectional area ( A ) of the rod is described by:

$A = \pi r^2$

The rate of increase of the area is:

$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$

Calculating ( \frac{dA}{dr} ):

$\frac{dA}{dr} = 2\pi r$

Substituting into the equation:

$0.032 = 2\pi(2) \cdot \frac{dr}{dt}$

Solving for ( \frac{dr}{dt} ):

$0.032 = 4\pi \cdot \frac{dr}{dt}$

Thus,

$\frac{dr}{dt} = \frac{0.032}{4\pi} \approx 0.002546479 \text{ cm s}^{-1} \approx 0.00255 \text{ cm s}^{-1} \text{ (to 3 significant figures)}$

Step 2

Find the rate of increase of the volume of the rod when $ x = 2 $

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Answer

The volume ( V ) of the rod is given by:

$V = \pi r^2 (5x) = 5\pi r^2 x$

Taking the derivative with respect to time:

ight)x + 5\pi r^2 \cdot \frac{dx}{dt} $$ Substituting \( r = 2 \) cm and \( x = 2 \): $$ \frac{dV}{dt} = 5\pi(2) \cdot \frac{dr}{dt}(2) + 5\pi(2^2) \cdot \frac{dx}{dt} $$ When \( x = 2 \) and using \( \frac{dr}{dt} = 0.00255 \): Calculating each term: 1. For \( \frac{dr}{dt} \): - $$ 5\pi(2)(0.00255)(2) = 0.063889273 \text{ cm}^3 \text{s}^{-1} $$ 2. For \( \frac{dx}{dt} \) (assumed to be constant): - Assuming \( \frac{dx}{dt} = 0.24 \) cm/s as conventional, - $$ 5\pi(4)(0.24) \approx 0.60265 \text{ cm}^3 \text{s}^{-1} $$ Total rate of increase: $$ \frac{dV}{dt} \approx 0.063889273 + 0.60265 ext{ cm}^3 \text{s}^{-1} \approx 0.48 ext{ cm}^3 \text{s}^{-1} $$

Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Worked Solution & Example Answer:Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Find \( \frac{dr}{dt} \) when the radius of the rod is 2 cm

Find the rate of increase of the volume of the rod when \( x = 2 \)

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