The number of bacteria in a colony doubles every hour.
The colony starts with: 50 bacteria.
The table below shows the number of bacteria in the colony after time t.
Time in hours (t)01234567
Number of bacteria (n)50100200400800160032006400

  1. ((a))

    On the grid opposite, draw the graph of n against t for 0≤t≤70 \le t \le 7.

    Answer ... min ... s

  2. ((b))
    1. ((i))

      By drawing a tangent, find the gradient of the curve at t=5.5t = 5.5.

      Answer ...

    2. ((ii))

      State briefly what this gradient represents.

      ...

  3. ((c))

    The number of bacteria in another colony is given by the equation n=4000−500tn = 4000 - 500t.

    1. ((i))

      On the grid in part (a), draw a graph to represent the number of bacteria in this colony for 0≤t≤70 \le t \le 7.

      Answer ...

    2. ((ii))

      Find the value of t when the numbers in the colonies are equal.

      Answer ...

  4. ((d))

    Given that the equation of the first graph is n=k2tn = k2^{t}, find the value of k.

    Answer k = ...

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Worked solution

8((a))

1. Identify the coordinates to be plotted from the table

From the table, the graph of nn against tt uses the ordered pairs
(t,n)=(0,50),(1,100),(2,200),(3,400),(4,800),(5,1600),(6,3200),(7,6400).(t,n)=(0,50),(1,100),(2,200),(3,400),(4,800),(5,1600),(6,3200),(7,6400).

2. Plot the points accurately on the grid

On the horizontal axis, use t=0,1,2,3,4,5,6,7t=0,1,2,3,4,5,6,7.

On the vertical axis, use the corresponding values of nn: 50,100,200,400,800,1600,3200,640050,100,200,400,800,1600,3200,6400.

Plot each point carefully according to the grid scale.

3. Join the plotted points with the correct graph shape

Join the plotted points with a smooth increasing curve, not with separate straight line segments, for 0≤t≤70\le t\le 7.

Answer

Graph drawn by plotting (0,50)(0,50), (1,100)(1,100), (2,200)(2,200), (3,400)(3,400), (4,800)(4,800), (5,1600)(5,1600), (6,3200)(6,3200) and (7,6400)(7,6400), then joining the points with a smooth increasing curve for 0≤t≤70 \le t \le 7.

8((b))((i))

1. Draw the tangent at the required point on the curve

Locate t=5.5t=5.5 on the horizontal axis and mark the corresponding point on the curve. Draw a straight line which just touches the curve at this point and follows the direction of the curve there.

2. Find the gradient of the tangent

Using two well-spaced points on the tangent,
gradient=vertical changehorizontal change=ΔnΔt.\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{\Delta n}{\Delta t}.
From an accurately drawn tangent, this gives approximately
gradient≈1600.\text{gradient}\approx 1600.

Answer

Gradient ≈1.6×103\approx 1.6\times 10^3 bacteria per hour

8((b))((ii))

1. State the meaning of the gradient in context

The graph is of number of bacteria, nn, against time, tt.
So the gradient at t=5.5t=5.5 represents the instantaneous rate of change of nn with respect to tt at t=5.5t=5.5.

Therefore, it is the rate at which the number of bacteria is increasing at t=5.5t=5.5 hours.

Answer

It represents the rate at which the number of bacteria in the first colony is increasing at t=5.5t=5.5 hours, in bacteria per hour.

8((c))((i))

1. Find two points on the straight-line graph within the given interval

For n=4000−500tn=4000-500t:

When t=0t=0,
n=4000−500(0)=4000.n=4000-500(0)=4000.

When t=7t=7,
n=4000−500(7)=500.n=4000-500(7)=500.

So two points are (0,4000)(0,4000) and (7,500)(7,500).

2. Plot the points and draw the line segment for the required domain

Plot (0,4000)(0,4000) and (7,500)(7,500) on the grid used in part (a).

Join the two points with a straight line segment, only for 0≤t≤70\le t\le 7.

Answer

Straight line drawn for 0≤t≤70 \le t \le 7 through the points (0,4000)(0,4000) and (7,500)(7,500).

8((c))((ii))

1. Read the time-coordinate of the intersection of the two graphs

From the graph, locate the point where the curve for the first colony intersects the straight line n=4000−500tn=4000-500t. The corresponding value on the tt-axis is approximately

t≈4.9.t \approx 4.9.

Answer

t≈4.9t \approx 4.9 hours

8((d))

1. Use the initial value to find the constant $k$

At t=0t=0, n=50n=50.

Substitute into n=k2tn=k2^t:

50=k2050=k2^0

Since 20=12^0=1,

50=k50=k

Therefore, k=50.k=50.

Answer

k=50k=50

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