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) 0 1 2 3 4 5 6 7 Number of bacteria (n) 50 100 200 400 800 1600 3200 6400
| Time in hours (t) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| Number of bacteria (n) | 50 | 100 | 200 | 400 | 800 | 1600 | 3200 | 6400 |
- ((a))
On the grid opposite, draw the graph of n against t for .
Answer ... min ... s
- ((b))
- ((i))
By drawing a tangent, find the gradient of the curve at .
Answer ...
- ((ii))
State briefly what this gradient represents.
...
- ((i))
- ((c))
The number of bacteria in another colony is given by the equation .
- ((i))
On the grid in part (a), draw a graph to represent the number of bacteria in this colony for .
Answer ...
- ((ii))
Find the value of t when the numbers in the colonies are equal.
Answer ...
- ((i))
- ((d))
Given that the equation of the first graph is , 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 against uses the ordered pairs
2. Plot the points accurately on the grid
On the horizontal axis, use .
On the vertical axis, use the corresponding values of : .
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 .
Graph drawn by plotting , , , , , , and , then joining the points with a smooth increasing curve for .
8((b))((i))
1. Draw the tangent at the required point on the curve
Locate 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,
From an accurately drawn tangent, this gives approximately
Gradient bacteria per hour
8((b))((ii))
1. State the meaning of the gradient in context
The graph is of number of bacteria, , against time, .
So the gradient at represents the instantaneous rate of change of with respect to at .
Therefore, it is the rate at which the number of bacteria is increasing at hours.
It represents the rate at which the number of bacteria in the first colony is increasing at hours, in bacteria per hour.
8((c))((i))
1. Find two points on the straight-line graph within the given interval
For :
When ,
When ,
So two points are and .
2. Plot the points and draw the line segment for the required domain
Plot and on the grid used in part (a).
Join the two points with a straight line segment, only for .
Straight line drawn for through the points and .
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 . The corresponding value on the -axis is approximately
hours
8((d))
1. Use the initial value to find the constant $k$
At , .
Substitute into :
Since ,
Therefore,
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