The daily growth rate, \(h\) millimetres per day, of Plant A is related to the number of hours of sunlight it receives each day, \(t\), by the formula \(h=\dfrac16t^2(4-t)\).
Complete the table of values for \(h=\dfrac16t^2(4-t)\). Values are given to 2 decimal places where appropriate.[1]
| \(t\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) |
|---|
| \(h\) | | \(0.5\) | \(0.94\) | \(1.33\) | \(1.56\) | \(1.5\) | \(1.02\) | \(0\) |
On the grid, draw the graph of \(h=\dfrac16t^2(4-t)\) for \(0\leq t\leq4\).[3]
Use your graph to find the maximum number of hours of sunlight Plant A can receive to achieve a daily growth rate of 1 millimetre per day.[1]
By drawing a tangent, find the gradient of the curve at \((3,1.5)\).[2]
What does the gradient in part (i) tell us about the effect of additional sunlight on the daily growth rate of Plant A at this point? Explain your answer.[1]
The daily growth rate, \(h\) millimetres per day, of another Plant B is related to the \(t\) hours of sunlight it receives each day by the formula \(2h=t\).
On the same grid, draw the graph of \(2h=t\) for \(0\leq t\leq4\).[1]
Find an equation in the form \(t^3+pt^2+qt+r=0\), where \(p\), \(q\) and \(r\) are constants to be found, such that the solutions represent the number of hours of sunlight when both Plant A and Plant B have the same daily growth rate.[2]