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If $x_0$ locates a stationary point of $f(x)$, so that $\\frac{dy}{dx} \\large \\vert_{x_0}=0$ then the stationary point is:
\na local minimum if $\\frac{d^2y}{dx^2} \\Large \\vert_{x_0}>0$
\na local maximum if $\\frac{d^2y}{dx^2} \\Large \\vert_{x_0}<0$
\ninconclusive if $\\frac{d^2y}{dx^2} \\Large \\vert_{x_0}=0$
\n", "advice": "We are asked to find the greatest height reached by a thrown ball whose height can be modelled by the function $h=\\simplify{ {C}+{CF2}t-{CF1}t^{2} }$
\nYou don't need to produce a graph to do this but right now, it may help us to visualise what it is that we are trying to do. Here is a graph of this function with the highest point indicated:
\n\nIt is easy to see that the highest point is where the graph \"turns\" and the gradient changes from positive to negative. That is the point that we are looking for.
\nAt this point the gradient will be zero. The gradient is given by the derivative of the function $\\frac{dh}{dt}$.
\nDifferentiate the function:
\n$\\frac{dh}{dt}=\\simplify{{CF2}-{CF1}*2*t}$
\nThe stationary points will be where this is zero, so:
\n$\\simplify{{CF2}-{CF1}*2*t}=0$
\nIf we now solve this equation it will give us the time to reach that height.
\n$\\simplify{{CF2}-{CF1}*2*t}=0$
\n$\\var{CF2}=\\simplify{{CF1}*2*t}$
\n$\\frac{\\var{CF2}}{\\simplify{{CF1}*2}}=t$
\n$t=\\var{t}$ seconds
\n\n\nNow that we know the time it takes for the ball to reach its highest point, we are ask to calculate how high it reaches.
\nWe simply need to use the original equation that models the height of the ball and substitute $t=\\var{t}$
\n$h=\\simplify{ {C}+{CF2}t-{CF1}t^{2} }$
\nbecomes:
\n$h= \\var{C}+\\var{CF2}(\\var{t})-\\var{CF1}(\\var{t})^{2} $
\n$h=\\var{h}$ m
\n\n\n\nIn this example it seems logically obvious that the stationary point is a maximum. After all, balls fly up and then fall down, they don't fly down and then float up again.
\nHowever, it is always good practice to check and it will help to stop you making silly mistakes - and it is quick and easy to do.
\n\nTo find out, we need the second derivative $\\frac{d^2h}{dt^2}$.
\nHere is the original function:
\n$h=\\simplify{ {C}+{CF2}t-{CF1}t^{2} }$
\nand its first derivative, that we worked out earlier:
\n$\\frac{dh}{dt}=\\simplify{{CF2}-{CF1}*2*t}$
\nDifferentiate a second time to give:
\n$\\frac{d^2h}{dt^2}=\\simplify{-{CF1}*2}$
\n\n\nWhich is negative, confirming that this point is a local maximum.
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The ball will reach its highest at the stationary (turning) point.
\nFirst differentiate the function:
\n$\\large \\frac{dh}{dt}=$ [[0]]
\n\n
At the stationary point, this gradient will be zero.
\nEnter the equation that shows this:
\n[[1]]
\n\n
Solve this equation to find the time when this happens:
\n$t=$ [[2]] seconds
\n\n
Substitute this values, into the original function to find the height reached at that time:
\n$h=$[[3]] m
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Whilst, in this case, it is not strictly necessary, it is good practice to check the nature of the stationary points that you find.
\n\nTo determine the nature of the stationary point we need the second derivative $\\frac{d^2h}{dt^2}$
\n\n
$\\frac{d^2h}{dt^2}=$ [[0]]
\n\n
The value of this second derivative is [[1]]
\nTherefore, this is [[2]]
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