// Numbas version: exam_results_page_options {"name": "Differentiation application-Geology questions", "timing": {"allowPause": true, "timedwarning": {"action": "none", "message": ""}, "timeout": {"action": "none", "message": ""}}, "metadata": {"licence": "All rights reserved", "description": ""}, "duration": 0, "question_groups": [{"pickingStrategy": "all-ordered", "name": "Group", "pickQuestions": 1, "questions": [{"name": "5: Differentiation - Temperature vs. Depth", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Sarah Turner", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/881/"}], "functions": {}, "variable_groups": [], "ungrouped_variables": ["zz"], "advice": "", "variablesTest": {"maxRuns": 100, "condition": ""}, "statement": "

The Earth’s theoretical internal temperature at depth $z$ is calculated from the equation:

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$T=(-8.225\\times10^{-5})z^2+1.05z+1110$

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where:
$T$ = Temperature ($^\\circ{C}$) 
$z$ = Depth ($km$)

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This produces a quadratic curve that shows the relationship between temperature and the depth within the earth.

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Find the rate of change of temperature with respect to depth.

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$\\frac{\\text{d} T}{\\text{d} z}$ = [[0]]

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What is the gradient at $z=\\var{zz}km$?

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[[0]]$^\\circ{C}km^{-1}$

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Give your answer to two decimal places.

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Differentitaion of a Qudratic function and substitution.

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Application of differentitaion in geology. A projectile problem.

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A volcanic bomb consists of a lump of lava ejected at high speed out of a volcanic crater. The volcanic bomb has an initial velocity as it is forced out of the volcano, but eventually the bomb begins to descend downwards. This point at which there is a directional change is when the velocity of the volcanic bomb is zero.

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(source: bbcbitesize)

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You are given the formula:

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$S=ut-\\frac{1}{2}gt^2$

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where;
$u$ = initial velocity ($ms^{-1}$)
$g$ = acceleration due to gravity ($9.81ms^{-2}$)
$t$ = time ($s$)
$S$ = height at time $t$ in metres ($m$).

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Please see the steps for the advice.

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Please note: The first derivative of distance (i.e. height) is velocity ($v=\\frac{\\text{d} s}{\\text{d} t}$) as differentiation simply means \"rate of change\".

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What is the maximum height the volcanic bomb reaches when initial velocity, u, $=\\var{u}ms^{-1}$?

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[[0]]$m$

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Give your answer to two decimal places.

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This question involves stationary points. You want to find the point where velocity is equal to zero as this is the stationary point (i.e. the maximum height the volcanic bomb reaches).

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You do not have velocity in the equation given, only initial velocity. However you know that velocity is the change in distance over the change in time.

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Therefore differentiate $S$ (distance) with respect to $t$ using the original equation in the question statement to give velocity,$v$, giving your answer as a mathematical expression.

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$v$ = $\\frac{\\text{d} S}{\\text{d} t}$ = 

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Now that you have an equation to work out velocity, you can rearrange to obtain an equation for when velocity is zero as this is a stationary point (i.e. maximum height as the bomb has no velocity). Give your answer as a mathematical expression, with $t$ as the subject.

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When $v = 0, t$ (in $s$) =

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Substitute known values into this equation to give an expression for time, $t$, in seconds as this is the time at which the volcanic bomb is at its maximum height (to 3 decimal places).

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$t$ (in $s$)=

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Substitute $g$, $t$ and $u$ into the original equation $S=ut-\\frac{1}{2}gt^2$, to give the maximum height the volcanic bomb reaches.

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