Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (g)
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Question 01 (g) 

ShortAnswer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. A vertical cylindrical tank with radius 3m is being filled with water at a rate of 5m^{3}/min. How fast is the height of the water increasing? 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

What is the relationship between the volume of water being added and the increase in the water's height? 
Hint 2 

Recall that the formula for a cylinder's volume is Where r is the radius of the base and h is the height. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. The two variables that are changing in this problem are the volume of water and the height of the water. These two variables are related in the formula for volume of a cylinder: For this problem, we know that the radius of the cylinder is 3, so we can put that into our formula to get: Since the question is asking about rates of change, we will differentiate the above formula with respect to t. In this new formula, dh/dt stands for the rate at which the height of the water is changing, which is what we're solving for. We also know that dV/dt stands for the rate at which the volume of water is changing, which is 5 m^{3}/min. So substituting in dV/dt = 5 we have: And solving for dh/dt we get: 