Science:Math Exam Resources/Courses/MATH110/April 2014/Question 06

MATH110 April 2014

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q3 (e)  •  Q3 (f)  •  Q3 (g)  •  Q4  •  Q5  •  Q6  •  Q7 (a)  •  Q7 (b)  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  Q9 (c)  •  

   Other MATH110 Exams

  •  December 2011  •  December 2010  •  April 2012  •  April 2011  •  December 2012  •  April 2013  •  December 2013  •  April 2014  •  December 2014  •  December 2015  •  April 2017  •  April 2016  •  December 2016  •  December 2017  •  April 2018  •  

Question 06
At noon, ship $A$ is $30$ kilometres west of ship $B$. Ship $A$ is sailing east at $20$ kilometres per hour and ship $B$ is sailing north at $30$ kilometres per hour. How fast is the distance between the ships changing at 4:00 p. m. on the same day? (You should indicate in your answer if the distance between the ships is increasing or decreasing.)

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
Draw a diagram to illustrate the setting, and label the diagram with variables that might be useful later.

Hint 2
What is the rate that we are trying to find?

Hint 3
Relating the rates directly is difficult. Try to relate the variables first.

Hint 4
Differentiate with respect to the correct variable to obtain a relation between the rates that we want.

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. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This is a related rates problem. So we go about it using the standard related rates approach. Draw a diagram Variables: Define the distance between the two ships to be $z$. We want to find $z'(t)$. Define the distance between the ships and the origin to be $a$ and $b$ respectively. Set the original location of ship B to be at the origin. After $4$ hrs, ship A is $50$ km east of the origin and ship B is $120$ km north of the origin. Equations: The Pythagorean theorem will be useful here. The distance between them is $z={\sqrt {50^{2}+120^{2}}}=130km$. Since A is going away from the origin and B is also going away from the origin, we have $a'=20$ and $b'=30$. Differentiating the distance formula: $z^{2}=a^{2}+b^{2}$, we have: $zz'=aa'+bb'$, substitute to get: $130z'=50\cdot 20+120\cdot 30$. Rearrange to get $z'={\frac {460}{13}}\approx 35.38km/hr$. The distance between the ships are increasing at ${\frac {460}{13}}\approx 35.38km/hr$.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Related rates, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag