Science:Math Exam Resources/Courses/MATH110/April 2019/Question 03

MATH110 April 2019

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Question 03
In a rectangle, the length of all sides of the rectangle are changing in such a way that the area of the rectangle remains constant and always equal to 12 m $^{2}$ . How fast is the diagonal of the rectangle changing when the base is 3 m long and is increasing at a rate of 2 m/min?

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
Since the area is constrained to be 12 m², the height and base are related. Find an expression for the height in terms of the base.

Hint 2
Let $h$ be the height of the rectangle and $b$ its base. The area of the rectangle is $bh=12$ , so $h={\frac {12}{b}}$ . Now if we can write down a formula for the length of the diagonal using $b$ and $h$ , it will be a single-variable formula, so we can take its derivative. Hint: draw a picture and remember Pythagoras.

Hint 3
If g(t) is some function and $f(t)={\sqrt {g(t)}},$ then ${\frac {{\text{d}}f}{{\text{d}}t}}={\frac {1}{2{\sqrt {g(t)}}}}\cdot {\frac {{\text{d}}g}{{\text{d}}t}}$

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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. Let $b,h$ be the base and height of the rectangle, respectively. Then $bh=12$ , and note that $b,h$ are both functions of time. By Pythagoras' theorem, the diagonal of the rectangle is $D={\sqrt {b^{2}+h^{2}}}={\sqrt {b^{2}+\left({\frac {12}{b}}\right)^{2}}}.$ Applying hint number 3, we have ${\frac {{\text{d}}D}{{\text{d}}t}}={\frac {1}{2{\sqrt {b^{2}+{\frac {144}{b^{2}}}}}}}\cdot {\frac {\text{d}}{{\text{d}}t}}\left(b^{2}+{\frac {144}{b^{2}}}\right)={\frac {1}{2{\sqrt {b^{2}+{\frac {144}{b^{2}}}}}}}\cdot \left(2b{\frac {{\text{d}}b}{{\text{d}}t}}+144\cdot {\frac {-2}{b^{3}}}\cdot {\frac {{\text{d}}b}{{\text{d}}t}}\right).$ At the moment when $b=3$ m and ${\frac {{\text{d}}b}{{\text{d}}t}}=2$ m/min, by putting these values in the above expression, we obtain $\color {blue}{\frac {{\text{d}}D}{{\text{d}}t}}={\frac {1}{10}}\cdot \left(12-{\frac {64}{3}}\right)=-{\frac {14}{15}}.$