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

MATH110 April 2019

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[hide] 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?

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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.

[show] 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.

[show] 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.

[show] 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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[show] 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 $\color {blue}{\frac {{\text{d}}D}{{\text{d}}t}}={\frac {1}{10}}\cdot \left(12-{\frac {64}{3}}\right)=-{\frac {14}{15}}.$