Answer :
Answer:
a) 15 [tex]\displaystyle{\frac{m^2}{s}}[/tex]
b) [tex]\approx 3.54 \displaystyle{\frac{m^2}{s}}[/tex]
c) [tex]\approx 7.1 \displaystyle{\frac{m^2}{s}}[/tex]
The equation relating the area of an isosceles right triangle, A, and the length of the legs of the triangle, x is [tex]\displaystyle{A=\frac{x^2}{2}}[/tex]
Step-by-step explanation:
The following properties hold for any isosceles right triangle:
- The non-base sides are equal
- The non-base angles are equal and their value is 45°
Now, as for any triangle, the area can be written as:
[tex]\displaystyle{A=\frac{x\cdot h}{2}}[/tex]
where [tex]x[/tex] is the length of one of the side and [tex]h[/tex] is the height relative to that side. In the case of an isosceles right triangle the area equation can be written as:
[tex]\displaystyle{A=\frac{x^2}{2}}[/tex] (1)
since the non-base sides are also relative heights to each other ([tex]h=x[/tex]). This answers the last part of the question.
a) We just need to take the derivative of the area equation (1), which gives:
[tex]\displaystyle{\frac{dA}{dt}=x\frac{dx}{dt}}[/tex] being, [tex]dx/dt[/tex] the rate at which the legs change. Therefore:
[tex]\displaystyle{\frac{dA}{dt}=3\cdot5 = \boxed{15\ \frac{m^2}{s}}}[/tex]
b) The relation between the hypotenuse and the legs is [tex]x = \sin{\alpha} \cdot H[/tex] and putting this on equation (1) we obtain:
[tex]\displaystyle{\frac{dA}{dt}=\sin{45^{\circ}}\cdot 1 \cdot 5 = \frac{5}{2}\cdot \sqrt{2}\ \boxed{\approx 3.54\ \frac{m^2}{s}}}[/tex]
c) Taking the derivatives with respect to the time in the equation [tex]\displaystyle{H=\frac{x}{\sin{\alpha}}}[/tex] (relation between the leg and the hypotenuse) we get:
[tex]\displaystyle{\frac{dH}{dt}=\frac{1}{\sin{\alpha}}\frac{dx}{dt}=\sqrt{2}\cdot5=\ \boxed{7.1\ \frac{m}{s}}}[/tex]