Question

2. Suppose that an isosceles right triangle is slowly expanding outwards with both legs of the triangle increasing at a rate of 10 cm/min. At the moment that the legs are 2 cm, how fast is the hypotenuse increasing in length? Fully simplify your answer

Answer 1

Answer:

$\frac{dh}{dt}=10\sqrt{2}$=14.14 [cm/min]

Explanation:

If we have an isosceles right triangle we can use the Pythagoras theorem to find the hypotenuse:

$h^{2}=x^{2}+x^{2}=2x^{2}$ (1)

$h=\sqrt{2x^{2}}=x\sqrt{2}$ (2)

From equation (2) let's take the derivative with respect to time (t):

$\frac {dh}{dt}=\frac {dx(\sqrt {2})}{dt}$

$\frac {dh}{dt}=\frac {dx}{dt}\sqrt {2}$

dx/dt is the increasing rate of the triangle legs, it is dx/dt = 10 [cm/min].

$\frac{dh}{dt}=10\sqrt{2}$=14.14 [cm/min] (3)

(3) is the hypotenuse increasing in length, and when x = 2 cm, using equation (2), h will be equal to 2.83 cm.

Hava a nice day!