Question

The legs of an isosceles right triangle increase in length at a rate of 2 m/s. Determine each. (a) At what rate is the area of the triangle changing when the legs are 2 m long? (b) At what rate is the area of the triangle changing when the hypotenuse is 1 m long? (c) At what rate is the length of the hypotenuse changing?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

a) The area of the triangle changing when the legs are 2 m long is 4 meter square.

b) The area of the triangle changing when the hypotenuse is 1 m long is $\sqrt2$ meter square.

c) The rate is the length of the hypotenuse changing is $2\sqrt2$ m/s.

Step-by-step explanation:

Given : The legs of an isosceles right triangle increase in length at a rate of 2 m/s.

Let the length of a leg of an isosceles right triangle be 'x'.

Let the length of the hypotenuse be 'h'.

The are of the triangle is given by,

$A=\frac{1}{2}\times \text{Base}\times \text{Height}$

$A=\frac{1}{2}\times x\times x$

$A(x)=\frac{1}{2}x^2$ ....(1)

Differentiate w.r.t t,

$\frac{dA}{dt}=\frac{dA}{dx}\times \frac{dx}{dt}$

We have given, $\frac{dx}{dt}= 2\ m/s$

$\frac{dA}{dx}=x$ (differentiate (1) w.r.t x)

$\frac{dA}{dt}=2x$ ....(2)

a) The legs are 2 m long  i.e. x=2 m

Substitute in (1),

$\frac{dA}{dt}=2\times 2$

$\frac{dA}{dt}=4\ m^2$

The area of the triangle changing when the legs are 2 m long is 4 meter square.

b) The hypotenuse is 1 m long.

Applying Pythagoras theorem,

$h^2=x^2+x^2$

$h^2=2x^2$

$x=\frac{h}{\sqrt2}$

Put h=1,

$x=\frac{1}{\sqrt2}$

Substitute the value of x in equation (2),

$\frac{dA}{dt}=2\times \frac{1}{\sqrt2}$

$\frac{dA}{dt}=\sqrt2\ m^2$

The area of the triangle changing when the hypotenuse is 1 m long is $\sqrt2$ meter square.

c) We know,

$x=\frac{h}{\sqrt2}$

or $h=\sqrt2 x$

Derivate w.r.t t,

$\frac{dh}{dt}=\sqrt2 \frac{dx}{dt}$

We have given, $\frac{dx}{dt}= 2\ m/s$

So, $\frac{dh}{dt}=\sqrt2\times 2$

$\frac{dh}{dt}=2\sqrt2$

The rate is the length of the hypotenuse changing is $2\sqrt2$ m/s.