Respuesta :
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 [tex]\sqrt2[/tex] meter square.
c) The rate is the length of the hypotenuse changing is [tex]2\sqrt2[/tex] 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,
[tex]A=\frac{1}{2}\times \text{Base}\times \text{Height}[/tex]
[tex]A=\frac{1}{2}\times x\times x[/tex]
[tex]A(x)=\frac{1}{2}x^2[/tex] ....(1)
Differentiate w.r.t t,
[tex]\frac{dA}{dt}=\frac{dA}{dx}\times \frac{dx}{dt}[/tex]
We have given, [tex]\frac{dx}{dt}= 2\ m/s[/tex]
[tex]\frac{dA}{dx}=x[/tex] (differentiate (1) w.r.t x)
[tex]\frac{dA}{dt}=2x[/tex] ....(2)
a) The legs are 2 m long i.e. x=2 m
Substitute in (1),
[tex]\frac{dA}{dt}=2\times 2[/tex]
[tex]\frac{dA}{dt}=4\ m^2[/tex]
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,
[tex]h^2=x^2+x^2[/tex]
[tex]h^2=2x^2[/tex]
[tex]x=\frac{h}{\sqrt2}[/tex]
Put h=1,
[tex]x=\frac{1}{\sqrt2}[/tex]
Substitute the value of x in equation (2),
[tex]\frac{dA}{dt}=2\times \frac{1}{\sqrt2}[/tex]
[tex]\frac{dA}{dt}=\sqrt2\ m^2[/tex]
The area of the triangle changing when the hypotenuse is 1 m long is [tex]\sqrt2[/tex] meter square.
c) We know,
[tex]x=\frac{h}{\sqrt2}[/tex]
or [tex]h=\sqrt2 x[/tex]
Derivate w.r.t t,
[tex]\frac{dh}{dt}=\sqrt2 \frac{dx}{dt}[/tex]
We have given, [tex]\frac{dx}{dt}= 2\ m/s[/tex]
So, [tex]\frac{dh}{dt}=\sqrt2\times 2[/tex]
[tex]\frac{dh}{dt}=2\sqrt2[/tex]
The rate is the length of the hypotenuse changing is [tex]2\sqrt2[/tex] m/s.
a) The rate of change of the area is 4 square meters per second.
b) The rate of change of the area is 2 square meters per second.
c) The rate of change of the length of the hypotenuse is [tex]2\sqrt{2}[/tex] meters per second.
The rate of change of the length of the hypotenuse is 4 meters per second.
How to find the rate of change in an isosceles triangle
a) The area of the right isosceles triangle ([tex]A[/tex]), in square meters, is described by the following expression:
[tex]A = \frac{1}{2}\cdot x\cdot y[/tex] (1)
Where [tex]x, y[/tex] are the legs of the right isosceles triangle, in meters.
The formula for the rate of change of the area ([tex]\dot A[/tex]), in square meters per second, is derived by differential calculus:
[tex]\dot A = \frac{1}{2}\cdot (\dot x \cdot y + x\cdot \dot y)[/tex] (2)
Where [tex]\dot x, \dot y[/tex] are the rates of change of the lengths of right isosceles legs, in meters per second.
If we know that [tex]x = y = 2\,m[/tex] and [tex]\dot x = \dot y = 2\,\frac{m}{s}[/tex], then the rate of change of the area is:
[tex]\dot A = \frac{1}{2}\cdot \left[\left(2\,\frac{m}{s} \right)\cdot (2\,m)+\left(2\,\frac{m}{s} \right)\cdot (2\,m)\right][/tex]
[tex]\dot A = 4\,\frac{m^{2}}{s}[/tex]
The rate of change of the area is 4 square meters per second. [tex]\blacksquare[/tex]
b) If we know that [tex]x = y = 1\,m[/tex] and [tex]\dot x = \dot y = 2\,\frac{m}{s}[/tex], then the rate of change of the area is:
[tex]\dot A = \frac{1}{2}\cdot \left[\left(2\,\frac{m}{s} \right)\cdot (1\,m)+\left(2\,\frac{m}{s} \right)\cdot (1\,m)\right][/tex]
[tex]\dot A = 2\,\frac{m^{2}}{s}[/tex]
The rate of change of the area is 2 square meters per second. [tex]\blacksquare[/tex]
c) The length of the hypotenuse ([tex]r[/tex]), in meters, is defined by Pythagorean theorem:
[tex]r^{2} = x^{2}+y^{2}[/tex] (3)
And by differential calculus we derive an expression for the rate of change of the length of the hypotenuse ([tex]\dot r[/tex]), in meters per second:
[tex]2\cdot r \cdot \dot r = 2\cdot x \cdot \dot x + 2\cdot y\cdot \dot y[/tex] (4)
[tex]\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{r}[/tex]
[tex]\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{\sqrt{x^{2}+y^{2}}}[/tex]
Now we proceed to calculate the rate of change of the length of the hypotenuse:
([tex]x = y= 2\,m[/tex], [tex]\dot x = \dot y = 2\,\frac{m}{s}[/tex])
[tex]\dot r = \frac{\left(2\,m\right)\cdot \left(2\,\frac{m}{s} \right)+\left(2\,m\right)\cdot \left(2\,\frac{m}{s} \right)}{\sqrt{(2\,m)^{2}+(2\,m)^{2}}}[/tex]
[tex]\dot r = 2\sqrt{2}\,\frac{m}{s}[/tex]
The rate of change of the length of the hypotenuse is [tex]2\sqrt{2}[/tex] meters per second. [tex]\blacksquare[/tex]
([tex]x = y = 1\,m[/tex], [tex]\dot x = \dot y = 2\,\frac{m}{s}[/tex])
[tex]\dot r = \frac{\left(1\,m\right)\cdot \left(2\,\frac{m}{s} \right)+\left(1\,m\right)\cdot \left(2\,\frac{m}{s} \right)}{\sqrt{(1\,m)^{2}+(1\,m)^{2}}}[/tex]
[tex]\dot r = 4\,\frac{m}{s}[/tex]
The rate of change of the length of the hypotenuse is 4 meters per second. [tex]\blacksquare[/tex]
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