Answer :
Answer:
the rate of changes of the area of the triangle
[tex]\frac{dA}{dt} = -11 cm^2 /sec[/tex]
Step-by-step explanation:
Explanation :-
The area of triangle( A) = [tex]\frac{1}{2} base X height[/tex] ..........(1)
Let 'b' be the base and 'l' be the length of the triangle
The base of a triangle is shrinking( means decreasing) at a rate of 11 cm/s
that is [tex]\frac{db}{dt} = -11cm/sec[/tex]
The height of a triangle is increasing at a rate of 11 cm/s
that is [tex]\frac{dh}{dt} = 11cm/sec[/tex]
Given h= 10cm and b = 8cm
The rate of change of triangle
applying uv formula [tex]\frac{d(uv)}{dx} = u( \frac{dv}{dx}) + v(\frac{du}{dx} )[/tex]
Differentiating equation (1) with respective to 't'
[tex]\frac{dA}{dt} =\frac{1}{2} ( b(\frac{dh}{dt} )+h ( \frac{db}{dt}))[/tex]
substitute all values in above equation, we get
h= 10cm , b = 8cm , [tex]\frac{db}{dt} = -11cm/sec[/tex] and [tex]\frac{dh}{dt} = 11cm/sec[/tex]
[tex]\frac{dA}{dt} = 10 (-11) + 8(11 )[/tex]
After simplification , we get
[tex]\frac{dA}{dt} =\frac{1}{2} (-110 +88) = -11 cm^2 /sec[/tex]
Answer:
The area of the triangle is decreasing at a rate 11 square centimeter per second
Step-by-step explanation:`
We are given the following in the question:
[tex]\dfrac{db}{dt} = -11\text{ cm per sec}\\\\\dfrac{dh}{dt} = 11\text{ cm per sec}[/tex]
Instant base = 8 cm
Instant height = 10 cm
Area of triangle =
[tex]A = \dfrac{1}{2}\times b \times h[/tex]
where b is the base of the triangle and h is the height of the triangle.
Rate of change of area =
[tex]\dfrac{dA}{at} = \dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})[/tex]
Putting values, we get,
[tex]\dfrac{dA}{dt} = \dfrac{1}{2}(8(11) + (10)(-11))\\\\\dfrac{dA}{dt}=-11[/tex]
Thus, the area of the triangle is decreasing at a rate 11 square centimeter per second