Question

The base of a triangle is shrinking at a rate of 11 cm/s and the height of the triangle is increasing at a rate of 11 cm/s. Find the rate at which the area of the triangle changes when the height is 10cm and the base is 8cm.

Answer 1

Answer:

the rate of changes  of the area of the triangle

$\frac{dA}{dt} = -11 cm^2 /sec$

Step-by-step explanation:

Explanation :-

The area of triangle( A) = $\frac{1}{2} base X height$    ..........(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 $\frac{db}{dt} = -11cm/sec$

The height of a triangle is increasing at a rate of 11 cm/s

that is $\frac{dh}{dt} = 11cm/sec$

Given h= 10cm and b = 8cm

The rate of change of triangle

applying uv formula  $\frac{d(uv)}{dx} = u( \frac{dv}{dx}) + v(\frac{du}{dx} )$

Differentiating equation (1) with respective to 't'

$\frac{dA}{dt} =\frac{1}{2} ( b(\frac{dh}{dt} )+h ( \frac{db}{dt}))$

substitute all values in above equation, we get

h= 10cm , b = 8cm , $\frac{db}{dt} = -11cm/sec$ and $\frac{dh}{dt} = 11cm/sec$

 $\frac{dA}{dt} = 10 (-11) + 8(11 )$

After simplification , we get

$\frac{dA}{dt} =\frac{1}{2} (-110 +88) = -11 cm^2 /sec$

Answer 2

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:

$\dfrac{db}{dt} = -11\text{ cm per sec}\\\\\dfrac{dh}{dt} = 11\text{ cm per sec}$

Instant base = 8 cm

Instant height = 10 cm

Area of triangle =

$A = \dfrac{1}{2}\times b \times h$

where b is the base of the triangle and h is the height of the triangle.

Rate of change of area =

$\dfrac{dA}{at} = \dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})$

Putting values, we get,

$\dfrac{dA}{dt} = \dfrac{1}{2}(8(11) + (10)(-11))\\\\\dfrac{dA}{dt}=-11$

Thus, the area of the triangle is decreasing at a rate 11 square centimeter per second