#### To determine

**To find:** How fast is the area of the triangle increasing when θ=π3rad.

#### Answer

The increasing area of triangle is dAdt=0.3 cm2/min_, when the angle θ=π3rad.

#### Explanation

**Given:**

The side length of triangle is a=2 cm and b=3 cm.

And the rate at which the angle θ increases is 0.2 rad/min.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) The area of the triangle with sides of length a and b: A=12ab sinθ.

(3) Product rule of differentiation:ddx(fg)=fddx(g)+gddx(f)

**Calculation:**

Let A be the area of the triangle with sides length a and b and θ is the angle between the sides a and b.

The area of the triangle and angle between the sides increases as the time t increases.

That is the area of the triangle and angle between the sides depends on the time variable t.

Thus, the area of the triangle and angle between the sides is a function of the time variable t.

Since the angle between the sides increases at the rate of 0.2 rad/min, dθdt=0.2 rad/min.

Obtain the derivative dAdt,when θ=π3.

First substitute a=2 and b=3 in the area of the triangle.

A=12(2)(3)sin θ=(3)sin θ=3sin θ

Differentiate A with respect to the time variable t.

dAdt=ddt(3sin θ)=(3)ddt(sin θ)

Applying chain rule of differentiation.

dAdt=(3)cos θdθdt

Substitutes π3 for θ and 0.2 for dθdt in dAdt.

dAdt=(3)(cos π3)(0.2)=(3)(12)(0.2) [Q cos π3=12]=(3)(0.1)=0.3 cm2/min

Therefore the required rate of change of the area increases is dAdt=0.3 cm2/min,when the angle θ=π3rad.

#### To determine

**To find:** How fast is the area of the triangle increasing when b=3 cm and θ=π3rad?

#### Answer

The required rate of change of the area increases is dAdt=0.3+343cm2/min [≈1.6] when b=3 cm and θ=π3 rad.

#### Explanation

**Given:**

The sides length of triangle a=2cm and b=3cm.

The rate at which the angle increases dθdt=0.2 /min.

The rate at which the side of the triangle increase dbdt=1.5 cm/min.

**Calculation:**

The area of the triangle and the sides of the triangle and the angle between the sides increase when the time t increases.

That is the area of the triangle and angle between the sides and side length depends on the time variable t.

Thus, the area of the triangle and angle between the sides and sides length are function of the time variable t.

Since the angle between the sides increases at the rate of 0.2 rad/min, dθdt=0.2 rad/min.

And the side length increases at the rate of 1.5 cm/min,dbdt=1.5 cm/min.

Obtain the derivative dAdt, when b=3 cm and θ=π3 rad.

First put a=2 in the area of the triangle.

A=12(2)bsin θ=bsin θ

Differentiate A with respect to the time variable t.

dAdt=ddt(b sin θ)=bddt(sin θ)+sin θddt(b) [Q ddx(fg)=fddx(g)+gddx(f)]=b(cos θ)dθdt+(sin θ)dbdt [Q dydx=dydu⋅dudx]

Substitutes the b=3, θ=π3, dθdt=0.2 and dbdt=1.5 in dAdt.

dAdt=(3)(cos π3)(.02)+(sin π3)(1.5)=(3)(12)(0.2)+(32)(1.5)=(3)(0.1)+(32)(32)=0.3+343 cm2/min [≈1.6]

Therefore the required rate of change of the area increases is dAdt=0.3+343 cm2/min [≈1.6].

#### To determine

**To find:** How fast is the area of the triangle increasing when? a=2 , b=3 and θ=π3.

#### Answer

The required rate of change of the area increases is dAdt=(2183+0.3) cm2/min,when a=2 , b=3 and θ=π3.

#### Explanation

**Given:**

The sides length of the triangle a=2 cm and b=3 cm.

And the rate at which the angle θ increases is 0.2 rad/min.

Rate at which the side length a increases is 2.5 cm/min.

And rate at which the side length b increase is 1.5 cm/min .

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) The area of the triangle with sides of length a and b:dAdt=12ab sin θ

(3) Product rule of differentiation:ddx(fgh)=fgddx(h)+fgddx(g)+ghddx(f).

**Calculation:**

Let A be the area of the triangle with sides length a and b and θ is the angle between the sides a and b.

The area of the triangle and the sides of the triangle and the angle between the sides increase when the time t increases.

That is the area of the triangle and angle between the sides and sides length depends on the time variable t.

Since the angle between the sides increases at the rate of 0.2 rad/min, dθdt=0.2 rad/min.

And the sides length increases at the rate of 2.5 cm/min and 1.5 cm/min,dadt=2.5 cm/min and dbdt=1.5 cm/min.

Obtain the derivative dAdt,when θ=π3, a=2 cm and b=3 cm .

Differentiate the area A with respect to time variable t .

dAdt=ddt(12ab sin θ)

dAdt=(ab)ddt(sin θ)+sin θddt(ab) [Q ddx(fg)=fddx(g)+gddx(f)]=12{(ab)ddt(sinθ)+sin θ[addt(b)+bddt(a)]} [Q ddx(fg)=fddx(g)+gddx(f)]=12{(ab)(cos θ)dθdt+a sin θdbdt+b sin θdadt } [Qdydx=dydu⋅dudx ]

Substitutes the value of a=2, b=3, θ=π3, dadt=2.5, dbdt=1.5, and dθdt=0.2 , in dAdt.

dAdt=12{(2)(3)(cos π3)(0.2)+(2)(sin π3)(1.5)+(3)(sin π3)(2.5)}=(6)(12)(0.2)+(2)(32)(1.5)+(3)(32)(2.5)=12{(6)(0.1)+323+(3)(32)(52)}=12(0.6+323+1543)

On further simplification, the increase in area is obtained as follows.

dAdt=12(0.6+2143)=(2183+0.3) cm2/min

Thus, the required rate of change of the area increases is dAdt=(2183+0.3) cm2/min,when a=2 , b=3 and θ=π3.