#### To determine

**To find:** The rate of change of the base of the triangle, when the altitude is 10 cm and area is 100 cm^{2}.

#### Answer

The base of the triangle decreases at a rate of 1.6 cm/min.

#### Explanation

**Given:**

The altitude of the triangle increases at the rate of dldt=1 cm/min.

The area of the triangle increases at the rate of dAdt=2 cm2/min.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) ddx[fg]=fdgdx+gdfdx

(3) The area of the triangle with base b and altitude l: A=12l×b

**Calculation:**

Let A be the area of the triangle and l be the altitude of the triangle and b be the base of the triangle.

Since area A and base b and altitude l changes with the time t, the area A and altitude l and base b are function of the time variable t.

Obtain dbdt when l=10 cm and A=100 cm2.

The value of b when l=10 cm and A=100 cm2.

100=12(10)×b200 =10bb =20010b =20 cm

Differentiate A with respect to the time t,

ddt[A]=ddt[12l×b]dAdt=12ddt[l×b]=12[bdldt+ldbdt] [Q dydx=dydu⋅dudx]

On the further simplification the required value of dbdt is obtained as follows.

2dAdt=bdldt+ldbdtldbdt=−bdldt+2dAdtdbdt=1l[2dAdt−bdldt]

Substitutes l=10 ,b=20,dAdt=2 and dldt=1 in dbdt,

dbdt=110[2×2−20×1]=110[4−20]=110×[−16]=−1.6 cm/min

Therefore, the base of the triangle decreases at a rate of 1.6 cm/min.