Q59. - Calculus

Question

Suppose the length a of one leg of a right triangle is increasing

at a rate of 4 inches per second while the length

b of its other leg is decreasing at a rate of 2 inches per

second. The triangle initially has legs of width a = 1 inch

and b = 10 inches.

(a) Find the rate of change of the area of the triangle over

time, in terms of its width and height.

(b) On what intervals do the variables a and b make

sense in this problem? On what time interval does

the problem make sense?

(c) When will the area of the triangle be increasing, and

when will it be decreasing? Answer these questions

both in terms of the width and height of the triangle

and in terms of time

Step-by-Step Solution

Answer

(a) Rate of change of the area of a triangle is $-a+2b.$

(b) $b=0,t=5$sec make sense in this problem.

(c)  Area of triangle increasing in $b>0$and decreasing if $b=0$.

Part (a) Step 1. Given Information

$\frac{da}{dt}=4, \frac{db}{dt}=-2$

Initially $a=1 \& b=10.$

Part (a) Step 2. Calculation

Since the area of the right angle triangle is 

$A=\frac{ab}{2}$

Differentiating both sides with respect to time $"t"$

$\frac{dA}{dt}=\frac{a}{2}.\frac{db}{dt}+\frac{b}{2}.\frac{da}{dt}$

$=\frac{a}{2}.-2+\frac{b}{2}.4$

$=-a+2b$

Part (b) Step 1. Given Information

$\frac{da}{dt }=4, \frac{db}{dt}=-2$

Initially, $a=1 \& b=10$

Part (b) Step 2. Calculation

Since $\frac{da}{dt}=4 \& \frac{db}{dt}=-2$

Integrating both equations we get 

$a=4t+c \& b=-2t+k$

Initially $a=1 \& b=10$

Therefore, $c=1 \& k=10$

Then the equations become 

$a=4t+1 \& b=-2t+10$

The problem make sense if $b=0$

i.e. if $t=5$sec.

Part (c) Step 1. Given Information

$\frac{da}{dt}=4 \& \frac{db}{dt}=-2$

Initially, $a=1 \& b=10$

Part (c) Step 2. Explanation

The area of the triangle increases when $b>0$ and decreases when $b=0.$