Question

The hypotenuse AB of a right triangle ABC is 5 ft, and one leg, AC, is decreasing at the rate of 2 ft/sec. The rate, in square feet per second, at which the area is changing when AC = 3 is?

Answer 1

Answer: $-\frac{7}{4} \quad \text{ft}^{2}/\sec$

Step-by-step explanation:

Since ABC is a right triangle, at any moment it holds that

$5^2=(AC)^2+(BC)^2$

Moreover, the area A of the triangle is given by

$A= \frac{1}{2}(AC)(BC)$

and we know that the rate of change of the length (AC) is

constant decreasing 2, which may be written using the Leibniz

notation as

$\dfrac{d(AC)}{dt}=-2$.

Using the chain rule and the product rule for derivation, the two

first equations tell us

that

$0 = 2 \dfrac{d(AC)}{dt}(AC) + 2 \dfrac{d(BC)}{dt}(BC)$

and

$\dfrac{dA}{dt} = \dfrac{1}{2} \left( \dfrac{d(AC)}{dt} \cdot (BC) + \dfrac{d(BC)}{dt} \cdot (AC)\right)$

Moreover, using the first of the last two equations we get

$(AC)\dfrac{d(AC)}{dt} = -(BC)\dfrac{d(BC)}{dt} \Rightarrow\\\\\\(AC)(-2) = -(BC) \dfrac{d(BC)}{dt} \quad \Rightarrow \quad \dfrac{d(BC)}{dt}=2 \dfrac{(AC)}{(BC)}$

Now, when (AC)=3, we have that

$25=(AC)^2 + (BC)^2 \quad \Rightarrow 25 = 9 + (BC)^2\\ \\\Rightarrow \quad 16=(BC)^2 \quad \Rightarrow (BC)=4$

and

$\dfrac{d(BC)}{dt} = 2 \dfrac{(AC)}{(BC)}=2 \dfrac{3}{4}=\dfrac{3}{2}$.

Hence, at this moment the rate of change of the area of the triangle is

$\dfrac{dA}{dt} = \dfrac{1}{2} \left( \dfrac{d(AC)}{dt} \cdot (BC) + \dfrac{d(BC)}{dt} \cdot(AC) \right)=\dfrac{1}{2}\left( -2 \cdot 4 + \dfrac{3}{2}\cdot 3\right ) = -\dfrac{7}{4}$

Answer 2

Okay 
we have the rate of change of AC = d(AC)/dt = -2 
the rate of change od BC = d(BC)/dt 
area = (1/2) *AC) (BC) 
taking differential on both sides we ge 
d(A)/dt = 1/2){ (BC) d(AC)/dt + (AC) d(BC)/dt)}....(1) 
again 
when AC= 3 
applying pythagorous thm 
we get 
(5)^2 =(3)^2 +(BC)^2 
hence we get BC = 4 
now we need to find d(BC)/dt 
we have 
(5)^2 = (AC)^2 +(BC)^2 
taking differenial 
0=2(AC) d(AC/dt) +2BC d(BC)/dt 
that is 
d(BC)/dt = -(3) *(-2)/4 ..(at AC =3) 
hence 
d(BC)/dt = 3/2 
substituting these values in equation (1) 
d(A)/dt = (1/2) {4 * -2 + 3 *3/2} 

which gives 
d(A)/dt = -7/4 

The rate, in square feet per second, at which the area is changing when AC = 3 is -7/4 ft/sec.

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