Mathematics: 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?

B) 4 inThe radius is half the size of the diameter...

The hypotenuse ab of a right triangle abc is 5

Ответ:

datzmypupppup

09.01.2022

Mathematics

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

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

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

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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?

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