Math / Calculus

Question3. Each side of a square is increasing at a rate of $6 \mathrm{~cm} / \mathrm{s}$ . At what rate is the area of the square increasing when the area of the square is $16 \mathrm{~cm}^{2}$ ?

Studdy Solution

STEP 1

1. The side length of the square is increasing at a constant rate of $6 \, \text{cm/s}$ .
2. We need to find the rate at which the area of the square is increasing when the area is $16 \, \text{cm}^2$ .

STEP 2

1. Express the area of the square as a function of its side length.
2. Use the chain rule to relate the rate of change of the area to the rate of change of the side length.
3. Calculate the rate of change of the area when the area is $16 \, \text{cm}^2$ .

STEP 3

Let $s$ be the side length of the square. The area $A$ of the square is given by: $A = s^2$

STEP 4

Differentiate the area with respect to time $t$ to find $\frac{dA}{dt}$ : $\frac{dA}{dt} = \frac{d}{dt}(s^2) = 2s \frac{ds}{dt}$

STEP 5

Given $\frac{ds}{dt} = 6 \, \text{cm/s}$ , substitute this value into the differentiated equation: $\frac{dA}{dt} = 2s \cdot 6$

STEP 6

When the area $A = 16 \, \text{cm}^2$ , solve for $s$ : $s^2 = 16$ $s = 4 \, \text{cm}$
Substitute $s = 4$ into the equation for $\frac{dA}{dt}$ : $\frac{dA}{dt} = 2 \times 4 \times 6 = 48 \, \text{cm}^2/\text{s}$
The rate at which the area of the square is increasing is:
$\boxed{48 \, \text{cm}^2/\text{s}}$

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Question3. Each side of a square is increasing at a rate of $6 \mathrm{~cm} / \mathrm{s}$ . At what rate is the area of the square increasing when the area of the square is $16 \mathrm{~cm}^{2}$ ?

Studdy Solution

STEP 1

1. The side length of the square is increasing at a constant rate of $6 \, \text{cm/s}$ .
2. We need to find the rate at which the area of the square is increasing when the area is $16 \, \text{cm}^2$ .

STEP 2

1. Express the area of the square as a function of its side length.
2. Use the chain rule to relate the rate of change of the area to the rate of change of the side length.
3. Calculate the rate of change of the area when the area is $16 \, \text{cm}^2$ .

STEP 3

Let $s$ be the side length of the square. The area $A$ of the square is given by: $A = s^2$

STEP 4

Differentiate the area with respect to time $t$ to find $\frac{dA}{dt}$ : $\frac{dA}{dt} = \frac{d}{dt}(s^2) = 2s \frac{ds}{dt}$

STEP 5

Given $\frac{ds}{dt} = 6 \, \text{cm/s}$ , substitute this value into the differentiated equation: $\frac{dA}{dt} = 2s \cdot 6$

STEP 6

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Question3. Each side of a square is increasing at a rate of 6 cm/s6 \mathrm{~cm} / \mathrm{s}6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm216 \mathrm{~cm}^{2}16 cm2 ?

Studdy Solution

STEP 1

1. The side length of the square is increasing at a constant rate of 6 cm/s6 \, \text{cm/s}6cm/s.2. We need to find the rate at which the area of the square is increasing when the area is 16 cm216 \, \text{cm}^216cm2.

STEP 2

1. Express the area of the square as a function of its side length.2. Use the chain rule to relate the rate of change of the area to the rate of change of the side length.3. Calculate the rate of change of the area when the area is 16 cm216 \, \text{cm}^216cm2.

STEP 3

Let s s s be the side length of the square. The area A A A of the square is given by: A=s2 A = s^2 A=s2

STEP 4

Differentiate the area with respect to time t t t to find dAdt\frac{dA}{dt}dtdA​: dAdt=ddt(s2)=2sdsdt \frac{dA}{dt} = \frac{d}{dt}(s^2) = 2s \frac{ds}{dt} dtdA​=dtd​(s2)=2sdtds​

STEP 5

Given dsdt=6 cm/s\frac{ds}{dt} = 6 \, \text{cm/s}dtds​=6cm/s, substitute this value into the differentiated equation: dAdt=2s⋅6 \frac{dA}{dt} = 2s \cdot 6 dtdA​=2s⋅6

STEP 6

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Question3. Each side of a square is increasing at a rate of $6 \mathrm{~cm} / \mathrm{s}$ . At what rate is the area of the square increasing when the area of the square is $16 \mathrm{~cm}^{2}$ ?

1. The side length of the square is increasing at a constant rate of $6 \, \text{cm/s}$ .
2. We need to find the rate at which the area of the square is increasing when the area is $16 \, \text{cm}^2$ .

1. Express the area of the square as a function of its side length.
2. Use the chain rule to relate the rate of change of the area to the rate of change of the side length.
3. Calculate the rate of change of the area when the area is $16 \, \text{cm}^2$ .

Let $s$ be the side length of the square. The area $A$ of the square is given by: $A = s^2$

Differentiate the area with respect to time $t$ to find $\frac{dA}{dt}$ : $\frac{dA}{dt} = \frac{d}{dt}(s^2) = 2s \frac{ds}{dt}$

Given $\frac{ds}{dt} = 6 \, \text{cm/s}$ , substitute this value into the differentiated equation: $\frac{dA}{dt} = 2s \cdot 6$