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The base of a triangle is increasing in length at a rate of
3cm per second, while its height is decreasing at a rate of
2cm per second. At the moment when the area of the triangle is
60cm^(2), its base is
20cm long. Determine the rate of change of the area at this time.

The base of a triangle is increasing in length at a rate of $3 \mathrm{~cm}$ per second, while its height is decreasing at a rate of $2 \mathrm{~cm}$ per second. At the moment when the area of the triangle is $60 \mathrm{~cm}^{2}$ , its base is $20 \mathrm{~cm}$ long. Determine the rate of change of the area at this time.

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Rate of travel: word problems

Full solution

Q. The base of a triangle is increasing in length at a rate of

3 \mathrm{~cm}

per second, while its height is decreasing at a rate of

2 \mathrm{~cm}

per second. At the moment when the area of the triangle is

60 \mathrm{~cm}^{2}

, its base is

20 \mathrm{~cm}

long. Determine the rate of change of the area at this time.

Area Formula Derivation: We know that the area of a triangle is given by the formula $A = \frac{1}{2} \times \text{base} \times \text{height}$ . Let's denote the base as $b$ and the height as $h$ . We are given that the base $b$ is increasing at a rate of $\frac{db}{dt} = 3 \, \text{cm/s}$ and the height $h$ is decreasing at a rate of $\frac{dh}{dt} = -2 \, \text{cm/s}$ (negative because it's decreasing). We need to find the rate of change of the area $\frac{dA}{dt}$ when the area $A$ is $60 \, \text{cm}^2$ and the base $b$ is $b$ $1$ .
Height Calculation: First, let's find the height $h$ of the triangle when the area $A$ is $60$ cm $^2$ and the base $b$ is $20$ cm. Using the area formula $A = \frac{1}{2} \times b \times h$ , we can solve for $h$ : $\newline$ $60 = \frac{1}{2} \times 20 \times h$ $\newline$ $h = \frac{60}{(\frac{1}{2} \times 20)}$ $\newline$ $h = \frac{60}{10}$ $\newline$ $h = 6$ cm $\newline$ So, the height of the triangle at this moment is $6$ cm.
Rate of Change Calculation: Now, let's use the formula for the area of a triangle to find the rate of change of the area with respect to time. We differentiate both sides of the equation $A = \frac{1}{2} \cdot b \cdot h$ with respect to time $t$ : $\frac{dA}{dt} = \frac{1}{2} \cdot (\frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt})$ We know $\frac{db}{dt} = 3$ cm/s, $\frac{dh}{dt} = -2$ cm/s, $b = 20$ cm, and $h = 6$ cm. Let's plug these values into the equation: $\frac{dA}{dt} = \frac{1}{2} \cdot (3 \cdot 6 + 20 \cdot -2)$
Final Result: Now we perform the calculations: $\newline$ $\frac{dA}{dt} = \frac{1}{2} \times (18 - 40)$ $\newline$ $\frac{dA}{dt} = \frac{1}{2} \times (-22)$ $\newline$ $\frac{dA}{dt} = -11 \, \text{cm}^2/\text{s}$ $\newline$ The negative sign indicates that the area of the triangle is decreasing at a rate of $11 \, \text{cm}^2$ per second at the moment when the base is $20 \, \text{cm}$ and the area is $60 \, \text{cm}^2$ .

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