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The base of a triangle is shrinking at a rate of
8cm//s and the height of the triangle is increasing at a rate of
2cm//s. Find the rate at which the area of the triangle changes when the height is
14cm and the base is
10cm.

The base of a triangle is shrinking at a rate of $8 \mathrm{~cm} / \mathrm{s}$ and the height of the triangle is increasing at a rate of $2 \mathrm{~cm} / \mathrm{s}$ . Find the rate at which the area of the triangle changes when the height is $14 \mathrm{~cm}$ and the base is $10 \mathrm{~cm}$ .

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Find instantaneous rates of change

Full solution

Q. The base of a triangle is shrinking at a rate of

8 \mathrm{~cm} / \mathrm{s}

and the height of the triangle is increasing at a rate of

2 \mathrm{~cm} / \mathrm{s}

. Find the rate at which the area of the triangle changes when the height is

14 \mathrm{~cm}

and the base is

10 \mathrm{~cm}

.

Find Rate of Change: We need to find the rate of change of the area of the triangle. The area of a triangle is given by the formula $A = \frac{1}{2} \times \text{base} \times \text{height}$ . We are given the rates at which the base and height are changing, so we will use related rates to find the rate of change of the area.
Denote Variables: Let's denote the base of the triangle as $b$ , the height as $h$ , and the area as $A$ . The rate of change of the base with respect to time is $\frac{db}{dt} = -8$ cm/s (negative because the base is shrinking), and the rate of change of the height with respect to time is $\frac{dh}{dt} = 2$ cm/s.
Differentiate Area Formula: Differentiate the area formula with respect to time to find the rate of change of the area, $\frac{dA}{dt}$ . Using the product rule for differentiation, we get $\frac{dA}{dt} = \frac{1}{2} \times (\text{base} \times \frac{d(\text{height})}{dt} + \text{height} \times \frac{d(\text{base})}{dt})$ .
Substitute Given Values: Now we substitute the given values into the differentiated formula. At the moment we are interested in, the base $b$ is $10$ cm and the height $h$ is $14$ cm. The rate of change of the base $\frac{db}{dt}$ is $-8$ cm/s and the rate of change of the height $\frac{dh}{dt}$ is $2$ cm/s. So we have $\frac{dA}{dt} = \frac{1}{2} \times (10$ cm $\times 2$ cm/s $10$ $0$ cm $10$ $1$ cm/s).
Perform Multiplication: Perform the multiplication for each term: $\frac{dA}{dt} = \frac{1}{2} \times (20 \, \text{cm}^2/s - 112 \, \text{cm}^2/s)$ .
Calculate Final Rate: Now, subtract the second term from the first term inside the parentheses and then multiply by $\frac{1}{2}$ to find $\frac{dA}{dt}$ . $\frac{dA}{dt} = \frac{1}{2} \times (-92 \, \text{cm}^2/\text{s}) = -46 \, \text{cm}^2/\text{s}.$
Interpret Result: The negative sign indicates that the area of the triangle is decreasing at the rate of $46 \, \text{cm}^2/\text{s}$ when the height is $14 \, \text{cm}$ and the base is $10 \, \text{cm}$ .

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