1 Answer

Thehennyy · Answer 1 · 2024-07-23T11:07:48+0000

Answer:

0.486 cm/min

Explanation:

Connected rates of change refers to the relationship between the rates at which two or more variables are changing with respect to a common independent variable.

A derivative represents the rate of change of a function with respect to its independent variable. So, when something changes over time, the derivative is d/dt of that variable.

The area of a triangle is given by:

A=(1)/(2)bh

where:

A is the area.
b is the base.
h is the height (altitude).

Given that the altitude (h) of a triangle is increasing at a rate of 2 cm/min, then this can be expressed as:

$\frac{\text{d}h}{\text{d}t}=2\; \sf cm/min$

Given that the area of the triangle is increasing at a rate of 4.5 cm²/min, then this can be expressed as:

$\frac{\text{d}A}{\text{d}t}=4.5\; \sf cm^2/min$

To determine the rate at which the base (b) of the triangle is changing, we need to find db/dt. To do this, begin by differentiating the equation for the area of a triangle with respect to t.

Place d/dt in front of each term of the equation for area:

$\frac{\text{d}}{\text{d}t}\left(A\right)=\frac{\text{d}}{\text{d}t}\left((1)/(2)bh\right)$

Take out the constant on the right side:

$\frac{\text{d}A}{\text{d}t}=(1)/(2)\cdot \frac{\text{d}}{\text{d}t}\left(bh\right)$

Use the product rule to differentiate bh with respect to t:

$\frac{\text{d}A}{\text{d}t}=(1)/(2)\cdot \left(b\cdot \frac{\text{d}h}{\text{d}t}+h\cdot \frac{\text{d}b}{\text{d}t}\right)$

Now, substitute the given values of dA/dt and dh/dt into the equation:

$4.5=(1)/(2)\cdot \left(b\cdot 2+h\cdot \frac{\text{d}b}{\text{d}t}\right)$

Simplify and rearrange to isolate db/dt:

$4.5=b+(h)/(2)\cdot \frac{\text{d}b}{\text{d}t}$

$4.5-b=(h)/(2)\cdot \frac{\text{d}b}{\text{d}t}$

$2(4.5-b)=h\cdot \frac{\text{d}b}{\text{d}t}$

$(2(4.5-b))/(h)=\frac{\text{d}b}{\text{d}t}$

$\frac{\text{d}b}{\text{d}t}=(2(4.5-b))/(h)$

$\frac{\text{d}b}{\text{d}t}=(9-2b)/(h)$

As we need db/dt in terms of h and A, we can rearrange the equation for the area of the triangle to isolate b, then substitute this into the equation for db/dt:

$A=(1)/(2)bh\implies b=(2A)/(h)$

Therefore:

$\frac{\text{d}b}{\text{d}t}=(9-2\left((2A)/(h)\right))/(h)$

$\frac{\text{d}b}{\text{d}t}=(9h-4A)/(h^2)$

Finally, to determine the rate at which the base (b) of the triangle is changing when the altitude is 11.5 cm and the area is 98 cm², substitute h = 11.5 and A = 9.8 into the equation for db/dt:

$\frac{\text{d}b}{\text{d}t}=(9(11.5)-4(9.8))/((11.5)^2)$