Question

the altitude (height) of a triangle is increasing at a rate of 1.5 cm/min while the area of the triangle is increasing at a rate of 2 sq. cm/min. recall the area formula for any triangle: ( a=\frac{1}{2}bh ) at what rate is the base of the triangle changing when the altitude is 12 cm and the area is 96 sq.cm? hints: use the product rule for the derivative step. side problem: use the area formula to find the base at the given moment. also, if the height increases quickly, but the area inscreases slowly, the base may be shrinking and the rate will be negative. ( \frac{db}{dt}=) cm/min

Explanation:

Step1: Find the base \(b\)

Given \(A=\frac{1}{2}bh\), when \(A = 96\) and \(h=12\).
Substitute into the formula: \(96=\frac{1}{2}b\times12\).
Solve for \(b\): \(b=\frac{96\times2}{12}=16\)

Step2: Differentiate the area formula with respect to \(t\)

Differentiate \(A=\frac{1}{2}bh\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = b\), \(v = h\).
\(\frac{dA}{dt}=\frac{1}{2}(\frac{db}{dt}h + b\frac{dh}{dt})\)

Step3: Substitute the known values

We know that \(\frac{dA}{dt}=2\), \(\frac{dh}{dt}=1.5\), \(h = 12\), and \(b = 16\)
\(2=\frac{1}{2}(\frac{db}{dt}\times12+16\times1.5)\)

Step4: Solve for \(\frac{db}{dt}\)

Multiply both sides by \(2\): \(4 = 12\frac{db}{dt}+24\)
Subtract \(24\) from both sides: \(12\frac{db}{dt}=4 - 24=-20\)
Divide both sides by \(12\): \(\frac{db}{dt}=-\frac{20}{12}=-\frac{5}{3}\)

Answer:

\(-\frac{5}{3}\)