Question

the altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 3 square centimeters/minute. at what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 93 square centimeters? cm/min

Explanation:

Step1: Recall the area formula of a triangle

The area formula of a triangle is \(A=\frac{1}{2}bh\), where \(A\) is the area, \(b\) is the base, and \(h\) is the altitude.
Differentiate both sides with respect to time \(t\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\). Here \(u = b\) and \(v = h\), so \(\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})\).

Step2: Find the base when \(A = 93\) and \(h=10.5\)

From \(A=\frac{1}{2}bh\), we can solve for \(b\). Substitute \(A = 93\) and \(h = 10.5\) into \(A=\frac{1}{2}bh\). Then \(93=\frac{1}{2}b\times10.5\), so \(b=\frac{93\times2}{10.5}=\frac{186}{10.5}=\frac{1860}{105}=\frac{124}{7}\) cm.

Step3: Substitute the known values into the differentiated formula

We know that \(\frac{dA}{dt}=3\) (area increasing rate), \(\frac{dh}{dt}=1\) (altitude increasing rate), \(h = 10.5\), and \(b=\frac{124}{7}\).
Substitute into \(\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})\):
\(3=\frac{1}{2}(\frac{124}{7}\times1 + 10.5\times\frac{db}{dt})\).
First, multiply both sides by \(2\): \(6=\frac{124}{7}+10.5\times\frac{db}{dt}\).
Then, \(10.5\times\frac{db}{dt}=6-\frac{124}{7}=\frac{42 - 124}{7}=\frac{- 82}{7}\).
Since \(10.5=\frac{21}{2}\), we have \(\frac{db}{dt}=\frac{-82}{7}\div\frac{21}{2}=\frac{- 82\times2}{7\times21}=\frac{-164}{147}\approx - 1.116\) cm/min.

Answer:

\(\frac{-164}{147}\approx - 1.12\) cm/min