Question

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the trinagle is increasing at a rate of \({\bf{2}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}{\rm{/}}{\bf{min}}\). At what rate is the base of the triangle changing when the altitude is 10 cm and the area is \({\bf{100}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\)?

Short answer

The base of the triangle is decreasing at a rate of \(1.6\;{\rm{cm/min}}\).

Step-by-step solution

Step 1:Find the expression for the rate of change of area

Let b is base and h is the height of the triangle. The area of the triangle is:

\(A = \frac{1}{2}bh\)

Differentiate the equation with respect to t.

\(\frac{{{\rm{d}}A}}{{{\rm{d}}t}} = \frac{1}{2}\left( {b\frac{{{\rm{d}}h}}{{{\rm{d}}t}} + h\frac{{{\rm{d}}b}}{{{\rm{d}}t}}} \right)\)

Find the rate of change of base

If \(h = 10\) and \(A = 100\), then by the expression for area:

\(\begin{aligned}A &= \frac{1}{2}bh\\100 &= \frac{1}{2}b\left( {10} \right)\\b &= 20\end{aligned}\)

Substitute 10 for h, 20 for b, 100 for A, 2 for \(\frac{{{\rm{d}}A}}{{{\rm{d}}t}}\), and 1 for \(\frac{{{\rm{d}}h}}{{{\rm{d}}t}}\) in the equation \(\frac{{{\rm{d}}A}}{{{\rm{d}}t}} = \frac{1}{2}\left( {b\frac{{{\rm{d}}h}}{{{\rm{d}}t}} + h\frac{{{\rm{d}}b}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned}2 &= \frac{1}{2}\left( {20\left( 1 \right) + 10\left( {\frac{{{\rm{d}}b}}{{{\rm{d}}t}}} \right)} \right)\\4 &= 20 + 10\left( {\frac{{{\rm{d}}b}}{{{\rm{d}}t}}} \right)\\\frac{{{\rm{d}}b}}{{{\rm{d}}t}} &= - \frac{{16}}{{10}}\\ &= - 1.6\end{aligned}\)

Thus, the base of the triangle is decreasing at a rate of \(1.6\;{\rm{cm/min}}\).