# The altitude of a triangle is increasing at a rate of $1\; cm/min$ while the area of the triangle is increasing at a rate of $2 \;cm^{2}/min$. At what rate is the base of the triangle changing when the altitude is $10$ cm and the area is $100\;cm^{2}$

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}=f'(x)$ is the rate of change of $y$. w.r.t $x$
• $\large\frac{dy}{dx_{(x_1,y_1}}$ is the slope of the tangent to the curve at the point $(x_1,y_1)$ on the curve. It is the slope of the curve at that point.
• The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
• It $y=f(x)$ and both depend on 't' then $\large\frac{dy}{dt}$$=f'(x).\large\frac{dx}{dt} Step 1: The area of a triangle is given by A=\large\frac{1}{2}$$bh$-----(i) Where b is the base, h is the altitude.
All three variables depend on time t
Step 2:
Rate of change of area is
$\large\frac{dA}{dt}=\frac{d}{dt}\bigg(\frac{1}{2}$$bh\bigg) \qquad=\large\frac{1}{2}$$\bigg(b\large\frac{dh}{dt}$$+h\large\frac{db}{dt}\bigg)-----(ii) Step 3: \large\frac{dA}{dt}$$=2 cm^2/min,$
$\large\frac{dh}{dt}$$=1 cm/min h=10 cm, A=100 cm^2 When h=10 cm, and A=100 cm^2 we have from (i) 100=\large\frac{1}{2}$$b \times 10$
$b=20 \;cm$
Step 4:
Substituting the values in (ii), we have
$2=\large\frac{1}{2}$$(20 \times 1+10 \large\frac{db}{dt}) 4=20+10 \large\frac{db}{dt} \large\frac{db}{dt}$$=-1.6 cm/sec$
The base of the triangle is decreasing at $-1.6 \;cm/s$