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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 $100\;cm^{2}$

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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 $-1.6 \;cm/s$
answered Jul 23, 2013 by meena.p
 

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