Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 $100\;cm^{2}$

Can you answer this question?

1 Answer

0 votes
  • 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
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App