Question: A 5-meter length ladder leans against a wall. At the beginning, the ladder is at rest. Then bottom of the ladder is sliding away from wall by a rate of 0.2 m/s. What is the rate of the top of ladder when the bottom is 3 meters away from the wall?

Solution: Assume the length of ladder is L. The distance from the bottom to the wall is D. The distance from the top to the ground is H. Because the angle of wall is a right triangle, we can use Pythagoras Theorem L^2=D^2+H^2. From the question, we know L=5, D=3, therefore, H=sqrt(25-9)=4. Then take the derivative of equation, we get 2LL’=2DD’+2HH’. We know the length of ladder does not change, so L’=0. Also from the question, we know D’=0.2. Now we can plug known values into equation, we get 2(5)(0)=2(3)(0.2)+2(4)H’. Do the calculation: -4H’=0.6 therefore H’=-0.15m/s which means the top of ladder is decreasing 0.15 meters per second.

Last week, people blew lots of balloons for decorating the celebration arena. Suppose air was being pumped into each balloon at a rate of 4 cm³/sec. What is the rate at which the radius of the balloon is increasing when the radius of the balloon is 6 cm.

We know the volume of a sphere is V=4/3πr³. Then we do the derivative which is dv/dt=(4πr²)(dr/dt). From the question, we get r=8, dv/dt=3. Plug these numbers into the equation, we get 4=(4π6²)(dr/dt). Therefore, dr/dt=0.00884 cm/sec.