This graphic presents the following problem: “Air is being pumped into a spherical balloon at a rate of 5 cubic centimeters per minute. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm. ” From reading this problem, you should recognize that the balloon is a sphere, so you will be dealing with the volume of a sphere. You should also recognize that you are given the diameter, so you should begin thinking how that will factor into the solution as well. Drawing a diagram of the problem can often be useful. In the case, you are to assume that the balloon is a perfect sphere, which you can represent in a diagram with a circle. Mark the radius as the distance from the center to the circle.
In the problem shown above, you should recognize that the specific question is about the rate of change of the radius of the balloon. Notice, however, that you are given information about the diameter of the balloon, not the radius. This will have to be adapted as you work on the problem. You should see that you are also given information about air going into the balloon, which is changing the volume of the balloon.
In this problem you should identify the following items: dVdt=5cm3/min{\displaystyle {\frac {dV}{dt}}=5cm^{3}/min} d=20cm. {\displaystyle d=20cm. } drdt={\displaystyle {\frac {dr}{dt}}=}unknown rate of radius change, to be solved Note that the data given to you regarding the size of the balloon is its diameter. However, planning ahead, you should recall that the formula for the volume of a sphere uses the radius. Therefore, you should identify that variable as well: r=10cm. {\displaystyle r=10cm. } (The radius is half the diameter. )
V=43πr3{\displaystyle V={\frac {4}{3}}\pi r^{3}}
dVdt=4πr2drdt{\displaystyle {\frac {dV}{dt}}=4\pi r^{2}{\frac {dr}{dt}}}
In this problem, you know the rate of change of the volume and you know the radius. The only unknown is the rate of change of the radius, which should be your solution. dVdt=4πr2drdt{\displaystyle {\frac {dV}{dt}}=4\pi r^{2}{\frac {dr}{dt}}} 5=4π∗102drdt{\displaystyle 5=4\pi *10^{2}{\frac {dr}{dt}}} 5=400πdrdt{\displaystyle 5=400\pi {\frac {dr}{dt}}} 5400π=drdt{\displaystyle {\frac {5}{400\pi }}={\frac {dr}{dt}}} 180π=drdt{\displaystyle {\frac {1}{80\pi }}={\frac {dr}{dt}}}
In this case, your solution is for drdt{\displaystyle {\frac {dr}{dt}}}, which is the rate of change of the radius. This is what the question asked for. You should then express your numerical answer with its units to present the final answer for the problem: drdt=180π{\displaystyle {\frac {dr}{dt}}={\frac {1}{80\pi }}} centimeters per minute.
A baseball diamond is 90 feet square. A runner runs from first base to second base at 25 feet per second. How fast is he moving away from home plate when he is 30 feet from first base? You can diagram this problem by drawing a square to represent the baseball diamond. Label one corner of the square as “Home Plate. "
One leg of the triangle is the base path from home plate to first base, which is 90 feet. The second leg is the base path from first base to the runner, which you can designate by length r{\displaystyle r}. You are asked to solve the problem when this distance is 30 feet. The rate of change of this distance, drdt{\displaystyle {\frac {dr}{dt}}}, is the runner’s speed. The hypotenuse of the right triangle is the straight line length from home plate to the runner (across the middle of the baseball diamond). Call this distance d{\displaystyle d}. You aren’t told this distance, but you can calculate it from the Pythagorean Theorem. If the two legs are 90 and 30, then the hypotenuse d{\displaystyle d} is 902+302{\displaystyle {\sqrt {90^{2}+30^{2}}}}. Thus, d=3010{\displaystyle d=30{\sqrt {10}}}. The actual question is for the rate of change of this distance, or how fast the runner is moving away from home plate. This will be the derivative, dddt{\displaystyle {\frac {dd}{dt}}}.
The first leg, a{\displaystyle a}, is the distance from home to first, 90 feet. The second leg, b{\displaystyle b}, is the distance from first to the runner. Use the variable r{\displaystyle r}. You are asked to solve the problem for the instant when r=30{\displaystyle r=30}. The hypotenuse, c{\displaystyle c}, is the distance from home to the runner, d{\displaystyle d}. Write out the new equation: d2=902+r2{\displaystyle d^{2}=90^{2}+r^{2}}
2ddddt=2rdrdt{\displaystyle 2d{\frac {dd}{dt}}=2r{\frac {dr}{dt}}} Note that the constant term, 902{\displaystyle 90^{2}}, drops out of the equation when you take the derivative.
2ddddt=2rdrdt{\displaystyle 2d{\frac {dd}{dt}}=2r{\frac {dr}{dt}}} 2∗3010dddt=2∗30∗25{\displaystyle 230{\sqrt {10}}{\frac {dd}{dt}}=23025} dddt=2∗30∗252∗3010{\displaystyle {\frac {dd}{dt}}={\frac {23025}{230{\sqrt {10}}}}} dddt=2510{\displaystyle {\frac {dd}{dt}}={\frac {25}{\sqrt {10}}}}
Water flows at 8 cubic feet per minute into a cylinder with radius 4 feet. How fast is the water level rising? Diagram this situation by sketching a cylinder. Make a horizontal line across the middle of it to represent the water height.
As the water fills the cylinder, the volume of water, which you can call V{\displaystyle V}, is increasing. The rate of the increase, dVdt{\displaystyle {\frac {dV}{dt}}}, is the amount of the water flow, or 8 cubic feet per minute. The height of the water, h{\displaystyle h}, is not given. The rate of change of the height, dhdt{\displaystyle {\frac {dh}{dt}}}, is the solution to the problem. You are also told that the radius of the cylinder r{\displaystyle r} is 4 feet.
V=πr2h{\displaystyle V=\pi r^{2}h}
V=πr2h{\displaystyle V=\pi r^{2}h} dVdt=πr2dhdt{\displaystyle {\frac {dV}{dt}}=\pi r^{2}{\frac {dh}{dt}}}
dVdt=πr2dhdt{\displaystyle {\frac {dV}{dt}}=\pi r^{2}{\frac {dh}{dt}}} 8=π(42)dhdt{\displaystyle 8=\pi (4^{2}){\frac {dh}{dt}}} 8=16πdhdt{\displaystyle 8=16\pi {\frac {dh}{dt}}} 816π=dhdt{\displaystyle {\frac {8}{16\pi }}={\frac {dh}{dt}}} 12π=dhdt{\displaystyle {\frac {1}{2\pi }}={\frac {dh}{dt}}}
