+159 I am extremely sorry for asking 10 question, but I had so much work and I had struggle and confusion on these questions. If you can could you please explain how to answer these questions? Also I have no clue why the computer writes the equations this way but I'm pretty sure when there is a \ like in the first question V=\frac{4}{3}\pi r^3 it is NOT a division sign I got confused on that for a sec so when it shows that before pi it doesn't mean divide by pi, just saying. And for question 8 when it says \cos \big(2xy\big)=y thats just cos(2xy)=y, again I have no clue why the computer writes it that way I guess thats how computers read it. Sorry for asking so much again and making you go through the trouble, but THANK YOU VERY MUCH for lookin into this and helping out, I really appreciate it!
1. The radius of a sphere is decreasing at a constant rate of 2 inches per second. At the instant when the volume of the sphere is 2343 cubic inches, what is the rate of change of the volume? The volume of a sphere can be found with the equation V=\frac{4}{3}\pi r^3. Round your answer to three decimal places.
2. The radius of a cone is decreasing at a constant rate of 5 inches per second. The volume remains a constant 52 cubic inches. At the instant when the radius of the cone is 6 inches, what is the rate of change of the height? The volume of a cone can be found with the equation V=\frac{1}{3}\pi r^2 h. Round your answer to three decimal places.
3. The volume of a sphere is increasing at a constant rate of 191 cubic centimeters per minute. At the instant when the volume of the sphere is 9 cubic centimeters, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation V=\frac{4}{3}\pi r^3 and the surface area can be found with S=4\pi r^2. Round your answer to three decimal places (if necessary).
4. The volume of a sphere is decreasing at a constant rate of 3393 cubic meters per minute. At the instant when the radius of the sphere is 7 meters, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation V=\frac{4}{3}\pi r^3 and the surface area can be found with S=4\pi r^2. Round your answer to three decimal places (if necessary).
5. The volume of a sphere is decreasing at a constant rate of 4635 cubic inches per second. At the instant when the radius of the sphere is 10 inches, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation V=\frac{4}{3}\pi r^3 and the surface area can be found with S=4\pi r^2. Round your answer to three decimal places (if necessary).
6. A right triangle has legs of 18 inches and 24 inches whose sides are changing. The short leg is decreasing by 3 in/sec and the long leg is shrinking at 2 in/sec. What is the rate of change of the area?
7. A rectangle has a length of 5 inches and a width of 8 inches whose sides are changing. The length is decreasing by 7 in/sec and the width is shrinking at 3 in/sec. What is the rate of change of the area?
8. Given \cos \big(2xy\big)=y, find \frac{dy}{dx} in terms of x and y.
9. If y^2+x^3-1=x^2y then find the equations of all tangent lines to the curve when x=-1.
10. An open cylindrical tank has a volume of 500 cubic feet. The inside surface requires an annual coat of paint. One gallon of paint covers 100 square feet. To the nearest gallon, how many gallons of paint are needed if the tank has been constructed to minimize the coat of paint on the inside surface?
Here is #1
Volume = 4/3 pi r3 When =2343 r = 8.239 inches
dV/dt = 4/3 pi dr/dt
dV/dt = 4/3 pi 3r2 dr/dt
= 4/3 pi 3(8.2392)(-2)
= - 568.732 in3 per second (decreasing at 568.732 in3 per second)
+159 Thank you very much for helping me out. But can someone please help me out on the rest of the questions?
Post them seperately......it looks like your question(s) are already answered when scrolling through.....people won't see that only one was answered....
though posting this many is a bit absurd....you should be able to attempt some of them yourself....we are not here to do your homework.....
