a friend is helping me (by that i mean he's doing the problems, giving me the answers, and then walking me through how to do them and giving me hints for the steps) and everytime he goes to explain a problem I die in the inside
thank god I'm switching to a BA so I don't have to worry about math in the future. :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad: :sad:
That's cool. Idk, math's pretty self explanatory and easy if you actually think about a problem. I'm pretty good at simple equations, idk
Sometimes it can be fun lol
Guts wroteThat's cool. Idk, math's pretty self explanatory and easy if you actually think about a problem. I'm pretty good at simple equations, idk
Sometimes it can be fun lol
Not simple :sad:
Yeah I don't like school math. I do enjoy real world math however. That stuff is fun.
b. 52 + 52 = c2?
a. didnt they already tell you the dimensions of the cylinder? 5h and 3.5 radius?
Aiba wroteb. 52 + 52 = c2?
a. didnt they already tell you the dimensions of the cylinder? 5h and 3.5 radius?
2 asian 4 me
Aiba wroteb. 52 + 52 = c2?
a. didnt they already tell you the dimensions of the cylinder? 5h and 3.5 radius?
b.)
That's not correct because we are given the height of the fence and the fence's distance to the wall with some arbitrary height. Our goal is to find the shortest possible length a ladder can have to pass over the fence and the wall.
a.)
We are given the dimensions of the cone, but we need to find the dimensions of the cylinder to accommodate a largest volume for the cylinder that's inscribed within the cone.
Not sure if you are still looking for the solution but the ladder problem is pretty straight forward, it just gets a little messy. It's been about 7 years since I've taken geometry but all you need to solve this is the similar triangles theorem and some calculus to optimize for the shortest length.
If we were to draw out the problem, the shortest ladder would be the one which touches the top of the ladder, right? So this creates two similar triangles, one on the outside of the fence and the other above the fence but in between the fence and the wall. To clarify, in between these triangles there is a 5x5 square formed by the height of the fence and the distance from the fence to the wall. Let's call the distance from the fence to the bottom of the ladder 'a' and the length from the top of the ladder at the wall, to the top of the square, 'b'. Then by similar triangles, b / 5 = 5 / a or b = 25 / a. Using Pythagorus for the length of the ladder, L = sqrt((5+b)2 + (5+a)2). Into this expression we can plug in the expression for b from similar triangles. Since we want to optimize this problem to find the shortest length, it is easier to isolate the quantity under the root, and call it 'theta'. Expand theta, and then take the first derivative to get theta' = 2(a+5)(1 - 5(25)/a3). To find the minimum, set theta' = 0 and then it is clear that 1 = 125 / a3 or a = 5. Plug this back into the expression for L and that is your length. I got something like 14 feet.
The cylinder problem is also a simple optimization. The trick is to derive a relationship between the radius, 'r', and the height in the cone, 'h'. When r = 0, h = 5, and when r = 3.5, h = 0. Putting this in point slope form yields, h - 5 = [(5 - 0)/(0 - 3.5)](r - 0) or h = -(10/7)r + 5. We can plug this into the volume of the cylinder, V = pir2h. Take the first derivative wrt radius, set it equal to zero. This will give you the radius which gives the maximum volume. Plug this into the expression for h, and then plug both of these back into the volume for the cylinder and you will have your answer.
Hope that helps.
Hecate74 wroteNot sure if you are still looking for the solution but the ladder problem is pretty straight forward, it just gets a little messy. It's been about 7 years since I've taken geometry but all you need to solve this is the similar triangles theorem and some calculus to optimize for the shortest length.
If we were to draw out the problem, the shortest ladder would be the one which touches the top of the ladder, right? So this creates two similar triangles, one on the outside of the fence and the other above the fence but in between the fence and the wall. To clarify, in between these triangles there is a 5x5 square formed by the height of the fence and the distance from the fence to the wall. Let's call the distance from the fence to the bottom of the ladder 'a' and the length from the top of the ladder at the wall, to the top of the square, 'b'. Then by similar triangles, b / 5 = 5 / a or b = 25 / a. Using Pythagorus for the length of the ladder, L = sqrt((5+b)2 + (5+a)2). Into this expression we can plug in the expression for b from similar triangles. Since we want to optimize this problem to find the shortest length, it is easier to isolate the quantity under the root, and call it 'theta'. Expand theta, and then take the first derivative to get theta' = 2(a+5)(1 - 5(25)/a3). To find the minimum, set theta' = 0 and then it is clear that 1 = 125 / a3 or a = 5. Plug this back into the expression for L and that is your length. I got something like 14 feet.
The cylinder problem is also a simple optimization. The trick is to derive a relationship between the radius, 'r', and the height in the cone, 'h'. When r = 0, h = 5, and when r = 3.5, h = 0. Putting this in point slope form yields, h - 5 = [(5 - 0)/(0 - 3.5)](r - 0) or h = -(10/7)r + 5. We can plug this into the volume of the cylinder, V = pir2h. Take the first derivative wrt radius, set it equal to zero. This will give you the radius which gives the maximum volume. Plug this into the expression for h, and then plug both of these back into the volume for the cylinder and you will have your answer.
Hope that helps.
This isn't a helping forum. You're not supposed to help people.
Well ffs, why didn't someone say so sooner? If you could point me in the general direction of a helping forum I will gladly be on my way.
That's why you don't do math.
