The blueprint of a rectangular house is 2.5 inches by 1.5 inches. If the scale is 1:20, what is the perimeter of the actual hous
1 answer:
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Answer:
Step-by-step explanation:
This question is incomplete; here is the complete question.
A closed cylindrical can of fixed volume V has radius r. (a) Find the surface area, S, as a function of r. (b) What happens to the value of S approaches to infinity? (c) Sketch a graph of S against r, if V=10 cm³.
A closed cylindrical can of volume V is having radius r and height h.
a). Surface area of a cylinder is given by
S = 2(Area of the circular sides) + Lateral area of the can
S = 2πr² + 2πrh
S = 2πr(r + h)
b). Since surface area is directly proportional to radius of the can
S ∝ r
Therefore, when r approaches to infinity (r → ∞)
c). If V = 10 cm³ Then we have to graph S against r.
From the formula V = πr²h
10 = πr²h
h =
By placing the value of h in the formula of surface area,
S =
Now we can get the points to plot the graph,
r -2 -1 0 1 2
S -13.72 -13.72 0 26.28 35.13
Complete Question:
The complete question is shown on the first uploaded image
Answer:
a) The graph of f(y) versus y. is shown on the second uploaded image
b) The critical point is at y = 0 and the solution is asymptotically unstable.
c)The phase line is shown on the third uploaded image
d) The sketch for the several graphs of solution in the ty-plane is shown on the fourth uploaded image
Step-by-step explanation:
Step One: Sketch The Graph of f(y) versus y
Looking at the given differential equation
for -∞ <
< ∞
We can say let = f(y) =
Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.
Step Two : Determine the critical point
To fin the critical point we have to set = 0
This means = 0
For this to be possible = 1
which means that =
which implies that y = 0
Hence the critical point occurs at y = 0
meaning that the equilibrium solution is y = 0
As t → ∞, our curve is going to move away from y = 0 hence it is asymptotically unstable.
Step Three : Draw the Phase lines
A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where f(y) = 0.
In each of the intervals bounded by the equilibrium draw an upward
pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.
This phase line would solely depend on y does not matter what t is
On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).
For below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper
Step Four : Draw a Solution Curve
A solution curve is a curve that shows the solution of a DE (deferential equation)
Here the solution curve would be drawn on the ty-plane
So the t-axis(x-axis) is its the equilibrium that is it is the solution
If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)
