The two triangles are similar. Identify the scale factor of the dilation from the larger triangle to the smaller triangle.
2 answers:
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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)
By running in a bearing of 127°, Paul motion is in the south eastern direction.
Correct responses:
a. Please find attached the required diagram created with MS Visio
b. Paul's distance travelled east is approximately 11.98 km
- Paul's distance travelled south is approximately 0.903 km
Given:
The distance Paul runs = 1.5 km
The bearing Paul is running = 127°
a. The bearing of path Paul runs, 127°, is the angle measured in the clockwise direction relative to the northern direction
Please find attached the diagram representing the situation, created with MS Visio.
b. Paul's distance travelled east relative to the starting point is given by the component of Paul's path in the x-axis direction which is found as follows;
- Paul's distance east = 15 km × cos(37°) ≈ <u>11.98 km</u>
Pau's distance travelled in the south direction relative to the starting point
is given by the vertical distance travelled relative to the x-axis, which is
given as follows;
Distance Paul travels south = 1.5 km × sin(37°) ≈ <u>0.903 km</u>
Learn more about bearings in trigonometry here:
