Studying help!
Flights of leaping animals typically have parabolic paths. The figure illustrates a frog jump superimposed on a coordinate plane. The length of the leap is 9 feet, and the maximum height off the ground is 3a feet. Find a standard equation for the path of the frog. Assume a = 1.
A doorway has the shape of a parabolic arch and is 16 feet high at the center and 8 feet wide at the base. If a rectangular box 7 feet high must fit through the doorway, what is the maximum width the box can have?
Your value of a should be -1 , or else your frog would jump up into space.
Assuming he starts at the origin, (I don't have your diagram), the equation would be
height = -(x -4.5)^2 + 3
2nd:
I am going to place the centre of the doorway at (0,0)
so the equation is
y = ax^2 + 16
but (4,0) lies on it
0 = a(16) + 16
a = -1
y = -x^2 + 16 is the equation of the arch.
so when y = 7
7 = -x^2 + 16
x^2 = 9
x = ± 3
the box would be 6 feet wide.
To find a standard equation for the path of the frog, we need to use the information given. The figure shows a parabolic path, which means we can use the standard equation for a parabola: y = ax^2 + bx + c.
In this case, the maximum height off the ground is given as 3a feet, where a = 1. So the maximum height is 3 feet.
We can also see that the length of the leap is 9 feet. Since the frog's path is symmetric, the x-coordinate of the vertex of the parabola is 0.
Now we can plug in the values we have to find the values of b and c.
At the vertex, x = 0 and y = 3. So we have:
3 = a(0)^2 + b(0) + c
3 = c
Now, to find the value of b, we can use another point on the path. Let's take the point (-4, 0), since the maximum height is symmetrically located. Plugging these values into the equation, we have:
0 = a(-4)^2 + b(-4) + 3
0 = 16a - 4b + 3
Since a = 1, we can simplify this equation to:
0 = 16 - 4b + 3
0 = -4b + 19
Solving for b:
4b = 19
b = 19/4
Now we can substitute the values of a, b, and c into the standard equation to get the final equation for the frog's path:
y = ax^2 + bx + c
y = x^2 + (19/4)x + 3
Therefore, the standard equation for the path of the frog is y = x^2 + (19/4)x + 3.
Moving on to the second question, we have a parabolic arch-shaped doorway that is 16 feet high at the center and 8 feet wide at the base. We are given that a rectangular box with a height of 7 feet needs to fit through this doorway.
To find the maximum width of the box, we need to determine the width of the doorway at a height of 7 feet. Since the parabolic arch is symmetric, we know that the height at the center of the arch is the maximum height.
So we need to find the width at a height of 7 feet. Using similar triangles, we can set up the following proportion:
(8/2) / 16 = x / 7
Simplifying this equation, we have:
4/16 = x/7
1/4 = x/7
To solve for x, we can cross-multiply:
7 = 4x
x = 7/4
Therefore, the maximum width the box can have is 7/4 feet, or 1.75 feet.
To find the standard equation for the path of the frog, we need to recall the standard equation of a parabolic function in the form: y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.
In this case, the vertex represents the maximum point of the frog's jump, which is located at (0, 3a) since the frog reaches a maximum height of 3a feet.
Additionally, we know that the length of the leap is 9 feet, so the x-coordinate of the maximum point (h) is half of that length, or h = 9/2 = 4.5 feet.
Since a = 1, the standard equation for the path of the frog would be: y = (x - 4.5)^2 + 3.
Now, let's move on to the second question.
To find the maximum width the box can have to fit through the doorway, we need to consider the width of the base of the doorway.
The base of the doorway is 8 feet wide. We know that a rectangular box with a height of 7 feet needs to fit through the doorway, so we need to find the maximum width that allows the box to pass through.
We can use similar triangles to establish a proportion between the width of the box and the width of the doorway.
Let's denote the width of the box as w. Using the similar triangles formed by the height of the door, the base of the door, and the width of the box, we have:
w/8 = 7/16
Cross-multiplying, we get:
16w = 8 * 7
16w = 56
Dividing both sides by 16, we find:
w = 3.5
Therefore, the maximum width the box can have to fit through the doorway is 3.5 feet.