Question 1175395: A spheroid (or oblate spheroid) is the surface obtained by rotating an
ellipse around its minor axis. The bowl in Figure 1.41 is in the shape of
the lower half of a spheroid; that is, its horizontal cross sections are
circles while its vertical cross sections that pass through the center are
semi-ellipses. If this bowl is 10 in wide at the opening and V10 in deep
at the center, how deep does a circular cover with diameter 9 in go into
the bowl?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step.
**1. Define the Ellipse**
* The bowl is a lower half of a spheroid, so its vertical cross-sections are semi-ellipses.
* The bowl is 10 inches wide at the opening, so the major axis of the ellipse is 10 inches. Thus, the semi-major axis (a) is 5 inches.
* The bowl is 10 inches deep at the center, so the semi-minor axis (b) is 10 inches.
* The equation of the ellipse, centered at the origin, is:
* x²/a² + y²/b² = 1
* x²/5² + y²/10² = 1
* x²/25 + y²/100 = 1
**2. Relate the Circular Cover to the Ellipse**
* The circular cover has a diameter of 9 inches, so its radius (r) is 4.5 inches.
* We want to find the depth (y) at which the cover fits into the bowl.
* When the cover fits into the bowl, the x-coordinate of the ellipse is equal to the radius of the circular cover.
* Therefore, x = 4.5 inches.
**3. Solve for the Depth (y)**
* Substitute x = 4.5 into the ellipse equation:
* (4.5)²/25 + y²/100 = 1
* 20.25/25 + y²/100 = 1
* 0.81 + y²/100 = 1
* y²/100 = 1 - 0.81
* y²/100 = 0.19
* y² = 19
* y = √19
* y ≈ 4.3589 inches
**4. Calculate the Depth from the Top**
* The total depth of the bowl is 10 inches.
* The y-coordinate we found (√19) is the distance from the bottom of the bowl.
* The depth from the top of the bowl is:
* 10 - √19 ≈ 10 - 4.3589 ≈ 5.6411 inches
**Answer:**
The circular cover with a diameter of 9 inches goes approximately 5.64 inches deep into the bowl.
|
|
|
