Centroid of parabola of height h and base b (y = Kx2) can be given a
s
The Correct Answer and Explanation is:
To find the centroid (Xˉ,Yˉ)(\bar{X}, \bar{Y}) of a parabolic segment defined by the curve y=Kx2y = Kx^2, with height hh and base bb, we use known standard results from calculus and geometry of parabolic segments.
1. Standard Setup
A standard upward-opening parabola is given by: y=Kx2y = Kx^2
Suppose the parabola spans from x=−ax = -a to x=ax = a, and its height at x=0x = 0 is hh. That means: h=K(0)2=0(wrong)⇒So maximum height occurs at x=0⇒let’s define y=h−Kx2h = K(0)^2 = 0 \quad \text{(wrong)} \Rightarrow \text{So maximum height occurs at } x = 0 \Rightarrow \text{let’s define } y = h – Kx^2
Now the parabola goes from x=−ax = -a to x=ax = a, and the vertex is at the top, y=hy = h, and it goes down to 0 at the ends, so: 0=h−Ka2⇒K=ha20 = h – K a^2 \Rightarrow K = \frac{h}{a^2}
Thus, the parabola is: y=h−ha2x2y = h – \frac{h}{a^2}x^2
2. Centroid of a Parabolic Segment
From calculus and geometric principles, the centroid of a parabolic segment (bounded by a parabola and a straight line from base to base) is located at: Xˉ=0(due to symmetry)\bar{X} = 0 \quad \text{(due to symmetry)} Yˉ=3h5\bar{Y} = \frac{3h}{5}
However, when considering just the area under a parabola from x=0x = 0 to x=ax = a (i.e., only one side), then the centroid is at: Xˉ=3a4,Yˉ=3h10\bar{X} = \frac{3a}{4}, \quad \bar{Y} = \frac{3h}{10}
This case appears to match the options provided.
3. Answer and Reasoning
Looking at the options:
- (a) Xˉ=3h4,Yˉ=3h10\bar{X} = \frac{3h}{4}, \bar{Y} = \frac{3h}{10}
- (b) Xˉ=3a4,Yˉ=3a10\bar{X} = \frac{3a}{4}, \bar{Y} = \frac{3a}{10}
- (c) Xˉ=3h4,Yˉ=3a10\bar{X} = \frac{3h}{4}, \bar{Y} = \frac{3a}{10}
- (d) Xˉ=3a4,Yˉ=3h10\bar{X} = \frac{3a}{4}, \bar{Y} = \frac{3h}{10}
Only (d) gives the correct coordinates: (d)Xˉ=3a4,Yˉ=3h10\boxed{(d) \quad \bar{X} = \frac{3a}{4}, \quad \bar{Y} = \frac{3h}{10}}
4. Explanation Summary
The centroid of a parabolic segment, particularly under a curve like y=h−ha2x2y = h – \frac{h}{a^2}x^2, where the curve spans from x=−ax = -a to x=ax = a and reaches height hh at x=0x = 0, is well-known from calculus. When considering the full symmetric segment, the centroid lies directly on the y-axis (i.e., Xˉ=0\bar{X} = 0) and a vertical position Yˉ=3h5\bar{Y} = \frac{3h}{5}. However, if we’re looking at the area under a parabolic arc from the vertex to the base edge (e.g., from x=0x = 0 to x=ax = a), the shape is not symmetric and we must consider the first moments of area to compute the centroid.
By using calculus, the centroid of a parabolic segment bound between the x-axis and a curve like y=h−ha2x2y = h – \frac{h}{a^2}x^2 from x=0x = 0 to x=ax = a, yields: Xˉ=3a4,Yˉ=3h10\bar{X} = \frac{3a}{4}, \quad \bar{Y} = \frac{3h}{10}
These results are consistent with integral-based centroid formulas. The correct choice among the multiple options is therefore: (d) Xˉ=3a4,Yˉ=3h10\boxed{\text{(d) } \bar{X} = \frac{3a}{4}, \bar{Y} = \frac{3h}{10}}
This matches known geometric centroid formulas for parabolic regions bounded by a horizontal axis and a parabola.
