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A square with area equal to 256 has its center at the origin
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18 Nov 2020, 10:20
18 Nov 2020, 10:20
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A square with area equal to 256 has its center at the origin, (0, 0), in the xy-coordinate plane. The graph of a function f defined by \(f(x) = cx^3\), for some positive constant c, intersects the square at two opposite vertices of the square. What is the value of c? (Enter your answer as a fraction.)
Show: ::
\(\frac{1}{64}\)
A square with area equal to 256 has its center at the origin
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21 Feb 2021, 08:54
21 Feb 2021, 08:54
6
void wrote:
plz explain 2nd half of the problem....
Hi!
Solution:
Firstly, the area of the square=256, Thus, the side of the square=16.
Next we are said that the origin (0,0) is the centre i.e. all the sides are equidistant. Hence, the vertices of the square are (8,8), (-8,-8), (8,-8) & (-8,8)
Now, the second half of the question says f(x)=c\(x^3\)
Whenever you see a f(x) in a coordinate geometry question simply substitute f(x) with y
So, we get, y=c\(x^3\)
The question then says this function passes through opposite vertices of the square i.e. it passes diagonally from point (8,8)to (-8,-8) or from point (8,-8) to (-8,8)
Lets substitute the value of x &y to get c
Considering (8,8)
8=c*\(8^3\)
c=\(\frac{1}{64}\)
Similarly, when you consider (-8,-8) you get C as \(\frac{1}{64}\)
And when we substitute the other two values we get \( \frac{-1}{64 } \). As c is a positive value we can negate the other two values
Thus, C=\(\frac{1}{64}\)
Hope this helps
General Discussion
Re: A square with area equal to 256 has its center at the origin
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18 Nov 2020, 23:00
18 Nov 2020, 23:00
1
here area of the square is 256=> the length of each side is 16 unit
as the center of this square is at the origin (0,0), so we can easily
determine the vertices of the square: (8,8);(-8,8); (-8,-8) and (8,-8)
in I, II, III, and IV quadrants respectively. vertices locate in I and III
quadrants are opposite, similarly, vertices locate in II and IV are opposit
to each other.
Here y= c*x^3 passes through the opposite vertices points. We get
c= 1/64, -1/64
as the c is positive, so we take c=1/64 (answer)
as the center of this square is at the origin (0,0), so we can easily
determine the vertices of the square: (8,8);(-8,8); (-8,-8) and (8,-8)
in I, II, III, and IV quadrants respectively. vertices locate in I and III
quadrants are opposite, similarly, vertices locate in II and IV are opposit
to each other.
Here y= c*x^3 passes through the opposite vertices points. We get
c= 1/64, -1/64
as the c is positive, so we take c=1/64 (answer)
