Question

A rectangle with edges parallel to the coordinate axeshas one vertex at the origin and the diagonally opposite vertex on the curve \(y=k x^{m}\) at the point where \(\mathrm{x}=\mathrm{b}(\mathrm{b}>0, \mathrm{k}>0\), and \(\mathrm{m} \geq 0)\). Show that the fraction of the area of the rectangle that lies between the curveand the \(x\)-axis depends on \(m\) but not on \(\mathrm{k}\) or \(\mathrm{b}\).

Short answer

Show that the fraction of the area of a rectangle with one vertex at the origin and the diagonally opposite vertex on the curve \(y=kx^m\) (where \(x=b\)), that lies between the curve and the x-axis depends on \(m\) but not on \(k\) or \(b\). Solution: We found the coordinates of the diagonal vertex to be \((b, k\cdot b^m)\), the area of the rectangle to be \(A_{rectangle} = bk^{1}b^m\), and the area under the curve within the interval \([0, b]\) to be \(A_{curve} = k\cdot\frac{b^{m+1}}{m+1}\). Upon calculating the ratio between the two areas, we found that the fraction of the rectangle area that lies between the curve and the x-axis is \(f = \frac{1}{m+1}\), which depends on \(m\) but not on \(k\) or \(b\).

Step-by-step solution

Find coordinates of the diagonal vertex

Since the diagonally opposite vertex is on the curve \(y=kx^m\) and we are given \(x=b\), we can find the y-coordinate of the vertex by plugging in the value of \(x=b\) into the equation: \(y=k(b)^m=k\cdot b^m\) So the coordinates of the diagonal vertex are \((b, k\cdot b^m)\).

Find the area of the rectangle

Since the rectangle has edges parallel to the coordinate axes, the sides of the rectangle are given by the difference between the coordinates of the diagonal vertices are \((0,0)\) and \((b, k\cdot b^m)\). The side lengths are \(b\) and \(k\cdot b^m\), and hence the area of the rectangle, \(A_{rectangle}\), is: \(A_{rectangle} = b \cdot k\cdot b^m = bk^{1}b^m\)

Find the area under the curve

To find the area under the curve \(y=kx^m\) within the interval \([0, b]\) (i.e., between the x-axis and the curve), we integrate the equation with respect to \(x\): \(\int_0^b kx^m dx\) Since \(k\) is a constant, we can take it out of the integral: \(k \int_0^b x^m dx\) Now, we find the antiderivative of \(x^m\): \(\frac{x^{m+1}}{m+1}\) Now, we evaluate the limits of the integral: \(k[\frac{b^{m+1}}{m+1} - \frac{0^{m+1}}{m+1}] = k\cdot\frac{b^{m+1}}{m+1}\) So the area under the curve is \(A_{curve} = k\cdot\frac{b^{m+1}}{m+1}\).

Calculate the ratio between the curve area and the rectangle area

Finally, we need to find the fraction of the rectangle area that lies between the curve and the x-axis, which is the ratio between the curve area and the rectangle area: \(f = \frac{A_{curve}}{A_{rectangle}}\) Substituting the values we found earlier: \(f = \frac{k\cdot\frac{b^{m+1}}{m+1}}{bk^{1}b^m} = \frac{b^{m+1}}{bk^{1}b^m(m+1)}\) Simplifying the expression, we can see that \(k\) and \(b\) elements cancel out: \(f = \frac{1}{m+1}\)

Conclusion

The fraction of the area of the rectangle that lies between the curve \(y=kx^m\) and the x-axis is \(\frac{1}{m+1}\), which depends only on \(m\) and not on \(k\) or \(b\).

Key concepts

Area Under a Curve

The concept of the area under a curve is a fundamental aspect of integral calculus and has numerous applications in physics, engineering, and economics. When a function is graphed on a coordinate plane, the area under the curve between two points can be interpreted as the accumulated total of the dependent variable. For instance, if the curve represents the velocity of a car over time, the area under the curve would represent the total distance traveled.

In our exercise, we determine this area for the curve given by the equation of the function, within specified limits. This application not only helps in determining spatial quantities but also aids in understanding other concepts such as work done by a force or the probability of events in certain distributions. By integrating the function over the range of interest, we capture this area and turn an abstract mathematical concept into a concrete quantity that has both meaning and utility.

Definite Integration

Definite integration is a critical tool for computing the area under a curve. It is definitive in the sense that it calculates the accumulated value of the function within a certain interval - from the lower to the upper limit. In the exercise, we applied definite integration to find the area under the curve represented by the equation y = kx^m over the interval from 0 to b.

This operation follows a systematic approach: the integrand, kx^m in this case, is integrated concerning the variable x. Moreover, the constant multiplier k simplifies the process, as constants can be factored out of integrals and handled separately. The resulting function after integration is evaluated at the upper and lower limits, providing the area between the curve and the x-axis. Having a strong grasp of the process of definite integration is essential for problem-solving in calculus.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometric figures through a coordinate system, such as the Cartesian plane. This system allows for a precise representation and analysis of geometrical shapes using algebraic equations. In the context of our IIT JEE problem, the rectangle is positioned within a coordinate system with one vertex at the origin, and its diagonal reaches the curve y = kx^m.

The coordinate geometry framework enables us to easily calculate the area of such rectangles by simply multiplying the lengths of their sides - these lengths are determined by the coordinates of the vertices. This fusion of algebra and geometry simplifies complex problems and is a powerful tool in many fields, including physics, engineering, and computer science. It is instrumental in taking abstract mathematical relationships and translating them into real-world visualizations that can be easily comprehended and applied.