Question

A curve passes through the point \((2,0)\) and the slope of the tangent at any point \((x, y)\) is \(x^{2}-2 x\) for all values of \(x\). The point of local maximum on the curve is (1) \(\left(0, \frac{4}{3}\right)\) (2) \(\left(0, \frac{2}{3}\right)\) (3) \(\left(1, \frac{2}{3}\right)\) (4) \(\left(2, \frac{4}{3}\right)\)

Short answer

The point of local maximum is (0, \(\frac{4}{3}\)). Option (1).

Step-by-step solution

Understand the given

The curve passes through the point (2, 0). The slope of the tangent at any point (x, y) is given by the function \(\frac{dy}{dx} = x^2 - 2x\).

Find the general solution of the differential equation

To find the equation of the curve, integrate the slope function: \[\frac{dy}{dx} = x^2 - 2x \implies y = \int (x^2 - 2x) \, dx\]

Integrate the function

Integrate the function: \[\int (x^2 - 2x) \, dx = \frac{x^3}{3} - x^2 + C\]So, the equation of the curve is given by \( y = \frac{x^3}{3} - x^2 + C \).

Apply the initial condition

The curve passes through (2,0), so we plug these values into the equation to find C: \[ 0 = \frac{2^3}{3} - 2^2 + C \implies 0 = \frac{8}{3} - 4 + C \implies 0 = \frac{8}{3} - \frac{12}{3} + C \] \[ \implies 0 = \frac{-4}{3} + C \implies C = \frac{4}{3}\]. Therefore, the equation of the curve is:\[ y = \frac{x^3}{3} - x^2 + \frac{4}{3}\].

Find the critical points

To find the critical points, set the derivative equal to zero and solve for x:\[ \frac{dy}{dx} = x^2 - 2x = 0 \implies x(x-2) = 0 \implies x = 0, x = 2 \].

Determine the nature of the critical points

Check the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^2-2x) = 2x-2 \].For \(x = 0\):\[ \frac{d^2y}{dx^2} = 2(0)-2 = -2 < 0 \], so x = 0 is a point of local maximum.For \(x = 2\):\[ \frac{d^2y}{dx^2} = 2(2)-2 = 2 > 0 \], so x = 2 is a point of local minimum.

Find the corresponding y-value for the local maximum

Plug x = 0 back into the equation of the curve:\[ y = \frac{(0)^3}{3} - (0)^2 + \frac{4}{3} = \frac{4}{3} \].Thus, the coordinates of the point of local maximum are \( (0, \frac{4}{3}) \).

Key concepts

Differential Equations

In this problem, the relationship between the variables is described by a differential equation. A differential equation is an equation involving the derivatives of a function. Here, we have the slope of the tangent line given by \(\frac{dy}{dx}=x^2-2x\). This means the rate of change of y with respect to x depends on the values of x. To find the actual equation of the curve, we integrate this differential equation, which is a key aspect of solving such problems. By integrating \(\frac{dy}{dx} = x^2-2x\), we get \(\frac{x^3}{3}-x^2 + C\), where C is the constant of integration.

Local Maximum

A local maximum is a point where the function reaches a peak value in its immediate vicinity. To find these points, we first look for critical points by setting the first derivative equal to zero \((x^2-2x=0)\). Solving this, we get critical points at x = 0 and x = 2. To confirm which of these points is a maximum, we use the second derivative test. If the second derivative is negative, the point is a local maximum. In this case, \(\frac{d^2y}{dx^2}=2x-2\). Plugging in x=0, we get -2, indicating a local maximum. At this point, the curve reaches the highest point in its neighborhood.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse process of differentiation. In the context of differential equations, integrating the given \(\frac{dy}{dx}=x^2-2x\) allows us to find the general form of y. The integral of \((x^2-2x)\) with respect to x is \(\frac{x^3}{3}-x^2 + C\). Integration enables us to build the function from its rate of change. To fully determine the function, we use given conditions (like the curve passing through (2,0)) to find the constant of integration (C). In this problem, solving for C gave us the specific equation of the curve, y=\(\frac{x^3}{3}-x^2 + \frac{4}{3} \).