A curve is such that \(\frac{dy}{dx} = \frac{12}{\sqrt{4x + a}}\), where \(a\) is a constant. The point \(P(2, 14)\) lies on the curve and the normal to the curve at \(P\) is \(3y + x = 5\).
(i) Show that \(a = 8\).
(ii) Find the equation of the curve.
The equation of a curve is such that \(\frac{d^2y}{dx^2} = 2x - 1\). Given that the curve has a minimum point at (3, -10), find the coordinates of the maximum point.
A curve is such that \(\frac{dy}{dx} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\). The curve passes through the point \((4, \frac{2}{3})\).
(i) Find the equation of the curve.
(ii) Find \(\frac{d^2y}{dx^2}\).
(iii) Find the coordinates of the stationary point and determine its nature.
A curve is defined for \(x > 0\) and is such that \(\frac{dy}{dx} = x + \frac{4}{x^2}\). The point \(P(4, 8)\) lies on the curve.
(i) Find the equation of the curve.
(ii) Show that the gradient of the curve has a minimum value when \(x = 2\) and state this minimum value.
The gradient of a curve at the point \((x, y)\) is given by \(\frac{dy}{dx} = 2(x + 3)^{\frac{1}{2}} - x\). The curve has a stationary point at \((a, 14)\), where \(a\) is a positive constant.
(a) Find the value of \(a\).
(b) Determine the nature of the stationary point.
(c) Find the equation of the curve.
A curve is such that \(\frac{d^2y}{dx^2} = -4x\). The curve has a maximum point at (2, 12).
(i) Find the equation of the curve.
A point \(P\) moves along the curve in such a way that the \(x\)-coordinate is increasing at 0.05 units per second.
(ii) Find the rate at which the \(y\)-coordinate is changing when \(x = 3\), stating whether the \(y\)-coordinate is increasing or decreasing.
A curve \(y = f(x)\) has a stationary point at \(P(3, -10)\). It is given that \(f'(x) = 2x^2 + kx - 12\), where \(k\) is a constant.
A curve is such that \(\frac{dy}{dx} = 5 - \frac{8}{x^2}\). The line \(3y + x = 17\) is the normal to the curve at the point \(P\) on the curve. Given that the \(x\)-coordinate of \(P\) is positive, find
A function f is defined for x ∈ ℝ and is such that f'(x) = 2x − 6. The range of the function is given by f(x) ≥ −4.
A curve is such that \(\frac{dy}{dx} = \frac{3}{(1 + 2x)^2}\) and the point \((1, \frac{1}{2})\) lies on the curve.
(i) Find the equation of the curve.
(ii) Find the set of values of \(x\) for which the gradient of the curve is less than \(\frac{1}{3}\).
The equation of a curve is such that \(\frac{dy}{dx} = \frac{6}{\sqrt{3x - 2}}\). Given that the curve passes through the point \(P(2, 11)\), find
(i) the equation of the normal to the curve at \(P\),
(ii) the equation of the curve.
A curve is such that \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 6\) and the point (9, 2) lies on the curve.
(i) Find the equation of the curve.
(ii) Find the \(x\)-coordinate of the stationary point on the curve and determine the nature of the stationary point.
A curve is such that \(\frac{dy}{dx} = k - 2x\), where \(k\) is a constant.
(i) Given that the tangents to the curve at the points where \(x = 2\) and \(x = 3\) are perpendicular, find the value of \(k\). [4]
(ii) Given also that the curve passes through the point (4, 9), find the equation of the curve. [3]
A curve is such that \(\frac{dy}{dx} = 4 - x\) and the point \(P(2, 9)\) lies on the curve. The normal to the curve at \(P\) meets the curve again at \(Q\). Find
A curve is such that \(\frac{dy}{dx} = \frac{4}{\sqrt{6 - 2x}}\), and \(P(1, 8)\) is a point on the curve.
(i) The normal to the curve at the point \(P\) meets the coordinate axes at \(Q\) and at \(R\). Find the coordinates of the mid-point of \(QR\).
(ii) Find the equation of the curve.
A function \(f\) is defined for \(x > \frac{1}{2}\) and is such that \(f'(x) = 3(2x-1)^{\frac{1}{2}} - 6\).
A curve is such that \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\) and \(P(3, 3)\) is a point on the curve.
(i) Find the equation of the normal to the curve at \(P\), giving your answer in the form \(ax + by = c\).
(ii) Find the equation of the curve.
A curve is such that \(\frac{dy}{dx} = 3x^2 - 4x + 1\). The curve passes through the point (1, 5).
(i) Find the equation of the curve.
(ii) Find the set of values of \(x\) for which the gradient of the curve is positive.
The gradient at any point \((x, y)\) on a curve is \(\sqrt{1 + 2x}\). The curve passes through the point \((4, 11)\). Find
(i) the equation of the curve,
(ii) the point at which the curve intersects the y-axis.
A curve is such that \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\) and \(P(1, 5)\) is a point on the curve.
(i) The normal to the curve at \(P\) crosses the x-axis at \(Q\). Find the coordinates of \(Q\).
(ii) Find the equation of the curve.
(iii) A point is moving along the curve in such a way that the \(x\)-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the \(y\)-coordinate when \(x = 1\).