Calculus：練習問題

Find the gradient of the tangent to the curve \(y = x^3 - 2x + 1\) at the point where \(x = 2\).

Calculate the area bounded by the curve \(y = 4 - x^2\), the \(x\)-axis, and the lines \(x=0\) and \(x=1\).

A rectangular field is to be enclosed by 200 m of fencing. If the dimensions of the field are \(x\) metres and \(y\) metres, find the maximum possible area of the field.

Differentiate \(y = (2x - 3)^5\) with respect to \(x\).

The velocity, \(v\) ms\(^{-1}\), of a particle moving in a straight line at time \(t\) seconds is given by \(v = t^3 - 6t^2 + 5t\). Find the acceleration of the particle when \(t = 3\).

Find the gradient of the curve \( y = 3x^2 - 5x + 2 \) at the point where \( x = 2 \).

Find the coordinates of the stationary point on the curve \( y = x^2 - 6x + 5 \) and determine its nature.

Differentiate \( f(x) = \sin(4x) + e^{2x} \) with respect to \( x \).

The curve \( y = (2x - 3)^5 \) intersects the \( y \)-axis at the point \( P \).
(a) Find the coordinates of \( P \).
(b) Find an expression for \( \frac{dy}{dx} \).
(c) Find the equation of the tangent to the curve at the point where \( x = 2 \).

A particle moves in a straight line such that, \( t \) seconds after passing a fixed point \( O \), its velocity, \( v \text{ m s}^{-1} \), is given by \( v = 8e^{2t} - 12 \).
(a) Find the initial velocity of the particle.
(b) Find the acceleration of the particle when \( t = \ln 2 \).
(c) Find the distance travelled by the particle in the third second (from \( t = 2 \) to \( t = 3 \)). Give your answer in terms of \( e \).