L2 Calculus Crash Course

This course is designed to rapidly build your core calculus skills. It focuses on the key concepts needed to confidently achieve Merit and Achieved grades in your exams.

Question Bank Deep Dive Course

Course Contents

  1. Finding Gradients
  2. Equations of Tangents
  3. Stationary Points
  4. Antidifferentiation (Integration)
  5. Kinematics (Motion)

1. Finding Gradients

The gradient of a curve at a specific point tells you how steep it is at that exact instant. To find it, you first need to find the derivative (also called the gradient function), which is a general formula for the gradient at any point $x$. The process of finding the derivative is called differentiation.

Achieved

Worked Example: A function is given by $f(x) = x^3 - 6x + 2$. Find the gradient of the function at the point where $x=4$.

Solution

Step 1 (Differentiate): Find the derivative, $f'(x)$, by applying the power rule to each term.

$$f'(x) = 3x^2 - 6$$

Step 2 (Substitute): Substitute the given x-value ($x=4$) into the derivative to find the gradient at that point.

$$f'(4) = 3(4)^2 - 6 = 3(16) - 6 = 48 - 6 = 42$$

Answer: The gradient at $x=4$ is 42.

Test Your Knowledge

Achieved

Question: For the function $h(x) = 0.5x^2 + 3x - 1$, find the x-coordinate of the point on the graph where the gradient is 5.

First, find the gradient function: $h'(x) = x + 3$.

Set the gradient function equal to 5 and solve for $x$: $x + 3 = 5 \implies x = 2$.

2. Equations of Tangents

A tangent is a straight line that just touches a curve at a single point and has the same gradient as the curve at that point. To find its equation, you need the gradient ($m$) and a point $(x_1, y_1)$ that it passes through.

Merit

Worked Example: Find the equation of the tangent to the curve $y = x^2 + 5x$ at the point $(2, 14)$.

Solution

Step 1 (Find Gradient Function): Differentiate to find the formula for the gradient.

$$\frac{dy}{dx} = 2x + 5$$

Step 2 (Find Gradient at Point): Substitute the x-coordinate of the point ($x=2$) into the gradient function to find the gradient of the tangent, $m$.

$$m = 2(2) + 5 = 9$$

Step 3 (Use Point-Gradient Formula): Use the formula $y - y_1 = m(x - x_1)$ with the point $(2, 14)$ and the gradient $m=9$.

$$y - 14 = 9(x - 2)$$

Step 4 (Simplify): Rearrange the equation into the form $y = mx + c$.

$$y - 14 = 9x - 18 \implies y = 9x - 4$$

Here is the graph of the curve and its tangent:

Graph of the curve y=x^2+5x and its tangent at (2,14)

Test Your Knowledge

Merit

Question: Use calculus to show that the line $y = x + 3.25$ is a tangent to the graph of the function $f(x) = 3x^2 - 2x + 4$.

The gradient of the line $y=x+3.25$ is $m=1$. We need to find where the curve has this gradient.

$f'(x) = 6x - 2$. Set this equal to 1: $6x - 2 = 1 \implies 6x = 3 \implies x = 0.5$.

Now find the y-value on the curve at $x=0.5$: $f(0.5) = 3(0.5)^2 - 2(0.5) + 4 = 0.75 - 1 + 4 = 3.75$. The point of tangency is $(0.5, 3.75)$.

Finally, check if this point lies on the line: $y = 0.5 + 3.25 = 3.75$. Since it does, the line is tangent to the curve at that point.

3. Stationary Points (Turning Points)

A stationary point is a point on a curve where the gradient is zero. These are the local maximums, minimums, and points of inflection. To find them, you set the derivative equal to zero and solve for $x$.

Merit

Worked Example: A skyrocket's height is given by $h(t) = 39.2t - 4.9t^2$. What is the maximum height the skyrocket will reach?

Solution

Step 1 (Differentiate): Find the derivative, which represents the velocity (rate of change of height).

$$h'(t) = 39.2 - 9.8t$$

Step 2 (Set to Zero): The maximum height occurs when the velocity is momentarily zero. Set the derivative to zero and solve for $t$.

$$39.2 - 9.8t = 0 \implies 9.8t = 39.2 \implies t = 4 \text{ seconds}$$

Step 3 (Find Maximum Value): Substitute this time back into the original height equation.

$$h(4) = 39.2(4) - 4.9(4)^2 = 156.8 - 78.4 = 78.4 \text{ metres}$$

Answer: The maximum height reached is 78.4 metres.

Test Your Knowledge

Excellence

Question: The function $f(x) = 2x^3 + kx^2 + 5$ has a minimum turning point when $x=1$. What are the coordinates of the maximum turning point?

First, find the derivative: $f'(x) = 6x^2 + 2kx$.

A turning point at $x=1$ means $f'(1) = 0$: $6(1)^2 + 2k(1) = 0 \implies 6 + 2k = 0 \implies k = -3$.

The derivative is $f'(x) = 6x^2 - 6x = 6x(x-1)$. The turning points are at $x=0$ and $x=1$.

Since $x=1$ is the minimum, the maximum must be at $x=0$.

The original function is $f(x) = 2x^3 - 3x^2 + 5$. Find the y-value at the maximum: $f(0) = 2(0)^3 - 3(0)^2 + 5 = 5$.

The maximum turning point is at (0, 5).

4. Antidifferentiation (Integration)

Antidifferentiation (or integration) is the reverse process of differentiation. If you have a gradient function, you can integrate it to find the original function. Remember to always add the constant of integration, "+ C".

Achieved

Worked Example: The gradient function of a curve is given by $\frac{dy}{dx} = 3x^2 - 5$. The curve passes through the point (1, 0). Find the equation of the curve.

Solution

Step 1 (Integrate): Find the indefinite integral of the gradient function. Raise the power of each term by one and divide by the new power.

$$y = \int (3x^2 - 5) \,dx = \frac{3x^3}{3} - 5x + C = x^3 - 5x + C$$

Step 2 (Find C): Use the given point $(1, 0)$ to find the value of the constant of integration, C. Substitute $x=1$ and $y=0$ into the equation.

$$0 = (1)^3 - 5(1) + C \implies 0 = 1 - 5 + C \implies 0 = -4 + C \implies C = 4$$

Step 3 (Write Final Equation): Substitute the value of C back into the integrated function.

$$y = x^3 - 5x + 4$$

5. Kinematics (Motion)

Kinematics is the study of motion. In calculus, we use a hierarchy: Displacement ($s$), Velocity ($v$), and Acceleration ($a$). You differentiate to move down the list and integrate to move up.

$s(t)$
$v(t)$
$a(t)$
Merit

Worked Example: An object has a constant acceleration of -4 cm/s². When timing begins, it is 12 cm from a point P and moving away from P with a velocity of 6 cm/s. Find the maximum distance of the object from P.

Solution

Step 1 (Find Velocity Function): Integrate acceleration to get velocity. $a(t) = -4$.

$$v(t) = \int -4 \,dt = -4t + C$$

At $t=0$, velocity is 6, so $6 = -4(0) + C \implies C=6$. The velocity function is $v(t) = -4t + 6$.

Step 2 (Find Time of Max Distance): Maximum distance occurs at the turning point, which is when velocity is zero.

$$-4t + 6 = 0 \implies 4t = 6 \implies t = 1.5 \text{ seconds}$$

Step 3 (Find Displacement Function): Integrate velocity to get displacement. $s(t) = \int (-4t + 6) \,dt = -2t^2 + 6t + D$.

At $t=0$, the object is 12 cm from P, so $12 = -2(0)^2 + 6(0) + D \implies D=12$. The displacement function is $s(t) = -2t^2 + 6t + 12$.

Step 4 (Calculate Max Distance): Substitute the time ($t=1.5$) into the displacement function.

$$s(1.5) = -2(1.5)^2 + 6(1.5) + 12 = -4.5 + 9 + 12 = 16.5 \text{ cm}$$