L2 Calculus Deep Dive
This course is for students aiming for Merit and Excellence. It explores complex applications of calculus, assuming a prior understanding of basic differentiation.
Course Contents
1. Advanced Tangent Problems
Beyond finding a simple tangent, Merit and Excellence questions often require you to work backwards. You might be given the gradient of a tangent and asked to find where it touches the curve, or you might be given a line and asked to prove it's a tangent. The key is always to use the derivative, $f'(x)$, as the link between the function and the gradient of the tangent.
Worked Example: A tangent to the curve $f(x) = x^2 + 2x - 1$ has a gradient of 6. Find the equation of this tangent.
Solution
Step 1 (Find where the gradient occurs): First, find the derivative of the function, which tells you the gradient at any point $x$.
$$f'(x) = 2x + 2$$We are told the gradient is 6, so we set the derivative equal to 6 and solve for $x$.
$$2x + 2 = 6 \implies 2x = 4 \implies x = 2$$This tells us the tangent touches the curve at the point where $x=2$.
Step 2 (Find the point of tangency): Now we find the y-coordinate of this point by substituting $x=2$ back into the original function.
$$f(2) = (2)^2 + 2(2) - 1 = 4 + 4 - 1 = 7$$The point of tangency is $(2, 7)$.
Step 3 (Find the equation of the line): We have the gradient $m=6$ and a point $(2, 7)$. Using the point-gradient formula $y - y_1 = m(x - x_1)$:
$$y - 7 = 6(x - 2) \implies y - 7 = 6x - 12 \implies y = 6x - 5$$Test Your Knowledge
Question: Find the coordinates of the point(s) on the graph of the function $y = 4x^3 - 4x + 4$ where the tangent to the curve is parallel to the line $y - 8x + 6 = 0$.
The line $y - 8x + 6 = 0$ can be rewritten as $y = 8x - 6$. Its gradient is 8. A parallel tangent must also have a gradient of 8.
Find the derivative of the curve: $\frac{dy}{dx} = 12x^2 - 4$.
Set the derivative equal to 8: $12x^2 - 4 = 8 \implies 12x^2 = 12 \implies x^2 = 1 \implies x = 1$ or $x = -1$.
Find the y-coordinates: When $x=1$, $y = 4(1)^3 - 4(1) + 4 = 4$. Point is $(1, 4)$. When $x=-1$, $y = 4(-1)^3 - 4(-1) + 4 = -4 + 4 + 4 = 4$. Point is $(-1, 4)$.
Question: There are two points on the graph of the function $f(x) = x^3 - 3x^2 - 4x$ where the tangent to the graph passes through the origin $(0,0)$. Find the coordinates of these two points.
Let a point of tangency be $(a, f(a))$. The gradient of the tangent at this point is $f'(a)$.
The gradient of the line from the origin $(0,0)$ to $(a, f(a))$ is $\frac{f(a) - 0}{a - 0} = \frac{a^3 - 3a^2 - 4a}{a} = a^2 - 3a - 4$.
The derivative is $f'(x) = 3x^2 - 6x - 4$. So, $f'(a) = 3a^2 - 6a - 4$.
At the point of tangency, these two gradients must be equal: $a^2 - 3a - 4 = 3a^2 - 6a - 4$.
This simplifies to $2a^2 - 3a = 0 \implies a(2a - 3) = 0$. So, $a=0$ or $a=1.5$.
Find the coordinates: When $a=0$, $f(0)=0$. Point is $(0,0)$. When $a=1.5$, $f(1.5) = (1.5)^3 - 3(1.5)^2 - 4(1.5) = -9.375$. Point is $(1.5, -9.375)$.
2. Stationary Points and Their Nature
Stationary points (or turning points) are crucial for understanding the shape of a graph. They occur where the gradient is zero ($f'(x) = 0$). To determine their nature (local maximum, local minimum, or point of inflection), we use the second derivative test. Find the second derivative, $f''(x)$, and substitute the x-value of the stationary point:
- If $f''(x) > 0$, the point is a local minimum (concave up).
- If $f''(x) < 0$, the point is a local maximum (concave down).
- If $f''(x) = 0$, the test is inconclusive (it could be a point of inflection).
Worked Example: The graph of the function $y = x^3 - 6x^2 + kx - 5$ has a turning point at $x=3$. Find the coordinates of both turning points and determine their nature.
Solution
Step 1 (Find k): The derivative is $\frac{dy}{dx} = 3x^2 - 12x + k$. Since there is a turning point at $x=3$, the gradient must be zero at that point.
$$3(3)^2 - 12(3) + k = 0 \implies 27 - 36 + k = 0 \implies k = 9$$Step 2 (Find all turning points): Now we know the derivative is $\frac{dy}{dx} = 3x^2 - 12x + 9$. Set it to zero to find all stationary points.
$$3(x^2 - 4x + 3) = 0 \implies 3(x-1)(x-3) = 0$$The turning points are at $x=1$ and $x=3$.
Step 3 (Determine their nature): Find the second derivative: $\frac{d^2y}{dx^2} = 6x - 12$.
At $x=1$, $\frac{d^2y}{dx^2} = 6(1) - 12 = -6$. Since it's negative, this is a local maximum.
At $x=3$, $\frac{d^2y}{dx^2} = 6(3) - 12 = 6$. Since it's positive, this is a local minimum.
Step 4 (Find coordinates): Substitute the x-values back into the original function $y = x^3 - 6x^2 + 9x - 5$.
At $x=1$, $y = 1 - 6 + 9 - 5 = -1$. The maximum is at $(1, -1)$.
At $x=3$, $y = 27 - 54 + 27 - 5 = -5$. The minimum is at $(3, -5)$.
Test Your Knowledge
Question: Find the local minimum value of the function $y = x^3(x-4)$. Justify your answer.
First, expand the function: $y = x^4 - 4x^3$.
Find the derivative: $\frac{dy}{dx} = 4x^3 - 12x^2$.
Set to zero: $4x^2(x-3) = 0$. Stationary points are at $x=0$ and $x=3$.
Find the second derivative: $\frac{d^2y}{dx^2} = 12x^2 - 24x$.
Test $x=3$: $\frac{d^2y}{dx^2} = 12(3)^2 - 24(3) = 108 - 72 = 36$. Since this is positive, $x=3$ is a local minimum.
The minimum value is $y = (3)^4 - 4(3)^3 = 81 - 108 = -27$.
Question: For the function $f(x) = \frac{2}{3}x^3 + \frac{3}{2}x^2 - 20x - 3$, find the values of $x$ for which the function is decreasing.
A function is decreasing when its derivative is negative, i.e., $f'(x) < 0$.
The derivative is $f'(x) = 2x^2 + 3x - 20$.
We need to solve $2x^2 + 3x - 20 < 0$. Factoring the quadratic gives $(2x-5)(x+4) < 0$.
The roots are at $x = 2.5$ and $x = -4$. Since the parabola of the derivative opens upwards, the function is negative between these roots.
The function is decreasing for $-4 < x < 2.5$.
3. Optimisation Problems
Optimisation is the process of finding the maximum or minimum value of a quantity (like area, volume, or profit) given certain constraints. The process involves creating a function for the quantity you want to optimise, differentiating it, setting the derivative to zero, and solving.
Worked Example: A parcel in the shape of a rectangular cuboid with a square cross section ($x$ by $x$) is to be sent. The sum of the length, $l$, and the perimeter of the square cross section is to be 100 cm. Find the maximum possible volume.
Solution
Step 1 (Formulate Equations): The perimeter of the cross-section is $4x$. The constraint is $l + 4x = 100$. From this, we can express the length in terms of $x$: $l = 100 - 4x$.
The volume of the cuboid is $V = (\text{width}) \times (\text{height}) \times (\text{length}) = x \cdot x \cdot l = x^2l$.
Step 2 (Create Volume Function): Substitute the expression for $l$ into the volume formula to get a function of a single variable, $x$.
$$V(x) = x^2(100 - 4x) = 100x^2 - 4x^3$$Step 3 (Differentiate and Solve): Find the derivative of the volume function and set it to zero to find stationary points (which correspond to max/min volume).
$$V'(x) = 200x - 12x^2$$ $$200x - 12x^2 = 0 \implies 4x(50 - 3x) = 0$$The solutions are $x=0$ (which gives zero volume) and $x = \frac{50}{3} \approx 16.67$ cm.
Step 4 (Find Maximum Volume): Substitute $x = \frac{50}{3}$ back into the volume function.
$$V(\frac{50}{3}) = 100(\frac{50}{3})^2 - 4(\frac{50}{3})^3 \approx 9259 \text{ cm³}$$Test Your Knowledge
Question: A rectangle ABCD measures 10 cm by 8 cm. A parallelogram EFGH is drawn inside it such that $AE=BF=CG=DH=x$. Find the smallest possible area of the parallelogram.
The area of the parallelogram is the area of the large rectangle minus the area of the four corner triangles.
Area of rectangle = $10 \times 8 = 80$.
The four triangles form two pairs. Area of two triangles = $2 \times \frac{1}{2}x(8-x) = 8x-x^2$. Area of other two = $2 \times \frac{1}{2}x(10-x) = 10x - x^2$.
Area of parallelogram $A(x) = 80 - (8x - x^2 + 10x - x^2) = 80 - 18x + 2x^2$.
To find the minimum, set the derivative to zero: $A'(x) = -18 + 4x = 0 \implies 4x = 18 \implies x = 4.5$.
Minimum area is $A(4.5) = 80 - 18(4.5) + 2(4.5)^2 = 80 - 81 + 40.5 = 39.5$ cm².