14I - Applications of differentiation
In this section we will cover the following applications of differentiation:
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■ Remember to always give answers in exact form unless otherwise stated. |
14I.1 - Tangents and normals
■ A tangent line touches a curve exactly once at the point \(P\). ■ A normal line is perpendicular to the tangent line at the point \(P\). To help determine the question of a tangent or normal line:
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14I.1 - Example 1: Calculating the gradients of tangent and normal lines
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14I.1 - Example 1: Video solution
14I.1 - Example 1: Practice
Question 1: Calculate the gradient of the tangent line at a point, given the gradient of the normal line is \(\frac{5}{3}\). Question 2: Calculate the gradient of the normal line at a point, given the gradient of the tangent line is \(-3\). 14I.1 - Example 1: Solutions
Question 1: \(m_T=\frac{-3}{5}\) Question 2: \(m_N=\frac{1}{3}\) |
14I.1 - Example 2: Determining the equation of a normal line to a curve (CAS)
Consider the curve \(f(x)=x^2-4x-12\). Determine the equation of the normal line to the curve when \(x=4\).
■ CAS Hint:The equation of a normal line to the curve \(y=f(x)\) at \(x=a\) can be found using Interactive → Calc → Line → normal. |
14I.1 - Example 2: Video solution
14I.1 - Example 2: Practice
Question 1: Determine the equation of a normal line to the curve of \(f(x)=x^2-6x\) at \(x=2\). Question 2: Determine the equation of a normal line to the curve of \(y=-\frac{x^3}{2}-\frac{3x^2}{2}\) at \(x=-1\). 14I.1 - Example 2: Solutions
Question 1: \(y=\frac{x}{2}-9\) Question 2: \(y=-\frac{2x}{3}-\frac{5}{3}\) |
14I.1 - Example 3: Determining the equations of tangents and normal lines
Consider the function \(f:[0,∞)→R,f(x)=2\sqrt{x}+1\).
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14I.1 - Example 3: Video solution
14I.1 - Example 3: Practice
Question 1:
The tangent line to the curve \(y=-x^2-6x-9\), at the point \(P\), is \(y=2x+7\).
Question 2:
Consider the function with rule \(f(x)=x^3-4x\).
14I.1 - Example 3: Solutions
Question 1:
Question 2:
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14I.2 - Optimisation
Optimisation problems involves maximising or minimising a variable in a relationship. This is often done by using differentiation to find stationary points for a function. Maximum and minimum applications build on the local maxima and minima we observed in Section 14G.
While many problems will be solved by finding maximum or minimum values associated with stationary points it is always important to test the endpoints of the practical domain as they may represent maximum or minimum values not calculated by using calculus. On the graphing screen of a CAS calculator we can use Analysis → G-Solve → fMin / fMax.
Setting the scene: Designing boxes (CAS exploration)
Open boxes can be folded out of 2D rectangles. In the CAS exploration we will examine the optimisation problem concerned with creating the largest possible box.
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Exploration Task
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Interactive model
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Other optimisations
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When tackling a maxima/minima question, use the following steps to help guide you through the problem:
- Wherever possible, draw a diagram and label all known information.
- Try to determine rules to represent the relationship between two variables. You may need to use known formula such as perimeter, area and volume. If appropriate, update you diagram with any new information.
- Identify the equation linking the variables to be maximised or minimised. Try to express in the form \(y=f(x)\).
- Find the derivative and then solve \(f'(x)=0\) for \(x\). This will provide information about stationary points.
- Verify that you have found the required maximum or minimum value. You can do this by sketching the graph of \(y=f(x)\) or by constructing a "sign" diagram (see Section 14G). At this point you should also check endpoints of the practical domain.
- Make sure you answer the question that has been asked. It is more than likely that you are asked for the maximum volume or area not just the value of \(x\) that results in this maximum or minimum point.
The following examples help to illustrate this process.
14I.2 - Example 1: Applications of differentiation to maximum and minimum problems
An open box is to be constructed from a piece of cardboard measuring 100 cm × 120 cm by cutting out four equal squares from each of the corners.
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14I.2 - Example 1: Video solution
14I.2 - Example 1: Practice
Question 1: ABC Question 2: ABC 14I.2 - Example 1: Solutions
Question 1: ABC Question 2: ABC |
14I.2 - Example 2: Applications of differentiation to maximum and minimum problems
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14I.2 - Example 2: Video solution
14I.2 - Example 2: Practice
Question 1: A 4 meter long piece of wire is cut and bent to form a circle and a square. How much wire should be in the square so that the combined area is minimised? 14I.2 - Example 2: Solutions
Question 1: Let \(x\) m be cut to form the square; therefore, \(4-x\) m is used to create the circle. \[\therefore A=(\frac{x}{4})^2+\pi (\frac{4-x}{2\pi})^2\] Next, solve \(A'(x)=0\) to find the maximum \[\therefore x=\frac{16}{\pi+4}\] Therefore, \(\frac{16}{\pi+4}\) metres for the square. |
14I.2 - Example 3: Applications of differentiation to maximum and minimum problems
Louie Gee launches a boat from point \(A\) on the bank of a uniform, straight river that is 4 km wide. He wants to reach point \(B\), which is located 10 km downstream on the opposite side of the river, as quickly as possible. He could row his boat directly across the river to point \(C\) and then run to \(B\), or he could row directly to \(B\), or he could row to some point \(D\), between \(C\) and \(B\), and then run to \(B\).
Louie Gee is able to row at a speed of 6 km/h and run at a speed of 8 km/h.
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14I.2 - Example 3: Video solution
14I.2 - Example 3: Practice
Question 1 [SQA, 2015]:
A crocodile is stalking prey located 20 meters further upstream on the opposite bank of a river. Crocodiles travel at different speeds on land and in water. The time taken for the crocodile to reach its prey can be minimised if it swims to a point \(P\), \(x\) meters upstream on the other side of the river as shown in the diagram below. The time taken, \(T\), measures in tenths of a second, is given by \[T=5\sqrt{36-x^2}+4(20-x)\]
14I.2 - Example 3: Solutions
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14I.2 - Example 4: Applications of differentiation to maximum and minimum problems
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14I.2 - Example 4: Video solution
14I.2 - Example 4: Practice
Question 1:
A rectangular prism is constructed with a square base, with side lengths \(x\) cm, and a height of \(h\) cm. The total volume of the rectangular prism is 1000 cubic centimeters. A diagram of the rectangular prism is shown to the right.
14I.2 - Example 4: Solutions
Question 1:
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14I.3 - Kinematics
Coming soon.
14I.4 - Rates of change
Rates of change were introduced previously in Section 13.
- The average rate of change is found by calculating the gradient of a line segment drawn between two points on a graph.
- The instantaneous rate of change is found by calculating the gradient of a tangent line drawn to a curve or function at a particular point.
14I.4 - Example 1: Applications of differentiation to rates of change (CAS)
A ball is dropped from a height of 12.25 metres. The height of the ball, \(h\) metres, after \(t\) seconds can be modelled by the equation \[h(t)=-4.9t^2+12.25\]
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14I.4 - Example 1: Video solution
14I.4 - Example 1: Practice
Question 1:
A spherical snowball is melting. As it melts, the surface area and volume decreases. The surface area of the snowball is \(S=4\pi r^2\) and the volume is modelled by the equation \(V=\frac{4}{3}\pi r^3\). The radius of the snowball, \(r\), is in centimeters.
14I.4 - Example 1: Solutions
Question 1:
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