AHL 2.11 Derivatives: exam-style practice · AI HL · AHL 2: Further Derivatives

AHL 2.11

Derivatives: exam-style practice

AHL 2.11

Derivatives: exam-style practice

By the end of this section

Key points

Models fitted to data

Use the calculator's quadratic regression for a, b and c. Give 3 significant figures.
R² close to 1 means the model fits the data well. It does not prove the model is valid outside the data.
Comment on validity with a number: a negative a means the graph turns down and later goes below 0; c = f(0) must make sense in context.
f′(x) is a rate of change. Interpret it with its units, such as "hundreds of birds per year".
To test a model, compare f′ with the rate in the table: change in value ÷ change in x.

Dimensions, volume and surface area

Write a dimension from the words ("3 more than" means add 3), then write volume = base area × height.
Put the volume equation as polynomial = 0 and use the polynomial solver. A missing power has coefficient 0.
Cuboid surface area: add the area of all six faces. Cylinder: 2πr² for the ends and 2πrh for the side.

Optimisation

Use the volume to remove h, so that A depends on r only. Write 1/r as r⁻¹ before you differentiate.
Minimum: set dA/dr = 0, solve for r, then put r back into A (not into dA/dr).
HL: if d²A/dr² > 0 for all r > 0, the graph is concave up, so a stationary point is a minimum. Give the reason, then the conclusion.

Further derivatives (HL)

Product rule: (uv)′ = u′v + uv′. The derivative of e^(kx) is ke^(kx).
A point of inflexion is where d²y/dx² = 0 and the sign changes. e^(kx) is never 0, so solve the other factor.
Logistic model: P = K/(1 + me^(−kn)) rises to the limit K. Differentiate with the chain rule.
x^(kx) = e^(kx ln x). Then differentiate with the chain rule and the product rule.

On the calculator

Quadratic regression

Enter the x-values and y-values in two lists.
Choose quadratic regression and read a, b and c.
Read R² from the same output.

Screenshot and video to come

Polynomial solver for a volume

Write the volume equation as polynomial = 0.
Enter the coefficients from the highest power down, with 0 for a missing power.
Keep the real, positive root.

Screenshot and video to come

Solving dA/dr = 0

Enter the equation dA/dr = 0 in the solver, or solve by hand and take the cube root.
Use the root as r, then evaluate A at that r.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.

Fit a model, then test it

A bird reserve counts its population of a rare species (hundreds of birds) every 10 years. The data are in the table. Let x be the number of years after 1960 and let f(x) = ax² + bx + c be the quadratic regression model. (a) Find a, b and c to 3 significant figures. (b) The coefficient of determination is R² = 0.9808. Does R² support a quadratic model? (c) Comment on the validity of the model with reference to c. (d) Find f′(60) and interpret it. Use the table to comment on the model.

Working

Regression
From the calculator: quadratic regression: a = −0.0357, b = 6.06, c = −5.57

Enter the x-values and the counts in two lists. Give 3 significant figures. The model is \(f(x)=-0.0357x^2+6.06x-5.57\).
R²
R² = 0.9808 is close to 1

Yes, it supports the model: almost all of the variation in the counts is explained by the quadratic. R² only says the model fits the points it was built from.
A parameter
c = −5.57, so f(0) = −5.57

The model gives a negative population in 1960, which is impossible. The data show 800 birds. The model is poor near the start.
Derivative
\(f'(x)=2ax+b=-0.0714x+6.06\); f′(60) = 1.77

Substitute x = 60 into the derivative.
Interpret
In 2020 the population is growing at 1.77 hundred birds per year.

Rate of change, with units: hundreds of birds per year.
Compare
Table, 2010 to 2020: (220 − 215) ÷ 10 = 0.5 hundred per year

The model's rate 1.77 is more than three times the rate in the data. The population is levelling off and the model still predicts fast growth, so it is not valid for later years.

Box dimensions from a volume

A lunch box has a square base of side l cm. Its height d cm is 3 cm more than the side of the base. The box holds 200 cm³. (a) Write down an expression for d in terms of l. (b) Find l and d. (c) Find the total external surface area of the box.

Working

The height
\(d=l+3\)

"3 cm more than the side" means add 3 to l.
The equation
\(l^2(l+3)=200\), so \(l^3+3l^2-200=0\)

Volume = base area × height. Write the equation with 0 on one side and enter 1, 3, 0, −200 (no l term).
Solve
Polynomial solver: l = 5 (the other two roots are complex)

Only a real, positive root can be a length.
The height
\(d=5+3=8\) cm

Put l back into d = l + 3.
Surface area
\(2(5\times5+5\times8+5\times8)=210\) cm²

Two squares (top and base) and four rectangles: 2(l² + 2ld).

The cheapest can

A tin holds 450 cm³ and is a closed cylinder of radius r cm and height h cm. Let A cm² be its total surface area. (a) Find h in terms of r. (b) Show that \(A=2\pi r^2+\frac{900}{r}\). (c) Find \(\frac{dA}{dr}\). (d) Find the value of r that minimises A, and the minimum value of A.

Working

Height
\(\pi r^2h=450\), so \(h=\frac{450}{\pi r^2}\)

Volume of a cylinder is πr²h. Make h the subject.
Area
\(A=2\pi r^2+2\pi rh=2\pi r^2+2\pi r\times\frac{450}{\pi r^2}=2\pi r^2+\frac{900}{r}\)

Two circular ends and the curved surface 2πrh. The π and one r cancel.
Differentiate
\(A=2\pi r^2+900r^{-1}\), \(\frac{dA}{dr}=4\pi r-900r^{-2}\)

Write 900/r as a power first.
Stationary point
\(4\pi r-\frac{900}{r^2}=0\), \(r^3=\frac{225}{\pi}\), r = 4.15

Set dA/dr = 0. Multiply by r², then take the cube root. Or use the calculator's solver.
Minimum area
A = 2π(4.15)² + 900 ÷ 4.15 = 325 cm²

Substitute r back into A, not into dA/dr.

Is it a minimum?

An open box has a square base of side x m and volume 500 m³. Its surface area is \(A=x^2+\frac{2000}{x}\) for x > 0. The area is stationary when x = 10. (a) Find \(\frac{d^2A}{dx^2}\). (b) Show that the graph of A is concave up for all x > 0, and say what this tells you about x = 10.

Working

First derivative
\(A=x^2+2000x^{-1}\), \(\frac{dA}{dx}=2x-2000x^{-2}\)

Write the fraction as a power.
Second derivative
\(\frac{d^2A}{dx^2}=2+4000x^{-3}=2+\frac{4000}{x^3}\)

Differentiate again. −2000 × (−2) = +4000.
Sign
x > 0, so 4000/x³ > 0, so d²A/dx² > 2 > 0

A positive number added to 2 is positive. This works for every x > 0, so the graph is concave up.
Conclusion
At x = 10: d²A/dx² = 2 + 4 = 6 > 0

A stationary point on a concave up graph is a minimum. State the reason, then the conclusion.

Product rule with an exponential

Consider the curve \(y=x(3-e^{2x})\). (a) Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\). (b) The curve has a point of inflexion at x = a. Find a.

Working

Set up
\(u=x,\; v=3-e^{2x},\; u'=1,\; v'=-2e^{2x}\)

The derivative of e^(2x) is 2e^(2x): the chain rule gives the 2.
First derivative
\(\frac{dy}{dx}=1(3-e^{2x})+x(-2e^{2x})=3-e^{2x}-2xe^{2x}\)

Product rule: u′v + uv′.
Second derivative
\(\frac{d^2y}{dx^2}=-2e^{2x}-2e^{2x}-4xe^{2x}=-4(1+x)e^{2x}\)

The term −2xe^(2x) needs the product rule again. Collect like terms.
Inflexion
\(-4(1+a)e^{2a}=0\), so \(a=-1\)

Put d²y/dx² = 0. e^(2a) is never 0, so 1 + a = 0. The sign of d²y/dx² changes at a = −1, so it is an inflexion.

A logistic population

Fish are put into a lake. The population, P thousand, n years later is modelled by \(P=\frac{12}{1+5e^{-0.4n}}\). (a) Find \(\frac{dP}{dn}\) and explain why the population always grows. (b) Find the initial population and the initial rate of growth. (c) Find the value of P as n becomes very large. (d) Sketch P against n.

Working

Derivative
\(P=12(1+5e^{-0.4n})^{-1}\), \(\frac{dP}{dn}=-12(1+5e^{-0.4n})^{-2}\times(-2e^{-0.4n})=\frac{24e^{-0.4n}}{(1+5e^{-0.4n})^2}\)

Chain rule: the inside is 1 + 5e^(−0.4n), whose derivative is −2e^(−0.4n).
Always growing
24e^(−0.4n) > 0 and the denominator is a square, so dP/dn > 0

A positive number divided by a positive number is positive, so P is always increasing.
At n = 0
P(0) = 12 ÷ (1 + 5) = 2 thousand; P′(0) = 24 ÷ 36 = 0.667 thousand per year

e⁰ = 1.
Large n
e^(−0.4n) → 0, so P → 12 ÷ 1 = 12

The population levels off at 12 thousand, the limit.
Sketch
Start at (0, 2), always rising, levelling off at P = 12.

Mark the starting value and the horizontal line P = 12.

A variable power

(a) Show that, for x > 0, \(x^{3x}=e^{3x\ln x}\). (b) Hence find the equation of the tangent to the curve \(y=x^{3x}\) at x = 1.

Working

Rewrite
\(x=e^{\ln x}\), so \(x^{3x}=(e^{\ln x})^{3x}=e^{3x\ln x}\)

Any positive number can be written as e to a power. Multiply the powers.
Differentiate
\(\frac{dy}{dx}=e^{3x\ln x}\times\left(3\ln x+3\right)\)

Chain rule: e^u has derivative e^u × u′. Product rule on u = 3x ln x gives u′ = 3 ln x + 3.
At x = 1
\(y=1^3=1\) and \(\frac{dy}{dx}=1\times(0+3)=3\)

ln 1 = 0 and e⁰ = 1.
Tangent
\(y-1=3(x-1)\), so \(y=3x-2\)

Gradient 3 through (1, 1).

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Further Derivatives

Derivatives: exam-style practice

By the end of this section

Key points

Models fitted to data

Dimensions, volume and surface area

Optimisation

Further derivatives (HL)

On the calculator

Quadratic regression

Polynomial solver for a volume

Solving dA/dr = 0

Worked examples

Fit a model, then test it

Working

Box dimensions from a volume

Working

The cheapest can

Working

Is it a minimum?

Working

Product rule with an exponential

Working

A logistic population

Working

A variable power

Working

Choose a skill

Quiz