AP Calculus BC Practice Exam 1: answers and scoring guide

The question

Question: find the slope at t = 2, when x = t² + 1 and y = t³ − 3t.

Both x and y move with a third letter, t, like time.

1 dx/dt and dy/dt

The power rule, with t: dx/dt = 2t.

And dy/dt = 3t² − 3.

2 dy/dx

The slope dy over dx is dy over dt, divided by dx over dt: the dt's cancel.

So dy/dx is (3t² − 3)/(2t).

3 At the point

Put in t = 2: dy/dt is 9 and dx/dt is 4.

The slope is 9/4.

Careful!

dy goes on top: dy over dt, divided by dx over dt.

Upside down gives 4/9: wrong.

So the slope at t = 2 is 9/4.

Draw the curve: for each t, the point (x, y).

The point at t = 2.

4 Check

Let's check with a calculator: move t a tiny bit around 2, and divide the change in y by the change in x.

About 2.25: the slope 2.25 our rule gives.

Right!

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: at the point.

Last: check.

dx/dt = 2t = 4 and dy/dt = 3t² − 3 = 9 at t = 2.

Speed = √(4² + 9²) = √97 ≈ 9.849.

Distance = ∫₀²√((2t)² + (3t² − 3)²) dt.

With a calculator: ≈ 7.605.

Straight up or down means dx/dt = 0 and dy/dt ≠ 0.

dx/dt = 2t = 0 at t = 0, where dy/dt = −3 ≠ 0.

The point is (x(0), y(0)) = (1, 0); the particle is moving straight down there.

The question

Question: find the area inside the curve r = 2 + 2cos θ.

The area inside a curve written with r and θ.

1 Draw it

A polar grid: circles around the center, and a line for each angle.

For each angle θ, go out a distance r.

The curve closes up.

2 One trip around

The curve goes around once as θ goes from 0 to 2π.

3 Write the integral

Cut the area into thin slices from the center.

Each is almost a thin triangle: one half r squared dθ.

A slice of a circle with angle θ has area one half r squared θ.

Add up the slices: A is one half of the integral of r squared.

Careful!

Do not forget the one half!

A slice is half of r squared dθ.

The right way: A = 1/2∫ from 0 to 2π of (2 + 2cos θ)² dθ.

4 Square it

Square the bracket: first squared, two times first times second, second squared.

5 Use the half-angle rule

The half-angle rule: cos squared is 1 plus cos of double the angle, over 2.

Use it, then add the plain numbers: 4 plus 2 is 6.

6 Integrate

Integrate: a number gives the number times θ.

cos gives sin, divided by the number in front of θ; sin gives minus cos.

7 Top minus bottom

Every sin part is 0 at these limits: sin of a whole number of half turns is 0.

Only the θ part is left: 6 times the length of the interval, 2π.

So the area is 6π square units, about 18.85.

8 Check

Let's check with a calculator: it adds up 1000 thin slices.

That gives about 18.85, the same as our answer.

Right!

Let's look back.

First: draw it.

Then: one trip around.

Then: write the integral.

Then: square it.

Then: use the half-angle rule.

Then: integrate.

Then: top minus bottom.

Last: check.

dr/dθ = −2sin θ, which is −2 at θ = π/2.

dr/dθ < 0, so r is getting smaller: the point is moving closer to the origin.

x = r cos θ and y = r sin θ. At θ = π/2: r = 2 and dr/dθ = −2.

dy/dθ = (dr/dθ)sin θ + r cos θ = −2 × 1 + 2 × 0 = −2.

dx/dθ = (dr/dθ)cos θ − r sin θ = 0 − 2 = −2.

dy/dx = −2−2 = 1.

C is outside r = 2 when cos θ > 0, for −π/2 < θ < π/2.

Area = ½∫_−π/2^π/2((2 + 2cos θ)² − 2²) dθ = 8 + π ≈ 11.142.

The question

Question: find the Maclaurin polynomial of degree 3 of e to the -1x.

The Maclaurin series: the nth derivative at 0, divided by n factorial, times x to the n.

Maclaurin means a = 0.

Why: we want a polynomial with the same value, slope, bend and so on at 0 as f.

Take the derivative of c times x to the n, n times: you get n factorial times c.

At 0 every other term is 0 (or gone after n derivatives), so only this one is left.

So c must be the nth derivative over n factorial.

1 The derivatives

We need f and its derivatives at 0.

Start with f: put in x = 0.

f(0) is 1.

The derivative of e to the kx is k times e to the kx.

2 Divide by n factorial

Divide each value by n factorial: 0!

= 1, 1!

= 1, 2!

= 2, 3!

= 6, 4!

= 24.

Careful!

Do not forget to divide by n factorial.

The x squared term is f″ over 2 factorial.

3 Write the series

Put them together: each number times x to its power.

Stop at the power 3.

4 Check

Let's check close to 0, at x = 0.1.

Our terms give about 0.90483.

A calculator gives 0.90484 for e to the -1x.

Almost the same.

Right!

Let's look back.

First: the derivatives.

Then: divide by n factorial.

Then: write the series.

Last: check.

P₃(0.1) = 1 − 0.1 + 0.012 − 0.0016 = 1 − 0.1 + 0.005 − 0.000167 ≈ 0.904833.

The error of an alternating series is at most the first term left out: (0.1)⁴/4! = 0.0001/24.

0.000124 ≈ 0.0000042, which is less than 0.00001.

Multiply each term of the series of e^−x by x.

x − x² + x³2 − x⁴/6.

The question

Question: use Euler's method on dy/dx = x + y, with y(0) = 1, step 0.5, to find y(1).

1 The idea

We cannot solve it exactly here.

So we walk: from a point, follow its slope for a short step h.

Then look at the slope again.

The slope is how much y goes up for each 1 to the right.

So for a step h, y goes up h times the slope: new y = old y + h times the slope.

2 The step size

From x = 0 to x = 1, in steps of 0.5: that is 2 steps.

3 Go to x = 0.5

The slope at our point (0, 1): put x and y into x + y.

It is 1.

New y = old y + h times the slope: 1 + 0.5 × 1, which is 1.5.

Now we are at x = 0.5.

Careful!

Times the step size h!

We only move 0.5 to the right, so y changes by 0.5 times the slope.

The right way: y₁ = 1 + 0.5 × 1 = 1.5.

4 Go to x = 1

The slope at (0.5, 1.5) is 2.

New y: 1.5 + 0.5 × 2 = 2.5.

All the points in a table.

On a graph:

Euler's steps: straight lines, each with the slope at its start.

The true solution, in green, bends away a little.

So y(1) ≈ 2.5.

5 Check

Let's check: a calculator with very tiny steps gives about 3.4366.

Close to our 2.5.

Euler is a bit low: the slope changes during each step, and each step uses only the slope at its start.

Smaller steps get closer.

Right!

Let's look back.

First: the idea.

Then: the step size.

Then: go to x = 0.5.

Then: go to x = 1.

Last: check.

d²y/dx² = 1 + dy/dx = 1 + x + y.

At (0, 1): 1 + 0 + 1 = 2, which is positive, so the curve is concave up there.

dy/dx = 2e^x − 1, and x + y = x + 2e^x − x − 1 = 2e^x − 1: the same, so it solves the equation.

At x = 0: y = 2 − 0 − 1 = 1, so f(0) = 1.

f(1) = 2e − 2 ≈ 3.437.

The estimate 2.5 is too small: the curve is concave up, so each Euler step stays under it.

The question

Question: find the integral of x cos x.

1 Choose u and dv

We will trade this integral for a new one with du in it.

So u should be the part that gets simpler when we take its derivative.

Choose u = x: its derivative is simpler, just a number.

dv is the rest: cos x dx, which we can integrate.

2 Find du and v

du: the derivative of u, times dx.

v: the integral of dv.

cos gives sin.

3 Use the rule

The rule: the integral of u dv is u v, minus the integral of v du.

u times v, minus the integral of v times du.

4 The new integral

The new integral is easy: just a number times sin.

sin integrates to minus cos.

Careful!

It is u v minus the integral of v du.

Not plus!

Open the bracket: a minus in front of a bracket changes the sign of every part inside.

Add C.

So the answer is x sin x + cos x + C.

5 Check

Let's check: take the derivative of our answer.

The product rule on each term: the first's slope times the second, plus the first times the second's slope.

The other terms cancel, and we get x cos x: the question again.

Right!

Let's look back.

First: choose u and dv.

Then: find du and v.

Then: use the rule.

Then: the new integral.

Last: check.

The question

Question: find the integral from 0 to infinity of e⁻³ˣ.

1 Why it is improper

Draw the curve.

The area under it goes on to the right for ever: the top limit is infinity.

Infinity is not a number we can put in.

So we stop at a letter, t, and then let t grow without end.

2 Use a limit

The integral is the limit of the integral from 0 to t, as t goes to infinity.

3 Integrate

e to the minus 3x integrates to itself over minus 3.

4 Let t go

Put in t, then take away the same with 0.

As t grows, e to the minus t is 1/a huge number: it goes to 0.

The first part goes to 0.

e to the 0 is 1.

5 Decide

The limit is a number: the integral converges to 1/3.

An area that goes on for ever, but adds up to 1/3!

Careful!

Never put infinity in like a number.

We let t grow and watch where the answer goes.

6 Check

Let's check with a calculator: the integral up to t = 1000 is about 0.3333.

Up to a million, about 0.3333.

Closer and closer to 0.333.

Right!

Let's look back.

First: why it is improper.

Then: use a limit.

Then: integrate.

Then: let t go.

Then: decide.

Last: check.

∫₁^b1x dx = ln b − ln 1 = ln b.

As b grows without bound, ln b grows without bound.

So the integral diverges.

The question

Question: find for which x the series converges: the sum of (x − 1)ⁿ over (n 2ⁿ).

A series with x in it: for which x does it converge?

1 The ratio

The terms have powers of x: the ratio test handles powers well.

L less than 1 converges, more than 1 diverges, exactly 1 cannot tell.

x can be negative, so we take sizes, with bars.

Divide the next term by this one.

(x − 1) to the n + 1, over (x − 1) to the n, leaves one (x − 1).

The powers of 2 leave 1/2.

The powers of n leave n/(n + 1).

2 The limit

n/(n + 1) goes to 1.

So L is 1/2 times the size of x minus 1.

3 The radius

It converges when L is less than 1.

Multiply both sides by 2.

So the radius of convergence is 2: x can be up to 2 away from 1.

Careful!

Not 1/2!

L is 1/2 times the size, so the size must be less than 2.

Open up the bars: x is between −1 and 3.

4 The ends

At x = 3: (x − 1) is 2, and its powers cancel the powers of 2.

That is the sum of 1/n: it diverges.

At x = −1: the same, but with signs that take turns.

The signs take turns and the sizes 1/n go to 0: it converges, by the alternating series test.

5 The interval

−1 is in: a filled dot.

3 is not in: an open dot.

Shade the x's in between.

In interval notation: [−1, 3).

A square bracket means the end is in; a round one, it is not.

6 Check

Let's check an x inside, 2, with a calculator: 40 terms and 80 terms add up to almost the same, 0.6931.

It converges there.

Right!

Let's look back.

First: the ratio.

Then: the limit.

Then: the radius.

Then: the ends.

Then: the interval.

Last: check.

At x = −1 the terms are (−2)ⁿ/(n × 2ⁿ) = (−1)ⁿ/n.

Σ (−1)ⁿ/n converges by the alternating series test (1n shrinks to 0).

Σ 1n, the harmonic series, diverges: so it converges conditionally.

At x = 2 the terms are 1/(n × 2ⁿ): 12, 18, 1/24.

12 + 18 + 124 = 1224 + 324 + 124 = 1624 = 2/3.