**Practice Problems: Calculus for Physics Solutions**

1. (easy) Determine the limit for each of the following:

a) lim (x - 8) as x → 4

As x gets closer to 4, x - 8 approaches -4

b) lim (x/2) as x → 10

As x gets closer to 10, x/2 approaches 5

c) lim (5x + 2) as x→ 3

As x gets closer to 3, 5x + 2 approaches 17

d) lim (4/x) as x → 0

As x gets closer to 0, 4/x approaches ∞

(This really means that there is no limit!)

2. (moderate) Determine the limit for each of the following:

a) lim [(x^{2} - 6x + 9)/(x - 3)] as x→3

Substitution causes an indeterminate form (0/0)

Factor the numerator:

[(x - 3)(x - 3)/(x - 3)]

cancel out the (x - 3)

new form: lim (x - 3) as x→3

As x gets closer to 3, (x - 3) approaches 0

b) lim[(x^{2} - 3x)/x^{2}] as x→0

Substitution causes an indeterminate form (0/0)

Factor the numerator:

x(x - 3)/x^{2}

cancel out an x

new form: lim (x - 3)/x as x→0

As x gets closer to 0 using positive numbers, (x - 3)/x approaches -∞.

As x gets closer to 0 using negative numbers, (x - 3)/x approaches ∞.

c) lim[(x^{2 }- 4)x/x^{2} as x→2

As x gets closer to 2, the numerator approaches 0 and the denominator approaches 4.

Thus the fraction gets closer and closer to the limit of 0

d) lim(3x^{3}/x^{4}) as x→∞

Simplify the fraction to be 3/x

As x gets closer to ∞, 3/x approaches the limit of 0.

3. (moderate) Find the derivative (dy/dx) of the following functions with respect to x.

a) y = 2x +5

dy/dx = 2

b) y = 3x^{2} + 7x

dy/dx = 6x + 7

c) y = 5cosx

dy/dx = -5sin(x)

d) y = 3/x^{2}dy/dx = -6/x^{3}e) y = (x + (1/x))(x - (1/x))

Simplify to: y = x^{2} - (1/x^{2})

dy/dx = 2x + (2/x^{3})

f) y = ln(x^{3})

dy/dx = (1/x^{3})(3x^{2}) = 3/x

g) y = 3e^{-2x }dy/dx = 3e^{-2x}(-2) = -6e^{-2x}

h) y = (6x)sin(2x)

Use the product rule:

dy/dx = (6x)(2cos(2x)) + sin(2x)(6)

Simplified:

dy/dx = 12xcos(2x) + 6sin(2x)

4. (moderate) Find the first, second, and third derivatives of the following functions wth respect to x:

a) y = 9x^{2} + 3x + 5

dy/dx = 18x + 3

d^{2}y/dx^{2} = 18

d^{3}y/dx^{3} = 0

b) y = 2/x^{5}dy/dx = -10/x^{6}d^{2}y/dx^{2} = 60/x^{7}d^{3}y/dx^{3} = -420/x^{8}

c) y = 10sin(6x)

dy/dx = 60cos(6x)

d^{2}y/dx^{2} = -360sin(6x)

d^{3}y/dx^{3} = -2160cos(6x)

5. (moderate) Given that y = x^{3} - 2x, find the slope one would measure on a graph of that function when:

a) x = 1

The slope is the value of the first derivative.

dy/dx = f'(x) = slope = 3x^{2} - 2

f'(1) = 3(1)^{2} - 2 = 1

b) x = -1

f'(1) = 3(-1)^{2} - 2 = 1

c) x = 2/3

f'(1) = 3(2/3)^{2} - 2 = -2/3

6. (moderate) Given that y = 3cos(4x), find the slope one would measure on a graph of that function when:

a) x = 2π radians

dy/dx = f'(x) = slope = -12sin(4x)

f'(2π) = -12sin(8π) = 0

b) x = 0

f'(0) = -12sin(0) = 0

c) x = π/8 radians

f'(π/8) = -12sin(π/2) = -12

(Note: Make sure that your calculator is in radians mode.)

7. (easy) Evaulate the indefinte integrals shown below:

a) ∫4xdx

∫4xdx = 2x^{2} + C

b) ∫(9x + 6)dx

∫(9x + 6)dx = 4.5x^{2} + 6x + C

c) ∫3x^{2}dx

∫3x^{2}dx = x^{3} + C

d) ∫(2/x)dx

∫(2/x)dx = 2ln(x) + C

e) ∫10e^{x}dx

∫10e^{x}dx = 10e^{x} + C

8. (moderate) Evaluate the indefinite integrals shown below:

a) ∫8e^{7x}dx

∫8e^{7x}dx = (8/7)e^{7x} + C

b) ∫(x^{3} + 6x^{2} + 4x + 8) dx

∫(x^{3} + 6x^{2} + 4x + 8) dx = x^{4}/4 + 2x^{3} + 2x^{2} + 8x + C

c) ∫(2x + 6)(x^{2} + 6x)^{2}dx (try u-substitution)

let u = x^{2} + 6x

du/dx = 2x + 6

du = (2x + 6)dx

Now substitute to get:

∫u^{2}du = u^{3}/3

Finally, reverse the substitution:

u^{3}/3 = (x^{2} + 6x)^{3}/3 + C

d) ∫(4 - x)^{4}dx (try u-substitution)

let u (4 - x)

du/dx = -1

du = -dx

Now substitute:

-∫(u)^{4}du = -u^{5}/5

Reverse the substitution:

-u^{5}/5 = -(4 - x)^{5}/5 + C

e) ∫(3x + 5)^{½} dx (try u-substitution)

let u (3x + 5)

du/dx = 3

du = 3dx

Now substitute:

(1/3)∫(u)^{½}du = (2/9)(u)^{3/2}

Reverse the substitution:

(2/9)(u)^{3/2}= (2/9)(3x + 5)^{3/2} + C

f) ∫e^{8x-6}dx (try u-substitution)

let u = 8x -6

du/dx = 8

du = 8dx

Now substitute:

(1/8)∫e^{u}du = (1/8)e^{u}

Reverse the substitution:

(1/8)e^{u }= (1/8)e^{8x - 6} + C

9. (easy) Evaluate the definite integrals shown below:

a) ∫5x^{2}dx (from x = 2 to x = 5)

∫5x^{2}dx = (5/3)x^{3} |^{5}_{2}

∫5x^{2}dx = (5/3)5^{3} - (5/3)2^{3} = 195

b) ∫12x^{-1}dx (from x = 3 to x = 10)

∫12x^{-1}dx = 12lnx |^{10}_{3}

∫12x^{-1}dx = 12(2.3 - 1.1)= 14.4

c) ∫5e^{x}dx (from x = -3 to x = -2)

∫5e^{x}dx = 5e^{x} |^{-2}_{-3}

∫5e^{x}dx = 5(e^{-2} - e^{-3}) = 0.43

d) ∫20sin(x)dx (from x = 0 to x = π/2) (radians mode on calculator, please)

∫20sin(x)dx = -20cos(x) |^{π/2}_{0}

∫20sin(x)dx = -20(cos(π/2) - cos0) = 20

10. (moderate) Evaluate the definite integrals shown below:

a) ∫xcos(x^{2} + 3)dx (from x = 4 to x = 6)

let u = x^{2} + 3

du/dx = 2x

du = 2xdx

Now substitute:

½∫cos(u)du = ½sin(u)

Reverse the substitution:

½sin(u) = ½sin(x^{2} + 3)|^{6}_{4}

½sin(x^{2} + 3)|^{6}_{4} = ½[sin(39) - sin(19)] = 0.41

b) ∫7cos(6x)dx (from x = 0 to x = π) (can use a u-substitution is you want)

let u = 6x

du/dx = 6

du = 6dx

Now substitute:

(1/6)∫7cos(u)du = (7/6)sin(u)

Reverse the substitution:

(7/6)sin(u) = (7/6)sin(6x) |^{π}_{0}

(7/6)sin(6x) |^{π}_{0 }= (7/6)(sin(6π) - sin0) = 0

c) ∫(sin(lnx))dx/x (from x = 10 to x = 15)

u = lnx

du/dx = 1/x

du = dx/x

Now substitute:

∫sin(u)du = -cos(u)

Reverse the substitution:

-cos(u)= -cos(lnx) |^{15}_{10}

-cos(lnx) |^{15}_{10} = -cos(ln15) - (-cos(ln10)) = 0.24

d) ∫(x + 4)dx/(x^{2} + 8x - 7) (from x = 2 to x = 3)

let u = x^{2} + 8x - 7

du/dx = 2x + 8

du = 2(x + 4) dx

By substitution:

∫(x + 4)dx/(x^{2} + 8x - 7) = ½∫(du/u) = ½ln(u)

Now, reverse the substitution:

½ln(u) = ½ln(x^{2} + 8x - 7)|^{3}_{2} = ½(3.26 - 2.56) = 0.35

11. (moderate) Find the change in y if x changes from x = 2 to x = 6 for these differential equations:

a) dy/dx = 3x^{2} + 7

dy = (3x^{2} + 7)dx

∫dy = ∫(3x^{2} + 7)dx

Δy = x^{3} + 7x |^{6}_{2}

Δy = 258 - 22 = 236

b) dy/dx = 1/x^{2}

dy = dx/x^{2}

∫dy = ∫dx/x^{2}

Δy = -1/x |^{6}_{2}

Δy = -(1/6) - (-1/2) = 0.33

c) dy/dx = (x + 1/x)(x - 1/x)

dy/dx = x^{2} - 1/x^{2}

dy = (x^{2} - 1/x^{2})dx

∫dy = ∫(x^{2} - 1/x^{2})dx

Δy = x^{3}/3 + 1/x |^{6}_{2}

Δy = 72.2 - 3.2 = 69.0

12. (moderate) Find y when x = 3 if y = 1 when x = 2 for:

dy/dx = x^{3}y^{2}

dy/y^{2} = x^{3}dx

∫dy/y^{2} = ∫x^{3}dx

-1/y |^{y2}_{1} = x^{4}/4 |^{3}_{2}

(-1/y_{2}) + 1 = (3^{4}/4 - 2^{4}/4)

y_{2} = -0.07