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 using positive numbers, 4/x approaches ∞.
As x gets closer to 0 using negative numbers, 4/x approaches -∞.
(This really means that there is no limit!)

2. (moderate) Determine the limit for each of the following:
a) lim [(x² - 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[(x² - 3x)/x²] as x→0
Substitution causes an indeterminate form (0/0)
Factor the numerator:
x(x - 3)/x²
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[(x² - 4)x/x² 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(3x³/x⁴) 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 = 3x² + 7x
dy/dx = 6x + 7
c) y = 5cosx
dy/dx = -5sin(x)
d) y = 3/x²
dy/dx = -6/x³
e) y = (x + (1/x))(x - (1/x))
Simplify to: y = x² - (1/x²)
dy/dx = 2x + (2/x³)
f) y = ln(x³)
dy/dx = (1/x³)(3x²) = 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 = 9x² + 3x + 5
dy/dx = 18x + 3
d²y/dx² = 18
d³y/dx³ = 0
b) y = 2/x⁵
dy/dx = -10/x⁶
d²y/dx² = 60/x⁷
d³y/dx³ = -420/x⁸
c) y = 10sin(6x)
dy/dx = 60cos(6x)
d²y/dx² = -360sin(6x)
d³y/dx³ = -2160cos(6x)

5. (moderate) Given that y = x³ - 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 = 3x² - 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 = 2x² + C
b) ∫(9x + 6)dx
∫(9x + 6)dx = 4.5x² + 6x + C
c) ∫3x²dx
∫3x²dx = x³ + C
d) ∫(2/x)dx
∫(2/x)dx = 2ln(x) + C
e) ∫10e^xdx
∫10e^xdx = 10e^x + C

8. (moderate) Evaluate the indefinite integrals shown below:
a) ∫8e^7xdx
∫8e^7xdx = (8/7)e^7x + C
b) ∫(x³ + 6x² + 4x + 8) dx
∫(x³ + 6x² + 4x + 8) dx = x⁴/4 + 2x³ + 2x² + 8x + C
c) ∫(2x + 6)(x² + 6x)²dx (try u-substitution)
let u = x² + 6x
du/dx = 2x + 6
du = (2x + 6)dx
Now substitute to get:
∫u²du = u³/3
Finally, reverse the substitution:
u³/3 = (x² + 6x)³/3 + C
d) ∫(4 - x)⁴dx (try u-substitution)
let u (4 - x)
du/dx = -1
du = -dx
Now substitute:
-∫(u)⁴du = -u⁵/5
Reverse the substitution:
-u⁵/5 = -(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)^½du = (2/9)(u)^3/2
Reverse the substitution:
(2/9)(u)^3/2= (2/9)(3x + 5)^3/2 + C
f) ∫e^8x-6dx (try u-substitution)
let u = 8x -6
du/dx = 8
du = 8dx
Now substitute:
(1/8)∫e^udu = (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) ∫5x²dx (from x = 2 to x = 5)
∫5x²dx = (5/3)x³ |⁵₂
∫5x²dx = (5/3)5³ - (5/3)2³ = 195
b) ∫12x^-1dx (from x = 3 to x = 10)
∫12x^-1dx = 12lnx |¹⁰₃
∫12x^-1dx = 12(2.3 - 1.1)= 14.4
c) ∫5e^xdx (from x = -3 to x = -2)
∫5e^xdx = 5e^x |^-2_-3
∫5e^xdx = 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/2₀
∫20sin(x)dx = -20(cos(π/2) - cos0) = 20

10. (moderate) Evaluate the definite integrals shown below:
a) ∫xcos(x² + 3)dx (from x = 4 to x = 6)
let u = x² + 3
du/dx = 2x
du = 2xdx
Now substitute:
½∫cos(u)du = ½sin(u)
Reverse the substitution:
½sin(u) = ½sin(x² + 3)|⁶₄ 
½sin(x² + 3)|⁶₄ = 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) |^π₀
(7/6)sin(6x) |^π₀ = (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) |¹⁵₁₀
-cos(lnx) |¹⁵₁₀ = -cos(ln15) - (-cos(ln10)) = 0.24
d) ∫(x + 4)dx/(x² + 8x - 7) (from x = 2 to x = 3)
let u = x² + 8x - 7
du/dx = 2x + 8
du = 2(x + 4) dx
By substitution:
∫(x + 4)dx/(x² + 8x - 7) = ½∫(du/u) = ½ln(u)
Now, reverse the substitution:
½ln(u) = ½ln(x² + 8x - 7)|³₂ = 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 = 3x² + 7
dy = (3x² + 7)dx
∫dy = ∫(3x² + 7)dx
Δy = x³ + 7x |⁶₂
Δy = 258 - 22 = 236
b) dy/dx = 1/x²
dy = dx/x²
∫dy = ∫dx/x²
Δy = -1/x |⁶₂
Δy = -(1/6) - (-1/2) = 0.33
c) dy/dx = (x + 1/x)(x - 1/x)
dy/dx = x² - 1/x²
dy = (x² - 1/x²)dx
∫dy = ∫(x² - 1/x²)dx
Δy = x³/3 + 1/x |⁶₂
Δy = 72.2 - 3.2 = 69.0

12. (moderate) Find y when x = 3 if y = 1 when x = 2 for:
dy/dx = x³y²
dy/y² = x³dx
∫dy/y² = ∫x³dx
-1/y |^y2₁ = x⁴/4 |³₂
(-1/y₂) + 1 = (3⁴/4 - 2⁴/4)
y₂ = -0.07