MCV4UZ Unit 2 Test

Let's tackle each question step by step.

Questions 1 to 3: Finding the First Derivative

1. For \( f(x) = (3x^2 + 2)(2x + 3) \)

   Using the product rule \( (uv)' = u'v + uv' \):

   • Let \( u = 3x^2 + 2 \) and \( v = 2x + 3 \).
   • Compute \( u' = 6x \) and \( v' = 2 \).

   Now applying the product rule: \[ f'(x) = u'v + uv' = (6x)(2x + 3) + (3x^2 + 2)(2) \] \[ = 12x^2 + 18x + 6x^2 + 4 = 18x^2 + 18x + 4 \]

   So, the first derivative is: \[ f'(x) = 18x^2 + 18x + 4 \]

2. For \( f(x) = \frac{x^2 + 2}{5x^3 + 3} \)

   Using the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \):

   • Let \( u = x^2 + 2 \) and \( v = 5x^3 + 3 \).
   • Compute \( u' = 2x \) and \( v' = 15x^2 \).

   Now applying the quotient rule: \[ f'(x) = \frac{(2x)(5x^3 + 3) - (x^2 + 2)(15x^2)}{(5x^3 + 3)^2} \] Simplifying the numerator: \[ = \frac{10x^4 + 6x - (15x^4 + 30x^2)}{(5x^3 + 3)^2} \] \[ = \frac{-5x^4 - 30x^2 + 6x}{(5x^3 + 3)^2} \] Hence, the first derivative is: \[ f'(x) = \frac{-5x^4 + 6x - 30x^2}{(5x^3 + 3)^2} \]

3. For \( f(x) = (2x^2 + 3x + 1)^2 \)

   Using the chain rule:

   • Let \( u = 2x^2 + 3x + 1 \).
   • Then \( f(x) = u^2 \) and \( f'(x) = 2u \cdot u' \).

   First, compute \( u' \): \[ u' = 4x + 3 \]

   Now substituting back: \[ f'(x) = 2(2x^2 + 3x + 1)(4x + 3) \] This gives us: \[ f'(x) = (8x^2 + 12x + 2)(4x + 3) \]

Question 4: Hammer Drop

The height of the hammer after \( t \) seconds is given by: \[ s(t) = 90 - 4.9t^2 \]

a. Velocity at 1 second: \[ v(t) = \frac{ds}{dt} = -9.8t \] So, at \( t = 1 \): \[ v(1) = -9.8 \text{ m/s} \]

b. Velocity at 4 seconds: \[ v(4) = -9.8(4) = -39.2 \text{ m/s} \]

c. When will the hammer hit the ground? Set \( s(t) = 0 \): \[ 90 - 4.9t^2 = 0 \implies 4.9t^2 = 90 \implies t^2 = \frac{90}{4.9} \implies t \approx 4.27 \text{ seconds} \]

d. Impact velocity (velocity at \( t = 4.27 \)): \[ v(4.27) \approx -9.8(4.27) \approx -41.0 \text{ m/s} \]

Question 5: Equal Accelerations

The positions are given by: \[ s_1(t) = 4t - t^2 \quad \text{and} \quad s_2(t) = 5t^2 - t^3 \]

Finding velocities: \[ v_1(t) = \frac{ds_1}{dt} = 4 - 2t \] \[ v_2(t) = \frac{ds_2}{dt} = 10t - 3t^2 \]

Finding accelerations: \[ a_1(t) = \frac{dv_1}{dt} = -2 \] \[ a_2(t) = \frac{dv_2}{dt} = 10 - 6t \]

Set accelerations equal: \[ -2 = 10 - 6t \implies 6t = 12 \implies t = 2 \]

Finding velocities at \( t = 2 \): \[ v_1(2) = 4 - 2(2) = 0 \] \[ v_2(2) = 10(2) - 3(2^2) = 20 - 12 = 8 \]

So, the results are:

• \( t = 2 \)
• \( v_1(2) = 0 \)
• \( v_2(2) = 8 \)

These are the calculations for the above questions. Please let me know if you need further assistance!