Question

MCV4UZ Unit 2 Test
For questions#1 to #3, find the first derivative of f(x).
1. 𝑓(𝑥) = (3𝑥 ^2 + 2)(2𝑥 + 3)
2. 𝑓(𝑥) = (𝑥^2 + 2) / (5𝑥^3+3)
3. 𝑓(𝑥) = (2𝑥^2 + 3𝑥 + 1)^2

4. A construction worker accidentally drops a hammer from a height of 90m while working on the roof of a new apartment building. The height, 𝑠, in metres, of the hammer after 𝑡 seconds can be modelled by: 𝑠(𝑡) = 90 − 4.9𝑡^2, t >0
a. Determine the velocity (vector quantity) at 1 second.
b. Determine velocity at 4 seconds.
c. When will the hammer hit the ground?
d. Determine the impact velocity (velocity of the hammer that hits the ground).

5. Two particles have positions at time 𝑡 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝑠1
(𝑡) = 4𝑡 − 𝑡 ^2 𝑎𝑛𝑑 𝑠2(𝑡) = 5𝑡^2 − 𝑡^3. Find the velocities 𝑣1(𝑡) 𝑎𝑛𝑑 𝑣2(𝑡) 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑠𝑡𝑎𝑛𝑡 𝑡ℎ𝑒 𝒂𝒄𝒄𝒆𝒍𝒆𝒂𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒘𝒐 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔 𝒂𝒓𝒆 𝒆𝒒𝒖𝒂𝒍.

Answer 1

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!