MCV4UZ Unit 4 Test 1

Question

Show your work for all the questions.
Find the first derivative on questions #1 to #3 and Show your work:
1) 𝑦 = 2𝑠𝑖𝑛^(3)𝑥 − 4𝑐𝑜𝑠^(2)x
2) 𝑦 = 𝑥𝑒^𝑥
3) 𝑦 = 7ln (𝑥 ^3 + 2)

4) An apple orchard now has 80 trees per hectare and the average yield is 400 apples per tree. For each additional tree planted per hectare the average yield per tree is reduced by approximately four apples. How many trees per hectare will give the largest crop of apples? Show your work.

5) The position of a particle as it moves horizontally is described by the given equation 𝑠 = 𝑠𝑖𝑛𝑡 − 𝑐𝑜𝑠𝑡 0 ≤ 𝑡 ≤ 2𝜋. If s is the displacement in meters and 𝑡 is the time in seconds, find the maximum and minimum displacements. Show your work.

6) Show that a rectangle with given perimeter in meters has maximum area in 𝑚𝑒𝑡𝑒𝑟^2 when it is a square. Show your work.

Answer 1

Sure! Let's go through each question step-by-step.

1. Derivative of $y=2{\sin }^3(x)-4{\cos }^2(x)$

To find the derivative, we'll use the chain rule and the power rule.

First, differentiate each term:

• For $2{\sin }^3(x)$:

  • Let $u=\sin (x)$, so $y=2u^3$
  • $\frac{dy}{du}=6u^2$
  • Using the chain rule, $\frac{du}{dx}=\cos (x)$
  • So, $\frac{dy}{dx}=6{\sin }^2(x)\cos (x)$

• For $-4{\cos }^2(x)$:

  • Let $v=\cos (x)$, so $y=-4v^2$
  • $\frac{dy}{dv}=-8v$
  • Using the chain rule, $\frac{dv}{dx}=-\sin (x)$
  • So, $\frac{dy}{dx}=-8\cos (x)(-\sin (x))=8\cos (x)\sin (x)$

Now, combine the derivatives: $$\frac{dy}{dx}=6{\sin }^2(x)\cos (x)+8\cos (x)\sin (x)=\cos (x)(6{\sin }^2(x)+8\sin (x))$$

2. Derivative of $y=x{e}^x$

To differentiate this, we'll use the product rule, which states $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$.

Let $u=x$ and $v=e^x$:

• $u^{\prime }=1$
• $v^{\prime }=e^x$

Using the product rule: $$\frac{dy}{dx}=u^{\prime }v+u{v}^{\prime }=1\cdot e^x+x\cdot e^x=e^x+x{e}^x=e^x(1+x)$$

3. Derivative of $y=7\ln (x^3+2)$

We'll use the chain rule again. The derivative of $\ln (u)$ is $\frac{1}{u}\cdot \frac{du}{dx}$.

Let $u=x^3+2$:

• $\frac{du}{dx}=3x^2$

So applying the chain rule: $$\frac{dy}{dx}=7\cdot \frac{1}{x^3+2}\cdot 3x^2=\frac{21x^2}{x^3+2}$$

4. Maximizing the Crop Yield

Let $n$ be the number of trees per hectare. The number of apples per tree decreases by 4 for each additional tree planted. The yield can be expressed as: $$\text{Yield}=n(400-4(n-80))$$

Expanding this: $$\text{Yield}=n(400-4n+320)=n(720-4n)$$

The yield function is a quadratic equation given by: $$A=720n-4n^2$$ This is a downward-opening parabola. The maximum occurs at the vertex, where $n=-\frac{b}{2a}=-\frac{720}{2\cdot -4}=90$.

The maximum yield occurs at 90 trees per hectare.

5. Finding Maximum and Minimum Displacements

Given $s(t)=\sin (t)-\cos (t)$.

To find the maximum and minimum, we need to differentiate and set the derivative to zero.

$$\frac{ds}{dt}=\cos (t)+\sin (t)$$ Setting this equal to zero: $$\cos (t)+\sin (t)=0\Rightarrow \sin (t)=-\cos (t)\Rightarrow \tan (t)=-1$$

This occurs for $t=\frac{3\pi }{4}+k\pi$ within $[0,2\pi ]$:

• At $t=\frac{3\pi }{4}$: $$s\left(\frac{3\pi }{4}\right)=\sin \left(\frac{3\pi }{4}\right)-\cos \left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$$

• At $t=\frac{7\pi }{4}$: $$s\left(\frac{7\pi }{4}\right)=\sin \left(\frac{7\pi }{4}\right)-\cos \left(\frac{7\pi }{4}\right)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=0$$

So the maximum is $\sqrt{2}$ meters and minimum is $0$ meters.

6. Maximum Area Rectangle with Given Perimeter

Let the sides of the rectangle be $x$ and $y$. The perimeter $P=2x+2y$ is given, so: $$y=\frac{P}{2}-x$$

To find the area $A$: $$A=xy=x(\frac{P}{2}-x)=\frac{P}{2}x-x^2$$

This is a quadratic equation in the form $A=-x^2+\frac{P}{2}x$, which opens downward (as coefficient of $x^2$ is negative), so the maximum occurs at the vertex: $$x=-\frac{b}{2a}=-\frac{\frac{P}{2}}{2\cdot -1}=\frac{P}{4}$$ Then, $$y=\frac{P}{4}$$

Thus, both sides are equal, meaning the rectangle is a square.

Conclusion: A rectangle with the given perimeter has maximum area when it is a square.