Solve all this calculus question

Here are the solutions to the calculus questions:

1. Evaluate the following limits:

b) $\lim_{x \to 0} \frac{\sin 3x}{\tan 5x}$

Step 1: Direct substitution yields $\frac{\sin(0)}{\tan(0)} = \frac{0}{0}$, an indeterminate form.

Step 2: Apply L'Hôpital's Rule by differentiating the numerator and denominator.
The derivative of $\sin 3x$ is $3 \cos 3x$.
The derivative of $\tan 5x$ is $5 \sec^2 5x$.

$$\lim_{x \to 0} \frac{3 \cos 3x}{5 \sec^2 5x}$$

Step 3: Substitute $x=0$ into the new expression.

$$\frac{3 \cos(0)}{5 \sec^2(0)} = \frac{3 \cdot 1}{5 \cdot 1^2} = \frac{3}{5}$$

The limit is $\boxed{\frac{3}{5}}$.

c) $\lim_{r \to \infty} \frac{3r^3 - 2r}{r^3 + 5r^2 + 1}$

Step 1: Divide every term in the numerator and denominator by the highest power of $r$ in the denominator, which is $r^3$.

$$\lim_{r \to \infty} \frac{\frac{3r^3}{r^3} - \frac{2r}{r^3}}{\frac{r^3}{r^3} + \frac{5r^2}{r^3} + \frac{1}{r^3}}$$

Step 2: Simplify the expression.

$$\lim_{r \to \infty} \frac{3 - \frac{2}{r^2}}{1 + \frac{5}{r} + \frac{1}{r^3}}$$

Step 3: Apply the limit. As $r \to \infty$, terms of the form $\frac{c}{r^n}$ approach 0.

$$\frac{3 - 0}{1 + 0 + 0} = \frac{3}{1} = 3$$

The limit is $\boxed{3}$.

2. Differentiate from first principles:

a) $y = x^2 + 3x$

Step 1: Write the definition of the derivative from first principles.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step 2: Substitute $f(x) = x^2 + 3x$ and $f(x+h) = (x+h)^2 + 3(x+h)$.

$$f(x+h) - f(x) = [(x+h)^2 + 3(x+h)] - [x^2 + 3x]$$ $$= [x^2 + 2xh + h^2 + 3x + 3h] - [x^2 + 3x]$$ $$= 2xh + h^2 + 3h$$

Step 3: Substitute this into the limit and simplify.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} = \lim_{h \to 0} \frac{h(2x + h + 3)}{h} = \lim_{h \to 0} (2x + h + 3)$$

Step 4: Apply the limit as $h \to 0$.

$$\frac{dy}{dx} = 2x + 0 + 3 = 2x + 3$$

The derivative is $\boxed{2x + 3}$.

b) $y = \cos x$

Step 1: Write the definition of the derivative from first principles.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step 2: Substitute $f(x) = \cos x$ and $f(x+h) = \cos(x+h)$.

$$f(x+h) - f(x) = \cos(x+h) - \cos x$$

Step 3: Use the trigonometric identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

$$\cos(x+h) - \cos x = -2 \sin\left(\frac{2x+h}{2}\right) \sin\left(\frac{h}{2}\right) = -2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$

Step 4: Substitute into the limit and rearrange.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} = \lim_{h \to 0} \left[ -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$$

Step 5: Apply the limit, using $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$$\frac{dy}{dx} = -\sin(x + 0) \cdot 1 = -\sin x$$

The derivative is $\boxed{-\sin x}$.

3. Find $dy/dx$:

a) $y = (x^2 + 1)^5$

Step 1: Use the chain rule. Let $u = x^2 + 1$, so $y = u^5$.

Step 2: Differentiate $y$ with respect to $u$ and $u$ with respect to $x$.

$$\frac{dy}{du} = 5u^4$$ $$\frac{du}{dx} = 2x$$

Step 3: Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

$$\frac{dy}{dx} = 5u^4 \cdot 2x = 5(x^2 + 1)^4 \cdot 2x = 10x(x^2 + 1)^4$$

The derivative is $\boxed{10x(x^2 + 1)^4}$.

b) $y = x^3 e^{-2x}$

Step 1: Use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = x^3$ and $v = e^{-2x}$.

Step 2: Find the derivatives of $u$ and $v$.

$$u' = 3x^2$$ $$v' = -2e^{-2x}$$

Step 3: Apply the product rule.

$$\frac{dy}{dx} = (3x^2)(e^{-2x}) + (x^3)(-2e^{-2x})$$ $$\frac{dy}{dx} = 3x^2 e^{-2x} - 2x^3 e^{-2x}$$

Step 4: Factor out common terms.

$$\frac{dy}{dx} = x^2 e^{-2x}(3 - 2x)$$

The derivative is $\boxed{x^2 e^{-2x}(3 - 2x)}$.

c) $y = \ln(\cos x)$

Step 1: Use the chain rule. Let $u = \cos x$, so $y = \ln u$.

Step 2: Differentiate $y$ with respect to $u$ and $u$ with respect to $x$.

$$\frac{dy}{du} = \frac{1}{u}$$ $$\frac{du}{dx} = -\sin x$$

Step 3: Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

$$\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x$$

The derivative is $\boxed{-\tan x}$.

d) $y = \arctan(2x)$

Step 1: Use the chain rule. The derivative of $\arctan u$ is $\frac{1}{1+u^2} \frac{du}{dx}$. Let $u = 2x$.

Step 2: Find the derivative of $u$ with respect to $x$.

$$\frac{du}{dx} = 2$$

Step 3: Apply the chain rule.

$$\frac{dy}{dx} = \frac{1}{1+(2x)^2} \cdot 2 = \frac{2}{1+4x^2}$$

The derivative is $\boxed{\frac{2}{1+4x^2}}$.

e) $x^2 y + y^3 = 5$ (implicit)

Step 1: Differentiate both sides of the equation with respect to $x$. Remember to use the product rule for $x^2 y$ and the chain rule for $y^3$.

$$\frac{d}{dx}(x^2 y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5)$$

Step 2: Apply differentiation rules.

$$(2x)y + x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$$

Step 3: Group terms containing $\frac{dy}{dx}$ and solve for $\frac{dy}{dx}$.

$$(x^2 + 3y^2) \frac{dy}{dx} = -2xy$$ $$\frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}$$

The derivative is $\boxed{\frac{-2xy}{x^2 + 3y^2}}$.

f) $x = e^t \cos t$, $y = e^t \sin t$ (parametric)

Step 1: Use the formula for parametric differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

Step 2: Find $\frac{dx}{dt}$ using the product rule.

$$\frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t + e^t (-\sin t) = e^t(\cos t - \sin t)$$

Step 3: Find $\frac{dy}{dt}$ using the product rule.

$$\frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t)$$

Step 4: Substitute into the formula for $\frac{dy}{dx}$.

$$\frac{dy}{dx} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)}$$

Step 5: Simplify by canceling $e^t$.

$$\frac{dy}{dx} = \frac{\sin t + \cos t}{\cos t - \sin t}$$

The derivative is $\boxed{\frac{\sin t + \cos t}{\cos t - \sin t}}$.

4. Find and classify all stationary points of $y = 2r^3 - 9r^2 + 12r - 3$.

Step 1: Find the first derivative, $\frac{dy}{dr}$.

$$\frac{dy}{dr} = 6r^2 - 18r + 12$$

Step 2: Set the first derivative to zero to find critical points.

$$6r^2 - 18r + 12 = 0$$

Divide by 6:

$$r^2 - 3r + 2 = 0$$

Factor the quadratic equation:

$$(r-1)(r-2) = 0$$

The critical points are $r=1$ and $r=2$.

Step 3: Find the corresponding $y$ values for these critical points.
For $r=1$: $y = 2(1)^3 - 9(1)^2 + 12(1) - 3 = 2 - 9 + 12 - 3 = 2$. Point: $(1, 2)$.
For $r=2$: $y = 2(2)^3 - 9(2)^2 + 12(2) - 3 = 16 - 36 + 24 - 3 = 1$. Point: $(2, 1)$.

Step 4: Find the second derivative, $\frac{d^2y}{dr^2}$, to classify the stationary points.

$$\frac{d^2y}{dr^2} = 12r - 18$$

Step 5: Evaluate the second derivative at each critical point.
For $r=1$: $\frac{d^2y}{dr^2} = 12(1) - 18 = -6$. Since $-6 < 0$, $(1, 2)$ is a local maximum.
For $r=2$: $\frac{d^2y}{dr^2} = 12(2) - 18 = 6$. Since $6 > 0$, $(2, 1)$ is a local minimum.

The stationary points are:

• $(1, 2)$, which is a local maximum.
• $(2, 1)$, which is a local minimum.

5. A rectangular box with no lid has a square base of side $x$ and height $h$. Given that the total surface area is $75 \text{ cm}^2$, find the dimensions for maximum volume.

Step 1: Write expressions for the volume ($V$) and surface area ($A$).
Volume: $V = x^2 h$
Surface Area (base + 4 sides): $A = x^2 + 4xh$

Step 2: Use the given surface area ($A=75 \text{ cm}^2$) to express $h$ in terms of $x$.

$$75 = x^2 + 4xh \implies 4xh = 75 - x^2 \implies h = \frac{75 - x^2}{4x}$$

Step 3: Substitute $h$ into the volume equation to get $V(x)$.

$$V(x) = x^2 \left(\frac{75 - x^2}{4x}\right) = \frac{x(75 - x^2)}{4} = \frac{75x - x^3}{4} = \frac{75}{4}x - \frac{1}{4}x^3$$

Step 4: Find the first derivative of $V(x)$ and set it to zero to find critical points.

$$\frac{dV}{dx} = \frac{75}{4} - \frac{3}{4}x^2$$ $$\frac{75}{4} - \frac{3}{4}x^2 = 0 \implies 75 - 3x^2 = 0 \implies 3x^2 = 75 \implies x^2 = 25$$

Since $x$ must be positive, $x = 5 \text{ cm}$.

Step 5: Find the corresponding height $h$.

$$h = \frac{75 - (5)^2}{4(5)} = \frac{75 - 25}{20} = \frac{50}{20} = 2.5 \text{ cm}$$

Step 6: Use the second derivative test to confirm it's a maximum.

$$\frac{d^2V}{dx^2} = -\frac{6}{4}x = -\frac{3}{2}x$$

At $x=5$: $\frac{d^2V}{dx^2} = -\frac{3}{2}(5) = -\frac{15}{2}$. Since this is negative, it's a maximum.

The dimensions for maximum volume are $x = \boxed{5 \text{ cm}}$ and $h = \boxed{2.5 \text{ cm}}$.

6. Water leaks from a conical tank (vertex down) at $3 \text{ m}^3/\text{min}$. The tank has radius $2 \text{ m}$ and height $5 \text{ m}$. Find the rate at which the water level falls when the depth is $3 \text{ m}$.

Step 1: Identify given information and what to find.
Given: $\frac{dV}{dt} = -3 \text{ m}^3/\text{min}$ (negative because water is leaking).
Tank dimensions: $R = 2 \text{ m}$, $H = 5 \text{ m}$.
Find: $\frac{dh}{dt}$ when $h = 3 \text{ m}$.

Step 2: Write the volume of a cone: $V = \frac{1}{3}\pi r^2 h$.

Step 3: Use similar triangles to relate $r$ and $h$ for the water in the cone.

$$\frac{r}{h} = \frac{R}{H} = \frac{2}{5} \implies r = \frac{2}{5}h$$

Step 4: Substitute $r$ into the volume formula to express $V$ in terms of $h$.

$$V = \frac{1}{3}\pi \left(\frac{2}{5}h\right)^2 h = \frac{1}{3}\pi \left(\frac{4}{25}h^2\right) h = \frac{4\pi}{75}h^3$$

Step 5: Differentiate $V$ with respect to time $t$.

$$\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \frac{dh}{dt} = \frac{4\pi}{25}h^2 \frac{dh}{dt}$$

Step 6: Substitute the known values ($\frac{dV}{dt} = -3$, $h=3$) and solve for $\frac{dh}{dt}$.

$$-3 = \frac{4\pi}{25}(3)^2 \frac{dh}{dt}$$ $$$$