Graph of f(x) = x²

The graph of f(x) = x² is a parabola that opens upwards, with its vertex at the origin (0, 0). It is symmetric about the y-axis. Key points include (0,0), (1,1), (-1,1), (2,4), and (-2,4).

• Domain: The domain of f(x) = x² is all real numbers, as there are no restrictions on the values x can take.
  Domain: R or (-∞, ∞)
• Range: Since x² is always non-negative, the smallest value f(x) can take is 0 (when x=0). The function increases as |x| increases.
  Range: [0, ∞)

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Evaluation of the Limit

To evaluate the limit:
lim (x→1) sin⁻¹((1 - √x) / (1 - x))

Let the inner function be g(x) = (1 - √x) / (1 - x).
First, evaluate the limit of the inner function as x approaches 1:
lim (x→1) [(1 - √x) / (1 - x)]

Direct substitution yields (1 - √1) / (1 - 1) = 0/0, which is an indeterminate form.
To resolve this, simplify the expression by factoring the denominator. The denominator (1 - x) can be recognized as a difference of squares: (1)² - (√x)² = (1 - √x)(1 + √x).

g(x) = (1 - √x) / [(1 - √x)(1 + √x)]

Since x → 1, x ≠ 1, so (1 - √x) ≠ 0. Therefore, we can cancel the common term (1 - √x) from the numerator and denominator:

g(x) = 1 / (1 + √x)

Now, evaluate the limit of the simplified inner function:
lim (x→1) [1 / (1 + √x)]
Substitute x = 1:
= 1 / (1 + √1)
= 1 / (1 + 1)
= 1/2

Finally, apply the outer function, sin⁻¹, which is continuous on its domain [-1, 1]. This allows us to pass the limit inside:

lim (x→1) sin⁻¹(g(x)) = sin⁻¹(lim (x→1) g(x))
= sin⁻¹(1/2)

The value y for which sin(y) = 1/2 in the principal range of sin⁻¹ (i.e., [-π/2, π/2]) is π/6.

Therefore:
lim (x→1) sin⁻¹((1 - √x) / (1 - x)) = π/6

The function f(x)=∣x∣ differentiable? Discuss.

To determine where the function f(x) = |x| is differentiable, we analyze its derivative using the definition and considering different intervals.

The function f(x) = |x| can be defined piecewise as:
f(x) = x, if x ≥ 0
f(x) = -x, if x < 0

1. Differentiability for x > 0:
For x > 0, f(x) = x.
The derivative is f'(x) = d/dx (x) = 1.
Thus, f(x) is differentiable for all x > 0.

2. Differentiability for x < 0:
For x < 0, f(x) = -x.
The derivative is f'(x) = d/dx (-x) = -1.
Thus, f(x) is differentiable for all x < 0.

3. Differentiability at x = 0:
To check differentiability at x = 0, we must evaluate the left-hand derivative and the right-hand derivative at this point. A function is differentiable at a point if and only if both one-sided derivatives exist and are equal.

• Left-hand derivative at x = 0:
  f'-(0) = lim (h→0-) [f(0+h) - f(0)] / h
  = lim (h→0-) [|h| - |0|] / h
  Since h→0-, h < 0, so |h| = -h.
  = lim (h→0-) [-h - 0] / h
  = lim (h→0-) [-h] / h
  = -1

• Right-hand derivative at x = 0:
  f'+(0) = lim (h→0+) [f(0+h) - f(0)] / h
  = lim (h→0+) [|h| - |0|] / h
  Since h→0+, h > 0, so |h| = h.
  = lim (h→0+) [h - 0] / h
  = lim (h→0+) [h] / h
  = 1

Since the left-hand derivative at x = 0 (which is -1) is not equal to the right-hand derivative at x = 0 (which is 1), the function f(x) = |x| is not differentiable at x = 0.

Conclusion:
The function f(x) = |x| is differentiable for all real numbers except at x = 0.
Graphically, the function f(x) = |x| has a sharp corner (or cusp) at x = 0, where a unique tangent line cannot be defined, indicating non-differentiability at that point.

A farmer has 1200 m of fencing and wants to fence off a rectangular field that borders a straight river. He does not need to fence along the river. What are the dimensions of the field that has the largest area?

1. Define Variables:
Let l be the length of the rectangular field parallel to the river.
Let w be the width of the rectangular field perpendicular to the river.

2. Formulate Equations:

• Fencing Constraint (Perimeter): The farmer has 1200 m of fencing. Since one side borders the river, only three sides need fencing (two widths and one length).
  2w + l = 1200

• Area Function: The area of a rectangle is given by:
  A = l * w

3. Express Area in Terms of One Variable:
From the fencing constraint, solve for l:
l = 1200 - 2w

Substitute this expression for l into the area function:
A(w) = (1200 - 2w)w
A(w) = 1200w - 2w^2

4. Find the Critical Points (Maximize Area):
To find the maximum area, take the derivative of A(w) with respect to w and set it to zero.
dA/dw = d/dw (1200w - 2w^2)
dA/dw = 1200 - 4w

Set dA/dw = 0 to find critical points:
1200 - 4w = 0
4w = 1200
w = 300 m

5. Second Derivative Test (Verify Maximum):
Calculate the second derivative of A(w):
d^2A/dw^2 = d/dw (1200 - 4w)
d^2A/dw^2 = -4

Since the second derivative is -4 (which is less than 0), the critical point w = 300 m corresponds to a local maximum.

6. Calculate the Dimensions:
Substitute w = 300 m back into the equation for l:
l = 1200 - 2(300)
l = 1200 - 600
l = 600 m

Conclusion:
The dimensions of the field that will have the largest area are:

• Width (perpendicular to the river) = 300 m
• Length (parallel to the river) = 600 m

Solution of the Initial Value Problem

The given initial value problem is $x^2y' + xy = 1$, with $y(1) = 2$ for $x > 0$.

1. Standard Form of the Differential Equation:
   Divide the entire equation by $x^2$ to bring it into the standard form of a first-order linear differential equation, $y' + P(x)y = Q(x)$:
   $y' + \frac{x}{x^2}y = \frac{1}{x^2}$
   $y' + \frac{1}{x}y = \frac{1}{x^2}$
   Here, $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{x^2}$.

2. Integrating Factor:
   The integrating factor $\mu(x)$ is given by $e^{\int P(x) dx}$.
   $\mu(x) = e^{\int \frac{1}{x} dx}$
   $\mu(x) = e^{\ln|x|}$
   Since the problem states $x > 0$, $|x| = x$.
   $\mu(x) = e^{\ln x} = x$

3. General Solution:
   Multiply the standard form of the differential equation by the integrating factor:
   $x \left(y' + \frac{1}{x}y\right) = x \left(\frac{1}{x^2}\right)$
   $xy' + y = \frac{1}{x}$
   The left side of the equation is the derivative of the product $(\mu(x)y)$:
   $\frac{d}{dx}(xy) = \frac{1}{x}$
   Integrate both sides with respect to $x$:
   $\int \frac{d}{dx}(xy) dx = \int \frac{1}{x} dx$
   $xy = \ln|x| + C$
   Since $x > 0$, we have:
   $xy = \ln x + C$
   Solve for $y$:
   $y(x) = \frac{\ln x + C}{x}$

4. Apply Initial Condition:
   Use the initial condition $y(1) = 2$ to find the value of $C$:
   $2 = \frac{\ln(1) + C}{1}$
   $2 = \frac{0 + C}{1}$
   $C = 2$

5. Final Solution:
   Substitute the value of $C$ back into the general solution:
   $y(x) = \frac{\ln x + 2}{x}$

Area Enclosed by a Line and a Parabola

The curves are given by $y = x - 1$ and $y^2 = 2x + 6$.

1. Find Intersection Points:
   Express both equations in terms of $x$:
   From the line: $x = y + 1$ (1)
   From the parabola: $2x = y^2 - 6 \Rightarrow x = \frac{1}{2}y^2 - 3$ (2)
   Set the expressions for $x$ equal to each other to find the points of intersection:
   $y + 1 = \frac{1}{2}y^2 - 3$
   Multiply by 2 to clear the fraction:
   $2(y + 1) = y^2 - 6$
   $2y + 2 = y^2 - 6$
   Rearrange into a quadratic equation:
   $y^2 - 2y - 8 = 0$
   Factor the quadratic equation:
   $(y - 4)(y + 2) = 0$
   The $y$-coordinates of the intersection points are $y = 4$ and $y = -2$.
   Substitute these $y$-values back into $x = y + 1$ to find the corresponding $x$-coordinates:
   For $y = 4$: $x = 4 + 1 = 5$. Intersection point: $(5, 4)$.
   For $y = -2$: $x = -2 + 1 = -1$. Intersection point: $(-1, -2)$.

2. Determine the Limits of Integration and Relative Position of Curves:
   The region is bounded by $y$ from $-2$ to $4$.
   We need to integrate with respect to $y$, so we use the forms $x_R(y) = y + 1$ (line) and $x_L(y) = \frac{1}{2}y^2 - 3$ (parabola).
   To determine which function is to the right and which is to the left, pick a test point between $y = -2$ and $y = 4$, for example, $y = 0$:
   For the line $x = y + 1$: $x = 0 + 1 = 1$.
   For the parabola $x = \frac{1}{2}y^2 - 3$: $x = \frac{1}{2}(0)^2 - 3 = -3$.
   Since $1 > -3$, the line $x = y + 1$ is to the right of the parabola $x = \frac{1}{2}y^2 - 3$ in the interval $[-2, 4]$.

3. Set up the Definite Integral for Area:
   The area $A$ is given by the integral of the right function minus the left function with respect to $y$:
   $A = \int_{y_1}^{y_2} (x_R(y) - x_L(y)) dy$
   $A = \int_{-2}^{4} \left((y + 1) - \left(\frac{1}{2}y^2 - 3\right)\right) dy$
   $A = \int_{-2}^{4} \left(y + 1 - \frac{1}{2}y^2 + 3\right) dy$
   $A = \int_{-2}^{4} \left(-\frac{1}{2}y^2 + y + 4\right) dy$

4. Evaluate the Integral:
   $A = \left[-\frac{1}{2}\frac{y^3}{3} + \frac{y^2}{2} + 4y\right]{-2}^{4}$
   $A = \left[-\frac{1}{6}y^3 + \frac{1}{2}y^2 + 4y\right]{-2}^{4}$
   Evaluate at the upper limit ($y=4$):
   $A(4) = -\frac{1}{6}(4)^3 + \frac{1}{2}(4)^2 + 4(4)$
   $A(4) = -\frac{64}{6} + \frac{16}{2} + 16$
   $A(4) = -\frac{32}{3} + 8 + 16$
   $A(4) = -\frac{32}{3} + 24 = \frac{-32 + 72}{3} = \frac{40}{3}$
   Evaluate at the lower limit ($y=-2$):
   $A(-2) = -\frac{1}{6}(-2)^3 + \frac{1}{2}(-2)^2 + 4(-2)$
   $A(-2) = -\frac{1}{6}(-8) + \frac{1}{2}(4) - 8$
   $A(-2) = \frac{8}{6} + 2 - 8$
   $A(-2) = \frac{4}{3} - 6 = \frac{4 - 18}{3} = -\frac{14}{3}$
   Subtract the value at the lower limit from the value at the upper limit:
   $A = A(4) - A(-2)$
   $A = \frac{40}{3} - \left(-\frac{14}{3}\right)$
   $A = \frac{40}{3} + \frac{14}{3}$
   $A = \frac{54}{3}$
   $A = 18$

The area enclosed by the line and the parabola is 18 square units.

To evaluate the definite integral $\int_0^{\sqrt{3}} \sqrt{1+x^2} \cdot x^3 , dx$:

1. Substitution:
   Let $u = 1+x^2$.
   Differentiating with respect to $x$, we get $du = 2x , dx$.
   This implies $x , dx = \frac{1}{2} du$.
   Also, from the substitution, $x^2 = u-1$.

2. Change of Limits:
   When $x=0$, $u = 1+(0)^2 = 1$.
   When $x=\sqrt{3}$, $u = 1+(\sqrt{3})^2 = 1+3 = 4$.

3. Rewrite the Integral:
   Substitute $u$ and $du$ into the integral:
   $\int_0^{\sqrt{3}} \sqrt{1+x^2} \cdot x^3 , dx = \int_0^{\sqrt{3}} \sqrt{1+x^2} \cdot x^2 \cdot x , dx$
   $= \int_1^4 \sqrt{u} \cdot (u-1) \cdot \frac{1}{2} du$
   $= \frac{1}{2} \int_1^4 (u^{1/2})(u-1) , du$
   $= \frac{1}{2} \int_1^4 (u^{3/2} - u^{1/2}) , du$

4. Integrate:
   Integrate term by term:
   $\frac{1}{2} \left[ \frac{u^{3/2+1}}{3/2+1} - \frac{u^{1/2+1}}{1/2+1} \right]_1^4$
   $= \frac{1}{2} \left[ \frac{u^{5/2}}{5/2} - \frac{u^{3/2}}{3/2} \right]_1^4$
   $= \frac{1}{2} \left[ \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right]_1^4$
   $= \left[ \frac{1}{5} u^{5/2} - \frac{1}{3} u^{3/2} \right]_1^4$

5. Evaluate at Limits:
   Substitute the upper limit ($u=4$):
   $\left( \frac{1}{5} (4)^{5/2} - \frac{1}{3} (4)^{3/2} \right)$
   $= \left( \frac{1}{5} (\sqrt{4})^5 - \frac{1}{3} (\sqrt{4})^3 \right)$
   $= \left( \frac{1}{5} (2)^5 - \frac{1}{3} (2)^3 \right)$
   $= \left( \frac{32}{5} - \frac{8}{3} \right)$
   $= \left( \frac{32 \cdot 3 - 8 \cdot 5}{15} \right) = \left( \frac{96 - 40}{15} \right) = \frac{56}{15}$

   Substitute the lower limit ($u=1$):
   $\left( \frac{1}{5} (1)^{5/2} - \frac{1}{3} (1)^{3/2} \right)$
   $= \left( \frac{1}{5} - \frac{1}{3} \right)$
   $= \left( \frac{3 - 5}{15} \right) = -\frac{2}{15}$

   Subtract the lower limit value from the upper limit value:
   $\frac{56}{15} - \left( -\frac{2}{15} \right)$
   $= \frac{56}{15} + \frac{2}{15}$
   $= \frac{58}{15}$

The final answer is $\boxed{\frac{58}{15}}$.

The Maclaurin series expansion of a function $f(x)$ is given by:
$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$

For $f(x) = \sin x$:

1. Calculate derivatives and evaluate at $x=0$:

   • $f(x) = \sin x \implies f(0) = \sin 0 = 0$
   • $f'(x) = \cos x \implies f'(0) = \cos 0 = 1$
   • $f''(x) = -\sin x \implies f''(0) = -\sin 0 = 0$
   • $f'''(x) = -\cos x \implies f'''(0) = -\cos 0 = -1$
   • $f^{(4)}(x) = \sin x \implies f^{(4)}(0) = \sin 0 = 0$
   • $f^{(5)}(x) = \cos x \implies f^{(5)}(0) = \cos 0 = 1$
     The pattern of derivative values at $x=0$ is $0, 1, 0, -1, 0, 1, \dots$

2. Substitute values into the Maclaurin series formula:
   $\sin x = 0 + (1)x + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5 + \dots$
   $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

3. General form of the Maclaurin series:
   The series contains only odd powers of $x$ with alternating signs.
   $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$

This expansion is valid for all $x \in \mathbb{R}$.

The unit normal and binormal vectors for the circular helix $\vec{r}(t) = \cos(t)\hat{i} + \sin(t)\hat{j} + t\hat{k}$ are determined as follows:

1. Velocity Vector
   The first derivative of $\vec{r}(t)$ gives the velocity vector:
   $\vec{r}'(t) = \frac{d}{dt}(\cos(t)\hat{i} + \sin(t)\hat{j} + t\hat{k})$
   $\vec{r}'(t) = -\sin(t)\hat{i} + \cos(t)\hat{j} + \hat{k}$

2. Speed
   The magnitude of the velocity vector is the speed:
   $|\vec{r}'(t)| = \sqrt{(-\sin(t))^2 + (\cos(t))^2 + (1)^2}$
   $|\vec{r}'(t)| = \sqrt{\sin^2(t) + \cos^2(t) + 1}$
   $|\vec{r}'(t)| = \sqrt{1 + 1} = \sqrt{2}$

3. Unit Tangent Vector ($\vec{T}$)
   The unit tangent vector is found by normalizing the velocity vector:
   $\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{-\sin(t)\hat{i} + \cos(t)\hat{j} + \hat{k}}{\sqrt{2}}$
   $\vec{T}(t) = -\frac{1}{\sqrt{2}}\sin(t)\hat{i} + \frac{1}{\sqrt{2}}\cos(t)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$

4. Derivative of Unit Tangent Vector
   $\vec{T}'(t) = \frac{d}{dt}\left(-\frac{1}{\sqrt{2}}\sin(t)\hat{i} + \frac{1}{\sqrt{2}}\cos(t)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\right)$
   $\vec{T}'(t) = -\frac{1}{\sqrt{2}}\cos(t)\hat{i} - \frac{1}{\sqrt{2}}\sin(t)\hat{j}$

5. Magnitude of $\vec{T}'(t)$
   $|\vec{T}'(t)| = \sqrt{\left(-\frac{1}{\sqrt{2}}\cos(t)\right)^2 + \left(-\frac{1}{\sqrt{2}}\sin(t)\right)^2}$
   $|\vec{T}'(t)| = \sqrt{\frac{1}{2}\cos^2(t) + \frac{1}{2}\sin^2(t)}$
   $|\vec{T}'(t)| = \sqrt{\frac{1}{2}(\cos^2(t) + \sin^2(t))} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

6. Unit Normal Vector ($\vec{N}$)
   The unit normal vector is found by normalizing $\vec{T}'(t)$:
   $\vec{N}(t) = \frac{\vec{T}'(t)}{|\vec{T}'(t)|} = \frac{-\frac{1}{\sqrt{2}}\cos(t)\hat{i} - \frac{1}{\sqrt{2}}\sin(t)\hat{j}}{\frac{1}{\sqrt{2}}}$
   $\vec{N}(t) = -\cos(t)\hat{i} - \sin(t)\hat{j}$

7. Binormal Vector ($\vec{B}$)
   The binormal vector is the cross product of the unit tangent and unit normal vectors:
   $\vec{B}(t) = \vec{T}(t) \times \vec{N}(t)$
   $\vec{B}(t) = \left(-\frac{1}{\sqrt{2}}\sin(t)\hat{i} + \frac{1}{\sqrt{2}}\cos(t)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\right) \times (-\cos(t)\hat{i} - \sin(t)\hat{j} + 0\hat{k})$
   $\vec{B}(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -\frac{1}{\sqrt{2}}\sin(t) & \frac{1}{\sqrt{2}}\cos(t) & \frac{1}{\sqrt{2}} \ -\cos(t) & -\sin(t) & 0 \end{vmatrix}$
   $\vec{B}(t) = \hat{i}\left(\frac{1}{\sqrt{2}}\cos(t) \cdot 0 - \frac{1}{\sqrt{2}} \cdot (-\sin(t))\right) - \hat{j}\left(-\frac{1}{\sqrt{2}}\sin(t) \cdot 0 - \frac{1}{\sqrt{2}} \cdot (-\cos(t))\right) + \hat{k}\left(-\frac{1}{\sqrt{2}}\sin(t) \cdot (-\sin(t)) - \frac{1}{\sqrt{2}}\cos(t) \cdot (-\cos(t))\right)$
   $\vec{B}(t) = \hat{i}\left(\frac{1}{\sqrt{2}}\sin(t)\right) - \hat{j}\left(\frac{1}{\sqrt{2}}\cos(t)\right) + \hat{k}\left(\frac{1}{\sqrt{2}}\sin^2(t) + \frac{1}{\sqrt{2}}\cos^2(t)\right)$
   $\vec{B}(t) = \frac{1}{\sqrt{2}}\sin(t)\hat{i} - \frac{1}{\sqrt{2}}\cos(t)\hat{j} + \frac{1}{\sqrt{2}}(\sin^2(t) + \cos^2(t))\hat{k}$
   $\vec{B}(t) = \frac{1}{\sqrt{2}}\sin(t)\hat{i} - \frac{1}{\sqrt{2}}\cos(t)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$

Final Answer:
The unit normal vector is $\vec{N}(t) = -\cos(t)\hat{i} - \sin(t)\hat{j}$.
The binormal vector is $\vec{B}(t) = \frac{1}{\sqrt{2}}\sin(t)\hat{i} - \frac{1}{\sqrt{2}}\cos(t)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.

The limit does not exist.

Justification:
To determine if $\lim_{(x,y) \to (0,0)} f(x,y)$ exists for $f(x,y) = \frac{xy}{x^2+y^2}$, we evaluate the limit along different paths approaching the origin $(0,0)$.

Consider paths of the form $y = mx$, where $m$ is a constant:

1. Substitute $y=mx$ into the function:
   $f(x, mx) = \frac{x(mx)}{x^2 + (mx)^2} = \frac{mx^2}{x^2 + m^2x^2}$

2. Factor out $x^2$ from the denominator:
   $f(x, mx) = \frac{mx^2}{x^2(1 + m^2)}$

3. For $x \neq 0$, simplify the expression:
   $f(x, mx) = \frac{m}{1 + m^2}$

4. Take the limit as $x \to 0$:
   $\lim_{x \to 0} f(x, mx) = \lim_{x \to 0} \frac{m}{1 + m^2} = \frac{m}{1 + m^2}$

Since the limit $\frac{m}{1 + m^2}$ depends on the slope $m$ of the path taken:

• Along the path $y=0$ (x-axis, $m=0$), the limit is $\frac{0}{1+0} = 0$.
• Along the path $y=x$ (line $y=x$, $m=1$), the limit is $\frac{1}{1+1} = \frac{1}{2}$.

Because the function approaches different values along different paths to $(0,0)$, the limit $\lim_{(x,y) \to (0,0)} \frac{xy}{x^2+y^2}$ does not exist.

The sequence $a_n = (-1)^n$ is divergent.

• Terms of the sequence:

  • For $n=1$, $a_1 = (-1)^1 = -1$
  • For $n=2$, $a_2 = (-1)^2 = 1$
  • For $n=3$, $a_3 = (-1)^3 = -1$
  • For $n=4$, $a_4 = (-1)^4 = 1$
    The sequence can be written as ${-1, 1, -1, 1, \dots}$.

• Behavior of the sequence: The terms of the sequence oscillate between two distinct values, -1 and 1. As $n \to \infty$, the terms do not approach a single finite limit.

• Conclusion: For a sequence to be convergent, its terms must approach a unique limit as $n$ tends to infinity. Since $a_n$ continually alternates between -1 and 1 and does not settle on a single value, the limit does not exist. Therefore, the sequence $a_n = (-1)^n$ is divergent.

Given the position vector of the object:
$r(t) = t^2 \vec{i} + t^3 \vec{j}$

Velocity

The velocity vector $v(t)$ is the first derivative of the position vector with respect to time:
$v(t) = \frac{dr}{dt} = \frac{d}{dt}(t^2)\vec{i} + \frac{d}{dt}(t^3)\vec{j}$
$v(t) = 2t \vec{i} + 3t^2 \vec{j}$

At $t=1$:
$v(1) = 2(1) \vec{i} + 3(1)^2 \vec{j}$
$v(1) = 2 \vec{i} + 3 \vec{j}$

Speed

The speed is the magnitude of the velocity vector:
Speed $= |v(t)| = \sqrt{(2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 9t^4}$

At $t=1$:
Speed $= |v(1)| = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}$

Acceleration

The acceleration vector $a(t)$ is the first derivative of the velocity vector (or the second derivative of the position vector) with respect to time:
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(2t)\vec{i} + \frac{d}{dt}(3t^2)\vec{j}$
$a(t) = 2 \vec{i} + 6t \vec{j}$

At $t=1$:
$a(1) = 2 \vec{i} + 6(1) \vec{j}$
$a(1) = 2 \vec{i} + 6 \vec{j}$

Geometric Illustration

At $t=1$:

• Position: $r(1) = 1 \vec{i} + 1 \vec{j}$, so the object is at point P(1,1).
• Velocity: $v(1) = 2 \vec{i} + 3 \vec{j}$. This vector is tangent to the path at P(1,1).
• Acceleration: $a(1) = 2 \vec{i} + 6 \vec{j}$. This vector originates from P(1,1).

<<<GRAPHVIZ_START>>>
digraph G {
rankdir="LR";
overlap=false;
splines=true;
node [shape=point, width=0.01, height=0.01, label=""];

// Define positions for key points
Origin [pos="0,0!"];
P_t1 [pos="1,1!"];
V_tip [pos="3,4!"]; // P_t1 + v(1) = (1+2, 1+3) = (3,4)
A_tip [pos="3,7!"]; // P_t1 + a(1) = (1+2, 1+6) = (3,7)

// Add labels to specific points for clarity
Origin_label [label="O(0,0)", pos="0,-0.2!", shape=none, fontcolor=black];
P_t1_label [label="P(1,1)", pos="1,1.2!", shape=none, fontcolor=black];

// Position vector
Origin -> P_t1 [label="r(1)", arrowhead="normal", color="blue", penwidth=2, fontcolor=blue];

// Velocity vector
P_t1 -> V_tip [label="v(1)", arrowhead="normal", color="red", penwidth=2, fontcolor=red];

// Acceleration vector
P_t1 -> A_tip [label="a(1)", arrowhead="normal", color="green", penwidth=2, fontcolor=green];

// Axis markers (optional)
X_axis_label [label="x", pos="4.5,0!", shape=none, fontcolor=black];
Y_axis_label [label="y", pos="0,7.5!", shape=none, fontcolor=black];
}
<<<GRAPHVIZ_END>>>

To show that $y = \frac{1 + Ce^{2x}}{1 - Ce^{2x}}$ is a solution to the differential equation $y' = y^2 - 1$:

1. Calculate the derivative $y'$:
   Given $y = \frac{1 + Ce^{2x}}{1 - Ce^{2x}}$.
   Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:
   Let $u = 1 + Ce^{2x}$, then $u' = \frac{d}{dx}(1 + Ce^{2x}) = 2Ce^{2x}$.
   Let $v = 1 - Ce^{2x}$, then $v' = \frac{d}{dx}(1 - Ce^{2x}) = -2Ce^{2x}$.

   $y' = \frac{(2Ce^{2x})(1 - Ce^{2x}) - (1 + Ce^{2x})(-2Ce^{2x})}{(1 - Ce^{2x})^2}$
   $y' = \frac{2Ce^{2x} - 2C^2e^{4x} + 2Ce^{2x} + 2C^2e^{4x}}{(1 - Ce^{2x})^2}$
   $y' = \frac{4Ce^{2x}}{(1 - Ce^{2x})^2}$

2. Calculate $y^2 - 1$:
   Substitute the expression for $y$ into $y^2 - 1$:
   $y^2 - 1 = \left(\frac{1 + Ce^{2x}}{1 - Ce^{2x}}\right)^2 - 1$
   $y^2 - 1 = \frac{(1 + Ce^{2x})^2}{(1 - Ce^{2x})^2} - \frac{(1 - Ce^{2x})^2}{(1 - Ce^{2x})^2}$
   $y^2 - 1 = \frac{(1 + Ce^{2x})^2 - (1 - Ce^{2x})^2}{(1 - Ce^{2x})^2}$

   Expand the numerator using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$, where $a = (1 + Ce^{2x})$ and $b = (1 - Ce^{2x})$:
   Numerator $= [(1 + Ce^{2x}) - (1 - Ce^{2x})][(1 + Ce^{2x}) + (1 - Ce^{2x})]$
   Numerator $= [1 + Ce^{2x} - 1 + Ce^{2x}][1 + Ce^{2x} + 1 - Ce^{2x}]$
   Numerator $= [2Ce^{2x}][2]$
   Numerator $= 4Ce^{2x}$

   Therefore, $y^2 - 1 = \frac{4Ce^{2x}}{(1 - Ce^{2x})^2}$

3. Compare $y'$ and $y^2 - 1$:
   From step 1, $y' = \frac{4Ce^{2x}}{(1 - Ce^{2x})^2}$.
   From step 2, $y^2 - 1 = \frac{4Ce^{2x}}{(1 - Ce^{2x})^2}$.

Since $y' = y^2 - 1$, the function $y = \frac{1 + Ce^{2x}}{1 - Ce^{2x}}$ is a solution to the differential equation $y' = y^2 - 1$.

Given the function $f(x, y, z) = 2x^4 + y^2 + z^2 - 4$.
The task is to find the partial derivatives with respect to $x$, $y$, and $z$, evaluated at the point $(2, 3, -1)$.

1. Calculate the partial derivative with respect to $x$:
   $f_x(x, y, z) = \frac{\partial}{\partial x}(2x^4 + y^2 + z^2 - 4)$
   $f_x(x, y, z) = 8x^3$

   Evaluate $f_x$ at the point $(2, 3, -1)$:
   $f_x(2, 3, -1) = 8(2)^3 = 8 \times 8 = 64$

2. Calculate the partial derivative with respect to $y$:
   $f_y(x, y, z) = \frac{\partial}{\partial y}(2x^4 + y^2 + z^2 - 4)$
   $f_y(x, y, z) = 2y$

   Evaluate $f_y$ at the point $(2, 3, -1)$:
   $f_y(2, 3, -1) = 2(3) = 6$

3. Calculate the partial derivative with respect to $z$:
   $f_z(x, y, z) = \frac{\partial}{\partial z}(2x^4 + y^2 + z^2 - 4)$
   $f_z(x, y, z) = 2z$

   Evaluate $f_z$ at the point $(2, 3, -1)$:
   $f_z(2, 3, -1) = 2(-1) = -2$

Therefore:

• $f_x(2, 3, -1) = 64$
• $f_y(2, 3, -1) = 6$
• $f_z(2, 3, -1) = -2$

To find the volume using the cylindrical shells method for rotation about the x-axis, the integral is set up with respect to $y$.

1. Express x in terms of y:
   Given $y = \sqrt{x}$, squaring both sides yields $x = y^2$.

2. Determine the y-bounds:
   When $x=0$, $y=\sqrt{0}=0$.
   When $x=1$, $y=\sqrt{1}=1$.
   Therefore, the integration limits for $y$ are from $0$ to $1$.

3. Identify radius and height of the cylindrical shell:

   • Radius (r): The distance from the x-axis to a point $(x,y)$ is $y$. So, $r = y$.
   • Height (h): For a horizontal shell, the height is the length of the horizontal strip. This length is the difference between the rightmost x-coordinate and the leftmost x-coordinate for a given $y$. The region is bounded by $y=\sqrt{x}$ (or $x=y^2$) and the line $x=1$. Thus, $x_{right} = 1$ and $x_{left} = y^2$.
     So, $h = 1 - y^2$.

4. Set up the integral for the volume:
   The formula for the volume using cylindrical shells for rotation about the x-axis is $V = \int_c^d 2\pi r h , dy$.
   Substituting the identified components:
   $V = \int_0^1 2\pi y (1 - y^2) , dy$

5. Evaluate the integral:
   $V = 2\pi \int_0^1 (y - y^3) , dy$
   $V = 2\pi \left[ \frac{y^2}{2} - \frac{y^4}{4} \right]_0^1$
   $V = 2\pi \left( \left( \frac{1^2}{2} - \frac{1^4}{4} \right) - \left( \frac{0^2}{2} - \frac{0^4}{4} \right) \right)$
   $V = 2\pi \left( \frac{1}{2} - \frac{1}{4} - 0 \right)$
   $V = 2\pi \left( \frac{2}{4} - \frac{1}{4} \right)$
   $V = 2\pi \left( \frac{1}{4} \right)$
   $V = \frac{\pi}{2}$

The volume of the solid is $\frac{\pi}{2}$ cubic units.