Paper Generator

The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r} = \vec{a} + t\vec{b}$. Here, $\vec{a} = (4, -3, 7)$ and $\vec{b} = (3, 1, 2)$. So, $(x, y, z) = (4, -3, 7) + t(3, 1, 2) = (4+3t, -3+t, 7+2t)$. Thus, $x = 3t+4, y = t-3, z = 2t+7$. This matches Option C.

The function is given by $y=x|x|$. We can define this piecewise: $$y = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$$ The area $A$ enclosed between the curve, the x-axis, $x=-2$ and $x=2$ is given by the integral of the absolute value of the function over the interval $[-2, 2]$: $$A = \int_{-2}^{2} |y| \, dx = \int_{-2}^{2} |x|x|| \, dx$$ Since $|x| \ge 0$, we have $|x|x|| = |x||x| = x^2$. So, the integral becomes: $$A = \int_{-2}^{2} x^2 \, dx$$ We can evaluate this integral: $$A = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{(2)^3}{3} - \frac{(-2)^3}{3}$$ $$A = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$$ The final answer is $\boxed{\frac{16}{3}}$.

The rate at which the height of sugar inside the cylindrical tank increases can be determined using the formula for the volume of a cylinder:

\(V = \pi r^2 h\)

Given:

• Radius, \(r = 10 \text{ cm}\)
• Rate of change of volume, \(\dfrac{dV}{dt} = 100\pi \text{ cm}^3/\text{s}\)

Since the radius remains constant, differentiate both sides of the volume equation with respect to time \(t\):

\(\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}\)

Substitute the known values:

\(100\pi = \pi (10)^2 \dfrac{dh}{dt}\)

\(100\pi = 100\pi \dfrac{dh}{dt}\)

Dividing both sides by \(100\pi\):

\(\dfrac{dh}{dt} = 1 \text{ cm/s}\)

Therefore, the height of the sugar in the tank is increasing at a rate of \(1 \text{ cm/s}\).

To determine the values of \(\lambda \) for which the function \(f(x)=\sin x-\cos x-\lambda x+C\) is always decreasing, we need to find the values of \(\lambda \) for which the derivative of \(f(x)\) is less than or equal to zero for all real \(x\).

The derivative of \(f(x)\) with respect to \(x\) is:

\(f^{\prime }(x)=\cos x+\sin x-\lambda \)

For \(f(x)\) to be a decreasing function for all real values of \(x\), we must have:\(f^{\prime }(x)\le 0\)\(\cos x+\sin x-\lambda \le 0\) that is

\(\cos x+\sin x\le \lambda \)

Note that for an expression of the form \(a\cos x+b\sin x\), the maximum value is \(\sqrt{a^{2}+b^{2}}\).

Here, \(a=1\) and \(b=1\). So the maximum value of \(\cos x+\sin x\) is:\(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)

Now the condition \(\cos x+\sin x\le \lambda \) for all real \(x\).implies that \(\lambda \) must be greater than or equal to the maximum possible value of \(\cos x+\sin x\).

Therefore, we must have:\(\lambda \ge \sqrt{2}\)

**Correct Option:** A **Reasoning:** * First derivative: \( \frac{dy}{dx} = -3x^2 + 6x + 8 \) * Set second derivative to zero: \( \frac{d^2y}{dx^2} = -6x + 6 = 0 \implies x = 1 \) * Substitute \( x=1 \) into original equation: \( y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -10 \). The point is (1, -10).

The diameter of the spherical ball is given by $D = \frac{5}{2}(3x+1)$. The radius of the ball is $r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Substitute the expression for $r$: $V = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)\right)^3$ $V = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3\right)$ $V = \frac{125\pi}{48}(3x+1)^3$ To find the rate of change of its volume with respect to $x$, we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}\left(\frac{125\pi}{48}(3x+1)^3\right)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^{3-1} \cdot \frac{d}{dx}(3x+1)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3$ $\frac{dV}{dx} = \frac{125\pi \cdot 9}{48}(3x+1)^2$ $\frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2$ $\frac{dV}{dx} = \frac{375\pi}{16}(3x+1)^2$ Now, we evaluate this rate of change when $x=1$: $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(4)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16} \cdot 16$ $\left.\frac{dV}{dx}\right|_{x=1} = 375\pi$ The final answer is $\boxed{\text{375\pi}}$.

Let the side of the square be $s$. The perimeter of the square is $P = 4s$. We are given that the rate of decrease of the side is $\frac{ds}{dt} = -1.5~cm/s$. We need to find the rate of decrease of its perimeter, which is $\frac{dP}{dt}$. Differentiate the perimeter equation with respect to time $t$: $\frac{dP}{dt} = \frac{d}{dt}(4s)$ $\frac{dP}{dt} = 4 \frac{ds}{dt}$ Substitute the given value of $\frac{ds}{dt}$: $\frac{dP}{dt} = 4 (-1.5~cm/s)$ $\frac{dP}{dt} = -6~cm/s$ The rate of decrease of its perimeter is $6~cm/s$. The final answer is $\boxed{6~cm/s}$.