Question
Find the slope of the curve $r=\cos 2 \theta$ at $\theta=\frac{\pi}{6}$.
(A) $\frac{\sqrt{3}}{7}$
(B) $\frac{\sqrt{3}}{4}$
(C) $-\frac{\sqrt{3}}{4}$
(D) $-\sqrt{3}$
▶️Answer/Explanation
Ans:A
Represent the coordinates parametrically as $(r \cos \theta, r \sin \theta)$. Then
$
\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{r \cos \theta+\frac{d r}{d \theta} \cdot \sin \theta}{-r \sin \theta+\frac{d r}{d \theta} \cdot \cos \theta}
$
Note that $\frac{d r}{d \theta}=-2 \sin 2 \theta$, and evaluate $\frac{d y}{d x}$ at $\theta=\frac{\pi}{6}$. (Alternatively, write $x=\cos 2 \theta \cos \theta$ and $y=\cos 2 \theta \sin$ $\theta$ to find $\frac{d y}{d x}$ from $\frac{d y / d \theta}{d x / d \theta}$ )
Note that $v$ is negative from $t=0$ to $t=1$ but positive from $t=1$ to $t=2$. Thus the distance traveled is given by
$
-\int_0^1\left(t^2-t\right) d t+\int_1^2\left(t^2-t\right) d t
$
Question
In which of the following series can the convergence or divergence be determined by using the Limit Comparison Test with $\sum_{n=1}^{\infty} \frac{1}{n^2}$ ?
(A) $\sum_{n_{=1}}^{\infty} \frac{5 n}{2 n+4}$
(B) $\sum_{n_{\bar{\infty}}}^{\infty} \frac{5 n}{2 n^2+4}$
(C) $\sum_{n_{\bar{\infty}}}^{\infty} \frac{5 n^2}{2 n^3+4 n}$
(D) $\sum_{n=1}^{\overline{\bar{\infty}}^1} \frac{5 n^2}{2 n^4+4 n}$
▶️Answer/Explanation
Ans:D
If we look at the end behavior of series (D), the numerator behaves like $n^2$ and the denominator behaves like $n^4$ as $n$ gets larger. So the nth term of the series behaves like $\frac{n^2}{n^4}=\frac{1}{n^2}$. Thus, we can try and compare to $\sum_{n=1}^{\infty} \frac{1}{n^2}$, a known convergent series.
Using the Limit Comparison Test: $\lim _{n \rightarrow \infty} \frac{\frac{5 n^2}{2 n^4+4 n}}{\frac{1}{n^2}}=\lim _{n \rightarrow \infty} \frac{5 n^4}{2 n^4+4 n}=\frac{5}{2}>0$. So both series converge. If you use the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ and the Limit Comparison Test for series (A), (B), or (C), the limit is infinite. So the Limit Comparison Test is inconclusive since the comparison is with a known convergent series.
Question
The parametric equations $x(t)=\sin \left(t^2+3\right)$ and $y(t)=2 \sqrt{t}$ give the position of a particle moving in the plane for $t \geq 0$. What is the slope of the tangent line to the path of the particle when $t=1$ ?
(A) 1
(B) $\frac{1}{2 \cos 4}$
(C) $\frac{2}{\sin 4}$
(D) $\frac{1}{\cos 4}$
▶️Answer/Explanation
Ans:B
The slope of the tangent line is $\frac{d y}{d x}$. Use the Chain Rule to find the slope with $\frac{d y}{d x}=\frac{d y / d t}{d x / d t} \cdot \frac{d y}{d t}=\frac{d}{d t}\left(2 t^{1 / 2}\right)=t^{-1 / 2}=\frac{1}{\sqrt{t}}$ and $\frac{d x}{d t}=\cos \left(t^2+3\right) \cdot 2 t$
Question
If $\frac{d y}{d x}=y \tan x$ and $y=3$ when $x=0$, then, when $x=\frac{\pi}{3}, y=$
(A) $2 \sqrt{3}$
(B) $\frac{3}{2}$
(C) $\frac{3 \sqrt{3}}{2}$
(D) 6
▶️Answer/Explanation
Ans:D
Separating variables yields $\frac{d y}{y}=\tan x d x$, so $\ln y=-\ln \cos x+C$. With $y=3$ when $x=0, C=\ln 3$
The general solution is therefore $y=\frac{3}{\cos x}$. When $x=\frac{\pi}{3}, y=\frac{3}{1 / 2}=6$.
Question
If $A=\int_0^1 e^{-x} d x$ is approximated using various sums with the same number of subdivisions, and if $L, R$, and $T$ denote, respectively, left Riemann Sum, right Riemann Sum, and trapezoidal sum, then it follows that
(A) $R \leq A \leq T \leq L$
(B) $R \leq T \leq A \leq L$
(C) $L \leq T \leq A \leq R$
(D) $L \leq A \leq T \leq R$
▶️Answer/Explanation
Ans:A