Answer to Question #133026 in Calculus for Promise Omiponle

Question

1)Find the limit.
a:)lim t->0 (e^(-3t)i+(t^2/sin^2 t)j|+ cos 2tk)
b:)lim t->1((t^2-t)/(t-1)i+sqrt(t+ 8)j+(sin pi t)/(ln t) k)

2) . Find a vector equation and parametric equations for the line segment that joins P(-1,2,2) and Q(-3,5,1).

Answer 1

1)

a) $lim_{ t \to 0} (e^{-3t}i+\frac{t^2}{sin^2 t}j+ cos 2tk)=(lim_{ t \to 0} e^{-3t})i+(lim_{ t \to 0}\frac{t^2}{sin^2 t})j+ (lim_{ t \to 0} cos 2t)k.$

Now we can find each limit:

$lim_{ t \to 0} (e^{-3t})=e^{-3\cdot 0}=e^0=1$.

To find the second limit we can use L'Hopital's rule twice:

$lim_{ t \to 0}(\frac{t^2}{sin^2 t})=[\frac{0}{0}]=lim_{ t \to 0}(\frac{(t^2)\prime}{(sin^2 t)\prime})=lim_{ t \to 0}(\frac{2t}{2sint cos t})=lim_{ t \to 0}(\frac{2t}{sin2t})=[\frac{0}{0}]=1$.

$lim_{ t \to 0} (cos 2t)=cos(2\cdot 0)=cos0=1.$

Finally:

$lim_{ t \to 0} (e^{-3t}i+\frac{t^2}{sin^2 t}j+ cos 2tk)=1i+1j+1k=i+j+k.$

b) $lim_{t \to 1}(\frac{t^2-t}{t-1}i+\sqrt{t+ 8}j+\frac{sin \pi t}{ln t} k)=(lim_{t \to 1}\frac{t^2-t}{t-1})i+(lim_{t \to 1}\sqrt{t+ 8})j+(lim_{t \to 1}\frac{sin \pi t}{ln t}) k.$

Now we can find each limit:

$lim_{t \to 1}(\frac{t^2-t}{t-1})=[\frac{0}{0}]=lim_{t \to 1}(\frac{t(t-1)}{t-1})=lim_{t \to 1}t=1.$

$lim_{t \to 1}\sqrt{t+ 8}=\sqrt{1+ 8}=\sqrt{9}=3.$

$lim_{t \to 1}\frac{sin \pi t}{ln t}=[\frac{0}{0}]=lim_{t \to 1}\frac{(sin \pi t)\prime}{(ln t)\prime}=lim_{t \to 1}\frac{\pi cos \pi t}{\frac{1}{t}}=lim_{t \to 1}(\pi t cos \pi t)=$

$=\pi \cdot 1 \cdot cos(\pi \cdot 1)=- \pi.$

Finally:

$lim_{t \to 1}(\frac{t^2-t}{t-1}i+\sqrt{t+ 8}j+\frac{sin \pi t}{ln t} k)=1i+3j+\pi k=i+3j-\pi k.$

2)

The vector equation of the line segment is given by $r(t)=(1-t)r_0+tr_1,$ where $0\leq t\leq 1$ and $r_0$ and $r_1$ are the vector equivalents of the endpoints:

$P(-1,2,2)$ becomes $r_0=⟨-1,2,2⟩,$

$Q(-3,5,1)$ becomes $r_1=⟨-3,5,1⟩.$

Plug $r_0$ and $r_2$ into the vector formula:

$r(t)=(1-t)⟨-1,2,2⟩+t⟨-3,5,1⟩=⟨-1+t-3t,2-2t+5t,2-2t+t⟩=$

$=⟨-1-2t,2+3t,2-t⟩$.

We have the vector equation: $r(t)=⟨-1-2t,2+3t,2-t⟩, 0 \leq t \leq 1,$ or $r(t)=(-1-2t)i+(2+3t)j+(2-t)k, 0 \leq t \leq 1.$

To get the parametric equations of the line segment, we have to take the coefficients by $i, j, k$:

$\begin{cases} x=-1-2t, \\ y=2+3t, \\ z=2-t, \end{cases}$ $0 \leq t \leq 1.$