Question

1. find two integer values of t so that the tangent line to the curve $mathbf{r}(t) = langle 4 - t, 2t^2, 5t + 1 \

angle$ contains the point $(3,-30,6)$.

Explanation:

Step1: Find point on curve at $t$

The point on $\mathbf{r}(t)$ is $\mathbf{r}(t) = \langle 4-t, 2t^2, 5t+1
angle$.

Step2: Find tangent vector

Differentiate $\mathbf{r}(t)$:
$\mathbf{r}'(t) = \langle \frac{d}{dt}(4-t), \frac{d}{dt}(2t^2), \frac{d}{dt}(5t+1)
angle = \langle -1, 4t, 5
angle$

Step3: Define vector from curve point to given point

Let $P=(3,-30,6)$. The vector $\mathbf{P} - \mathbf{r}(t)$ is:
$\langle 3-(4-t), -30-2t^2, 6-(5t+1)
angle = \langle t-1, -30-2t^2, 5-5t
angle$

Step4: Set vectors linearly dependent

$\mathbf{P} - \mathbf{r}(t)$ is parallel to $\mathbf{r}'(t)$, so there exists scalar $k$ such that:

$$\begin{cases} t-1 = -k \\ -30-2t^2 = 4tk \\ 5-5t = 5k \end{cases}$$

Step5: Solve for $k$ from first equation

From $t-1=-k$, we get $k=1-t$.

Step6: Substitute $k$ into second equation

Substitute $k=1-t$ into $-30-2t^2=4tk$:
$-30-2t^2 = 4t(1-t)$
Expand right side: $-30-2t^2 = 4t - 4t^2$
Rearrange into quadratic: $2t^2 -4t -30=0$
Divide by 2: $t^2 -2t -15=0$

Step7: Factor quadratic equation

$(t-5)(t+3)=0$

Step8: Verify solutions

For $t=5$: $\mathbf{r}(5)=\langle -1,50,26
angle$, $\mathbf{r}'(5)=\langle -1,20,5
angle$, $\mathbf{P}-\mathbf{r}(5)=\langle 4,-80,-20
angle = -4\mathbf{r}'(5)$ (parallel).
For $t=-3$: $\mathbf{r}(-3)=\langle 7,18,-14
angle$, $\mathbf{r}'(-3)=\langle -1,-12,5
angle$, $\mathbf{P}-\mathbf{r}(-3)=\langle -4,-48,20
angle = 4\mathbf{r}'(-3)$ (parallel).

Answer:

$t = -3$ and $t = 5$