Find a vector function that represents the curve of intersection of the two surfaces: the top half of the ellipsoid $x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49$ and the parabolic cylinder $y = x ^ { 2 }$.
A) $\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 4 } \mathbf { j } + \sqrt { \frac { 49 - t ^ { 2 } + 7 t } { 7 } } \mathbf { k }$
B) $\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 49 + t ^ { 2 } - 7 t ^ { 4 } } { 7 } } \mathbf { k }$
C) $r ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 49 - t ^ { 2 } - 7 t } { 7 } } \mathbf { k }$
D) $\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + \sqrt { \frac { 7 - t ^ { 2 } - 7 t ^ { 4 } } { 7 } } \mathbf { k }$
E) $\mathbf { r } ( t ) = t \mathbf { i } - t ^ { 2 } \mathbf { j } - \sqrt { \frac { 49 - t ^ { 2 } - 7 t } { 7 } } \mathbf { k }$

Question

Quizplus · Accepted Answer

We solve the problem by setting the two given equations equal to each other to get
\begin{align*}
x^2 + 7y^2 + 7z^2 &= 49 \
y &= x^2.
\end{align*}
Substituting the second equation into the first equation gives
$$x^2 + 7x^4 + 7z^2 = 49,$$
which can be simplified to
$$7x^4 + x^2 + 7z^2 - 49 = 0.$$
To represent this curve with a vector function, we can parametrize it with $x$ and $z$, and express $y$ in terms of $x$:
\begin{align*}
x &= t, \
y &= t^2, \
z &= \sqrt{\frac{49 - t^2 - 7t}{7}}.
\end{align*}
Therefore, the vector function that represents the curve of intersection is
$$\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + \sqrt{\frac{49 - t^2 - 7t}{7}}\mathbf{k}.$$

Find a Vector Function That Represents the Curve of Intersection x2+7y2+7z2=49x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49x2+7y2+7z2=49

Find a Vector Function That Represents the Curve of Intersection $x ^ { 2 } + 7 y ^ { 2 } + 7 z ^ { 2 } = 49$