At what points does the helix $\mathbf{r}(t) = \langle \sin(t), \cos(t), t \rangle$ intersect the sphere $x^2 + y^2 + z^2 = 17$? (Round your answers to three decimal places. If an answer does not exist, enter DNE.)

• smaller t-value $(x, y, z) = \boxed{\ }$
• larger t-value $(x, y, z) = \boxed{\ }$

Neetesh Kumar | December 14, 2024

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This is the solution to Math 1C
Assignment: 13.1 Question Number 9
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Step-by-step solution:

The vector equation for the helix is: $\mathbf{r}(t) = \langle \sin(t), \cos(t), t \rangle$

Thus, we have:

• $x(t) = \sin(t)$
• $y(t) = \cos(t)$
• $z(t) = t$

The equation of the sphere is: $x^2 + y^2 + z^2 = 17$

Substituting the components of the helix into the equation of the sphere: $\sin^2(t) + \cos^2(t) + t^2 = 17$

Using the Pythagorean identity $\sin^2(t) + \cos^2(t) = 1$, we get: $1 + t^2 = 17$

Solving for $t^2$: $t^2 = 16$

Taking the square root of both sides: $t = \pm 4$

Step 1: Find the points corresponding to $t = 4$ and $t = -4$

For $t = 4$:

• $x(4) = \sin(4) \approx -0.756$
• $y(4) = \cos(4) \approx -0.654$
• $z(4) = 4$

Thus, the point for $t = 4$ is approximately $(-0.756, -0.654, 4)$.

For $t = -4$:

• $x(-4) = \sin(-4) \approx 0.756$
• $y(-4) = \cos(-4) \approx 0.654$
• $z(-4) = -4$

Thus, the point for $t = -4$ is approximately $(0.756, 0.654, -4)$.

Final Answer:

smaller t-value $(x, y, z) = \boxed{(0.756, 0.654, -4)}$

larger t-value $(x, y, z) = \boxed{(-0.756, -0.654, 4)}$

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