The curves r1 = < 3t, t2, t3 > and r2 = < sin(t), sin(5t), t > intersect at the origin. Find their angle of intersection, è correct to the nearest degree.

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Well, isn't this a "curvy" situation we have here? Let's tackle it head-on!

To find the angle of intersection, we need to find the vector of each curve at the point of intersection (which is the origin). Once we have those vectors, we can use the dot product to find the angle between them.

First, let's find the vector of r1 at the origin:
r1(0) = < 3(0), (0)^2, (0)^3 > = < 0, 0, 0 >

Now, let's find the vector of r2 at the origin:
r2(0) = < sin(0), sin(5(0)), 0 > = < 0, 0, 0 >

Well, well, it seems like both vectors turn out to be the zero vector at the origin. That means they are parallel (or coincident) and don't form an angle!

So, in this case, the angle of intersection is actually 0 degrees! They don't cross paths in the conventional sense.

But hey, don't be too disappointed. At least our calculation didn't go "off the curve," right? Keep your chin up, my friend!

To find the angle of intersection between the curves r1 = < 3t, t^2, t^3 > and r2 = < sin(t), sin(5t), t > at the origin, we can use the dot product.

Step 1: Find the common point of intersection:
Setting the x, y, and z components of r1 and r2 equal to zero gives us the following equations:
3t = 0
t^2 = 0
t^3 = 0

From the first equation, we get t = 0. Substituting this value into the second and third equations, we also get t = 0. Therefore, the curves intersect at the origin, (0, 0, 0).

Step 2: Calculate the tangent vectors of the curves at the origin:
To find the tangent vectors, we can take the derivatives with respect to t of each curve and then evaluate them at t = 0.

For r1:
r1' = < 3, 2t, 3t^2 >
At t = 0, r1' = < 3, 0, 0 >

For r2:
r2' = < cos(t), 5cos(5t), 1 >
At t = 0, r2' = < 1, 5, 1 >

Step 3: Calculate the dot product of the tangent vectors:
The dot product of two vectors is given by the formula: A · B = |A| * |B| * cos(θ), where A and B are vectors, |A| and |B| are their magnitudes, and θ is the angle between them.

In this case, we have A = r1' and B = r2', and we want to find θ. Since we are interested in the angle of intersection, we can set the dot product equal to zero:
r1' · r2' = 0

Substituting the values of the tangent vectors, we have:
< 3, 0, 0 > · < 1, 5, 1 > = 0

This simplifies to:
3 * 1 + 0 * 5 + 0 * 1 = 0
3 = 0

Since the equation 3 = 0 is not true, this means that the dot product is not equal to zero, and the angle of intersection does not exist.

To find the angle of intersection between two curves, we need to find the angle between their tangent vectors at the point of intersection. Let's find the tangent vectors first.

The tangent vector of a curve can be found by taking the derivative of the position vector with respect to the parameter.

For the first curve r1 = < 3t, t^2, t^3 >:
- Taking the derivative with respect to t, we get: r1' = < 3, 2t, 3t^2 >

For the second curve r2 = < sin(t), sin(5t), t >:
- Taking the derivative with respect to t, we get: r2' = < cos(t), 5cos(5t), 1 >

Now let's find the point of intersection by setting the two position vectors equal to each other:

< 3t, t^2, t^3 > = < sin(t), sin(5t), t >

From the x-components, we get: 3t = sin(t)
From the y-components, we get: t^2 = sin(5t)
From the z-components, we get: t^3 = t

Solving these equations simultaneously is not straightforward, so we can use numerical methods or graphing software to find an approximate value for t where the curves intersect. In this case, we are given that they intersect at the origin, so we can directly conclude that t = 0.

Substituting t = 0 into the tangent vectors:
r1'(0) = < 3, 0, 0 >
r2'(0) = < 1, 5, 1 >

To find the angle between these two vectors, we can use the dot product and the formula:

cos(θ) = (r1'(0) · r2'(0)) / (||r1'(0)|| ||r2'(0)||)

Finding the dot product:
r1'(0) · r2'(0) = (3)(1) + (0)(5) + (0)(1) = 3

Finding the magnitudes of the vectors:
||r1'(0)|| = sqrt(3^2 + 0^2 + 0^2) = sqrt(9) = 3
||r2'(0)|| = sqrt(1^2 + 5^2 + 1^2) = sqrt(27) = 3√3

Substituting these values into the formula:
cos(θ) = (3) / (3 ∙ 3√3) = 1 / √3

To find the angle θ, we can take the inverse cosine of the above value using a calculator:

θ ≈ acos(1 / √3) ≈ 54.74 degrees

Therefore, the angle of intersection between the curves r1 and r2, correct to the nearest degree, is approximately 55 degrees.