Thread: Graphs of Sine and Cosine

1. Graphs of Sine and Cosine

Hello again!

Using a weight on a string called a plumb bob, it is possible to erect a pole that is exactly vertical, which means that the pole points directly toward the center of the earth. Two such poles are erected one hundred miles apart. If the the poles were extended they would meet at the center of the earth at an angle of $1.4333^o$. Compare the radius of the earth.

So that's the question and I'm just totally lost because I cant seem to picture it.

The equation for Arc length is $S = r\theta$

Thanks!

2. Hello again!

Using a weight on a string called a plumb bob, it is possible to erect a pole that is exactly vertical, which means that the pole points directly toward the center of the earth. Two such poles are erected one hundred miles apart. If the the poles were extended they would meet at the center of the earth at an angle of . Compare the radius of the earth.

So that's the question and I'm just totally lost because I cant seem to picture it.

The equation for Arc length is

Thanks!

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basically imagine two poles that meet at the center of the earth, on the surface the poles are a distance of 100miles away but it is a slightly curved path as the earth is naturally curved, we will call this distance S for arc length. S=100, Q(theta)= 43/30 and we need r. well? this is a straightfoward solution. using s=rq. r=s/q. r=100/43/30. r=3000/43.

3. Originally Posted by purebladeknight

basically imagine two poles that meet at the center of the earth, on the surface the poles are a distance of 100miles away but it is a slightly curved path as the earth is naturally curved, we will call this distance S for arc length. S=100, Q(theta)= 43/30 and we need r. well? this is a straightfoward solution. using s=rq. r=s/q. r=100/43/30. r=3000/43.

Awesome! Thanks!

4. Unfortunately, purebladeknight's argument is incorrect. The radius of the earth is considerably larger than 3000/43= 70 miles. If it were that radius, the circumference of the earth would be $140\pi= 438$ miles and you could drive around the earth in about 6 hours!

The "equation for arclength is $S= r\theta$" only if $\theta$ is given in radians! So you need to start writing the angle in radians. 180 degrees corresponds to $\pi$ radians so $1.4333^o$ degrees corresponds to $1.4333/180= 0.0079611\pi= 0.025016$ radians.

And a radian is defined so 1 radian cuts an arc equal to the radius. Here the arc is 100 miles= 0.025016 times the radius of the earth so the earth's radius is 100/0.025016= 3997 miles, approximately.

If you wanted to do this in degrees, you could argue by proportions. An entire arc of $360^o$ covers the entire circumference, $2\pi r$ and angle of $1.4333^o$ covers $\frac{1.4333}{360}(2\pi r)$ of the circumference. So we have $\frac{1.4333}{360}(2\pi r)= 100$ to solve for r. The first thing you would do, of course, if find that coefficient: [tex]\frac{1.4333}{360}(2\pi)= 0.025016[tex] (the radian measure we got before). Now your equation is 0.0250156 r= 100 so $r= \frac{100}{0.0250156}= 3997$ miles, again.

5. Originally Posted by HallsofIvy
Unfortunately, purebladeknight's argument is incorrect. The radius of the earth is considerably larger than 3000/43= 70 miles. If it were that radius, the circumference of the earth would be $140\pi= 438$ miles and you could drive around the earth in about 6 hours!

The "equation for arclength is $S= r\theta$" only if $\theta$ is given in radians! So you need to start writing the angle in radians. 180 degrees corresponds to $\pi$ radians so $1.4333^o$ degrees corresponds to $1.4333/180= 0.0079611\pi= 0.025016$ radians.

And a radian is defined so 1 radian cuts an arc equal to the radius. Here the arc is 100 miles= 0.025016 times the radius of the earth so the earth's radius is 100/0.025016= 3997 miles, approximately.

If you wanted to do this in degrees, you could argue by proportions. An entire arc of $360^o$ covers the entire circumference, $2\pi r$ and angle of $1.4333^o$ covers $\frac{1.4333}{360}(2\pi r)$ of the circumference. So we have $\frac{1.4333}{360}(2\pi r)= 100$ to solve for r. The first thing you would do, of course, if find that coefficient: [tex]\frac{1.4333}{360}(2\pi)= 0.025016[tex] (the radian measure we got before). Now your equation is 0.0250156 r= 100 so $r= \frac{100}{0.0250156}= 3997$ miles, again.
thanks for the fix up, silly me, i never realised that the theta was in radians =p.