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## Reviewing 9.2 Homework....HELP

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### Reviewing 9.2 Homework....HELP

I am going through 9.2 homework and was doing great until i got to question 26 and 33. Its been a while since I've done these types of problems so please be patient and explain EVERYTHING to me.

26. Find the magnitude of the resultant force and the angle it makes with the positive -axis.
On page 647.

33.Find the unit vectors that are parallel to the tangent line to the parabola y=x^2 at the point (2,4).
brinebaby

Posts: 5
Joined: Wed Jan 11, 2012 12:43 am

### Re: Reviewing 9.2 Homework....HELP

You should really post these issues in the "Calculus III" topic and not the "Miscellaneous" topic. Anywho, here goes:
9.2...

#26: You want to start by breaking these vectors down into their components. We'll call the 20lb vector that makes a 45 degree angle above the positive x axis vector A. We'll call the 16lb vector that makes a 30 degree angle below the positive x axis vector B. Now, start by breaking vector A into x and y components. You know the magnitude, and you know the angle. That's all you need. To find the x component, what trigonometric function do we use? (cosine). And to find the y component? (sine). Multiply each of these by the magnitude and you'll have your x & y components of force:

Vector A:
Magnitude is 20lb, angle is 45 degrees. The math goes like this (make sure your calculator is in degree mode):
20cos(45) = x component = 14.14lb in the x direction
20sin(45) = y component = 14.14lb in the y direction
That gives you the vector <14.14,14.14>.

Now for the second:

Vector B:
Magnitude is 16lb, angle is -30 degrees (because it is below x axis).
16cos(-30) = x component = 13.86lb in the x direction
16sin(-30) = y component = -8lb in the y direction.
That gives you the vector <13.86,-8>

Vector R (resultant):
Now add the two vectors you found together: <14.14,14.14> + <13.86,-8> = <28,6.14>
That is your resultant vector: <28, 6.14>
Find the magnitude using pythagorean's theorem: √(28² + 6.14²) = |28.67|
This is the magnitude of the resultant force. To find the angle it makes with the positive x axis, think trigonometrically again. Inverse trigonometric functions give you an angle: arctan(y/x) = angle above positive x axis.

Therefore, arctan(6.14/28) = 12.4 degrees above positive x axis (which is what the problem asked for).

There you have it! Your resultant vector has magnitude |28.67| with an angle 12.4 degrees above the positive x axis.

I must go to class, but I will answer #33 as soon as possible.
seth

Posts: 10
Joined: Wed Jan 11, 2012 2:08 pm

### Re: Reviewing 9.2 Homework....HELP

As for #33:

It's given that the parabola is y = x². Find the vectors parallel to the tangent line at the point (2,4). This is asking for the slope of the parabola y = x² at the point (2,4). As you know from Calculus I, the slope is given by the derivative. Our derivative here is 2x. Plugging 2 into x, we get a slope of 4. Slope is rise over run, so rise (y) = 4, run (x) = 1. Converting into vector notation, we have <1,4> or (i, 4j). Don't forget their "hats." We know that a unit vector is a vector whose length is 1. In order to convert a vector into a unit vector, we must divide that vector by its magnitude. If we have <1,4>, then we can use Pythagorean's theorem to find the magnitude: √(1²+4²). This gives us a magnitude of √(17). We must divide the given vectors by √(17) (or multiply by 1/√(17)) to find the unit vectors we're looking for. That gives us 1/√(17)×<1,4>, or <1/√(17), 4/√(17)>, or 1/√(17)×(i, 4j). Finally, we must note that the question asked us for any vectors that are PARALLEL to this tangent line. A vector parallel to a line could point in the same or opposite direction, hence we add a ±. The final answer comes to ±1/√(17)×(i + 4j).

P.S. This question was already answered in the Calculus III forum under the topic "HWK 9.2 EX 33."
seth

Posts: 10
Joined: Wed Jan 11, 2012 2:08 pm

### Re: Reviewing 9.2 Homework....HELP

Thanks! I will post to calc III next time, sorry. I understand how everything works for #26. I do have a question. If the resultant force is the whole triangle then why did we do the arctan of the whole angle if we were only trying to find above the x-axis? Just a little confused. And also to clarify, with all problems like these we should start off by finding the components first?
brinebaby

Posts: 5
Joined: Wed Jan 11, 2012 12:43 am

### Re: Reviewing 9.2 Homework....HELP

The resultant force is not the whole triangle. The magnitude of the resultant force is the hypotenuse of the x and y components of the two vectors we added together. Arctan only gives you angles in reference to the positive x axis. The question asked how many degrees above positive x axis the resultant force was. That's why we used arctan. We didn't use the arctan OF the angle, we used the arctan of the COMPONENTS (y/x) to FIND the angle. Hope that makes sense.

Also, for problems like these, I'd recommend breaking the vectors down into their components. If you know a vector's magnitude and the angle it makes with an axis, you can easily determine its components.
seth

Posts: 10
Joined: Wed Jan 11, 2012 2:08 pm