Physics Test Questions Essay

Submitted By Saket-Bhargava
Words: 886
Pages: 4

1- IS MISSING and THERE ARE TWO NUMBER 2s
YIZHI FANG(LARA)
I have hard time to fix mistakes in your work since that you have not used Microsoft equation 3.0
2.for r(t)=
a)find the tangential and normal component of acceleration. Since r(t)= V=r’(t)= A= r’’(t)=

t j+k

b)find the osculating circle at t=2.
According to the formula: K(t)=
K(2)= ==
Saket Bhargava

2) - For
a) Find the tangential and normal component of acceleration
b) Find the osculating circle at t = 2. Wrong Differentiations

a)

Yoel Pikari
3- Prove the formula for curvature in rectangular form and find the curvature of at
Let the equation of the curve be . And let be the angle which the tangent at any point makes with the x axis. Differentiating with respect to s gives:

Where come from they were not defined
, because

The curvature of at :
,

Fareeha Very Good
4. a) Find radius of curvature on where is the radius and is the curvature of r such that and When taking the magnitudes of these vectors, using the trig identity , the equation simplifies to

b) Show that the radius is increasing for all times
Because the radius is inversely proportional to the curvature, as the curvature decreases, the radius increases. We can see through the graph of the function as well as a table that the curvature decreases throughout time, which means that the radius is increasing as t goes to infinity.

In addition, by taking the derivative of the radius of curvature as a function of time, we find that (using the product rule)

increases Here, we can see that the radius has no critical points, meaning it increases always
Gouri Gupta
#5. Verify that for all the unit vectors are orthogonal to each other.

Since are unit vectors and perpendicular, You have not showed that T and N are perpendicular yetalso being a unit vector, is perpendicular to

Bijan Very Good
6. Prove the formula for curvature in polar and find the curvature for

Let us begin this derivation by making some statements about the above trigonometry.

If we can find out what are equal to, we can solve for the curvature.Referring to the above,
, where r’=Taking a derivative with respect to θ yields,
, where r’’= Now, we know that
Now that we have both pieces of the puzzle, we can solve for the curvature.

Kseniya Konova Very Good
7- Show that

Edrees Very Good
# 8 - Prove that
First of all, we know that the unit tangent vector is equal to r’ (t) over the magnitude of r’(t). In other words: . Then we multiply each side by the magnitude of r’ (t) so we end up with: . We also know that the magnitude of r’ (t) is equal to (ds/dt) so we get: . The next step is to differentiate our current equation and to do so we must use the product rule so we get: . Next we use the chain rule to get: , which simplifies to , where Κ is the curvature of the function and is the unit normal vector. If we wish to rewrite the equation in a different notation we get .
9- IS MISSING BUT I HAVE SENT YOU TH HINT AND SOME STEPS ON SATURDAYs
Chase Yamate Very Good (Fix it)

10) –a) Find the oscillating circle to the curveat time T =1 sec. b) And Find how fast the area of the circle increases as the time increases.
a)the curve is a parabola. So: and
And from the lessons explained by Professor Mosh, the curvature
Substituting what we found: Now if t=1 sec then x=1, therefore we have: so the curvature…