1. Find the arclength of the curve y = ex from
x = 0 to x = 1.
Since y is a function of x here, we can represent the arclength of the curve in
terms of an integral with respect to x, i.e,
To evaluate the integral, we need to make a substitution:
. To write integral with respect to u, we need to write dx
in terms of u. From the substitution, it follows e2x = u2 − 1, hence by taking ln at
The integral becomes
in which the integrand is a rational function. It is not a
proper rational function,
we need to do long division to apply the partial fractions. The long division here
is very simple: u2 = (u2 − 1) * 1 + 1. hence
It follows (note u2 − 1 = (u + 1)(u − 1))
2. Find the solution to the initial value problem
The RHS of the equation is a function of y.hence by
seperating the variable we
Integrating both sides
The initial condition y(1) = 0 yields
Hence c = kπ − 1 where k is an integer. Hence the
solution to the initial value
Since tan(x + kπ − 1) = tan(x − 1) holds for any x.
the solution to the I.V.P.
3.Find the slope of the tangent line to the curve
at the point corresponding to t = 1.
As we know, the slope m at one point is defined to be at that point. i.e,
On the other hand
When t = 1
Hence the slope at the given point is
4. Consider the point with the Cartesian coordinates . Find the
polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π
where k is an integer. On the other hand, it is obvious
the point is in the third
quadrant (Please plot the point and you will see it) . Hence
since r > 0. ( What if r < 0?) It follows k = 1, i.e,
Hence the required polar coordinates for the point are
5. Identify the curve by finding the cartesian equation for the curve
Multiply both sides by r gives
Hence the curve is a circle with center and radius .