The price p(t) of a share at a time t satisfi�es the equation
p + t = 1 + e^-p sin(�pi*t). Justify if p(t) is increasing or decreasing at t = 0.
To determine whether the price function p(t) is increasing or decreasing at t = 0, we need to find the first derivative of p(t) with respect to t, and then evaluate it at t = 0.
Let's start by differentiating the given equation with respect to t:
dp/dt + 1 = e^(-p) * (-p'(t)) * sin(πt) + e^(-p) * cos(πt) * π
Simplifying this equation gives us:
dp/dt = e^(-p) * (1 - p'(t) * sin(πt) - cos(πt) * π)
Now, we can evaluate this derivative at t = 0:
dp/dt | t=0 = e^(-p) * (1 - p'(0) * sin(0) - cos(0) * π)
Since sin(0) = 0 and cos(0) = 1, the equation becomes:
dp/dt | t=0 = e^(-p) * (1 - p'(0) * 0 - 1 * π)
Simplifying further:
dp/dt | t=0 = e^(-p) * (1 - π)
Now, we have an expression for the derivative at t = 0. To determine whether the price is increasing or decreasing at this point, we need to examine the sign of dp/dt | t=0.
If dp/dt | t=0 > 0, then the price is increasing at t = 0.
If dp/dt | t=0 < 0, then the price is decreasing at t = 0.
If dp/dt | t=0 = 0, then we cannot determine whether the price is increasing or decreasing at t = 0 using this method.
To determine the sign of dp/dt | t=0 = e^(-p) * (1 - π), we need to consider the value of e^(-p).
If e^(-p) > 0, then the sign of dp/dt | t=0 depends on the sign of (1 - π).
If e^(-p) < 0, then the sign of dp/dt | t=0 is opposite to the sign of (1 - π).
Note that e^(-p) is always greater than zero since it represents the exponential function with a negative exponent.
Now, we look at the sign of (1 - π). Since π is a positive constant (approximately 3.14159), (1 - π) is a negative value. Thus, (1 - π) < 0.
Therefore, for all values of p, e^(-p) > 0, and (1 - π) < 0. Hence, dp/dt | t=0 is positive.
In summary, the price function p(t) is increasing at t = 0.