Question

Given that f ′(t)=sec(t)(sec(t)tan(t)) and f (4) =−4 for −2A) f(t)=sec(t)+C

B) f(t)=−sec(t)+C
C) f(t)=−tan(t)+C
D) f(t)=tan(t)+C

Answer 1

To find the particular solution to the differential equation, we integrate f ′(t) and use the initial condition f (4) = -4 to find the constant C. The correct solution is f(t) = sec(t) + C.

Step-by-step explanation:

The student's question involves finding the particular solution to a differential equation where f ′(t) is given, and an initial condition f (4) =−4 is provided. Given that f ′(t)=sec(t)(sec(t)tan(t)), we can find f(t) by integrating f ′(t). The antiderivative of sec(t)tan(t) is sec(t), and since we are multiplying by another sec(t), the antiderivative we need is sec^2(t). This leads us to an antiderivative of the form f(t) = sec(t) + C. Using the initial condition f (4) = -4, we can solve for C.

Substituting 4 into ‘t’ and ‘−4’ into f(t), we get ‘−4 = sec (4) + C’. Solving for C, we find that C is equal to ‘−4 - sec (4)’.

So, the function that satisfies the given differential equation and the initial condition is A) f(t)=sec(t)+C.