Math Problem Statement
solve question number 9
If the following function is continuous everywhere and a and k are constants, then what is the value of a?
f(t) =
{
2t + k, if t < 0
2 cos(t) - 3, if 0 ≤ t ≤ π/2
a sin(t) + 5k, if t > π/2
}
Solution
The problem asks to determine the value of such that the given piecewise function is continuous everywhere. The function is defined as:
\begin{cases} 2t + k & \text{if } t < 0 \\ 2 \cos(t) - 3 & \text{if } 0 \leq t \leq \frac{\pi}{2} \\ a \sin(t) + 5k & \text{if } t > \frac{\pi}{2} \end{cases}$$ To ensure the function is continuous everywhere, we need to check for continuity at the points where the pieces of the function change, specifically at $$t = 0$$ and $$t = \frac{\pi}{2}$$. ### Step 1: Continuity at $$t = 0$$ The function must satisfy: $$\lim_{t \to 0^-} f(t) = \lim_{t \to 0^+} f(t) = f(0)$$ - For $$t < 0$$, $$f(t) = 2t + k$$. As $$t \to 0^-$$, we get: $$\lim_{t \to 0^-} f(t) = 2(0) + k = k$$ - For $$0 \leq t \leq \frac{\pi}{2}$$, $$f(0) = 2 \cos(0) - 3 = 2(1) - 3 = -1$$. For continuity at $$t = 0$$: $$k = -1$$ ### Step 2: Continuity at $$t = \frac{\pi}{2}$$ Next, the function must be continuous at $$t = \frac{\pi}{2}$$, so: $$\lim_{t \to \left( \frac{\pi}{2} \right)^-} f(t) = \lim_{t \to \left( \frac{\pi}{2} \right)^+} f(t)$$ - For $$0 \leq t \leq \frac{\pi}{2}$$, $$f(t) = 2 \cos(t) - 3$$. As $$t \to \frac{\pi}{2}^-$$: $$f\left(\frac{\pi}{2}\right) = 2 \cos\left(\frac{\pi}{2}\right) - 3 = 2(0) - 3 = -3$$ - For $$t > \frac{\pi}{2}$$, $$f(t) = a \sin(t) + 5k$$. As $$t \to \frac{\pi}{2}^+$$, $$\sin\left(\frac{\pi}{2}\right) = 1$$, so: $$\lim_{t \to \left(\frac{\pi}{2}\right)^+} f(t) = a(1) + 5k = a + 5k$$ For continuity at $$t = \frac{\pi}{2}$$: $$a + 5k = -3$$ Substituting $$k = -1$$ into the equation: $$a + 5(-1) = -3 \implies a - 5 = -3 \implies a = 2$$ Thus, the value of $$a$$ is $$\boxed{2}$$. Would you like further clarification or details? --- Here are 5 questions related to this solution: 1. Why is checking the continuity at $$t = 0$$ and $$t = \frac{\pi}{2}$$ important? 2. How would the solution change if $$k$$ had a different value? 3. What does it mean for a function to be continuous at a point? 4. How would we check differentiability at the transition points? 5. Can the same approach be applied to piecewise functions with more complex definitions? **Tip:** Always remember that for continuity at a point, both the left-hand limit and right-hand limit must equal the function value at that point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Formulas
lim t -> a f(t)
sin(t)
cos(t)
Theorems
Continuity at a point
Limit Definition
Suitable Grade Level
Grades 11-12 (Pre-Calculus/Calculus)