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Math  /  Calculus

QuestionFind the minimum and maximum values of y=10θ−5sec⁡θy=\sqrt{10} \theta-\sqrt{5} \sec \theta on the interval [0,π3]\left[0, \frac{\pi}{3}\right] fmin⁡=−5fmax⁡=□\begin{array}{l} f_{\min }=-\sqrt{5} \\ f_{\max }=\square \end{array}

Studdy Solution

STEP 1

1. The function y=10θ−5sec⁡θ y = \sqrt{10} \theta - \sqrt{5} \sec \theta is defined on the interval [0,π3][0, \frac{\pi}{3}].
2. We need to find the critical points within the interval and evaluate the function at these points as well as at the endpoints of the interval.
3. The secant function, sec⁡θ\sec \theta, is defined as 1cos⁡θ\frac{1}{\cos \theta} and is undefined where cos⁡θ=0\cos \theta = 0, but this does not occur within the given interval.

STEP 2

1. Differentiate the function y y with respect to θ \theta .
2. Find the critical points by setting the derivative equal to zero.
3. Evaluate the function y y at the critical points and the endpoints of the interval.
4. Determine the minimum and maximum values of y y .

STEP 3

Differentiate the function y=10θ−5sec⁡θ y = \sqrt{10} \theta - \sqrt{5} \sec \theta with respect to θ \theta .
The derivative is:
dydθ=10−5⋅ddθ(sec⁡θ) \frac{dy}{d\theta} = \sqrt{10} - \sqrt{5} \cdot \frac{d}{d\theta} (\sec \theta)
Using the derivative of sec⁡θ\sec \theta, which is sec⁡θtan⁡θ\sec \theta \tan \theta, we have:
dydθ=10−5sec⁡θtan⁡θ \frac{dy}{d\theta} = \sqrt{10} - \sqrt{5} \sec \theta \tan \theta

STEP 4

Find the critical points by setting the derivative equal to zero:
10−5sec⁡θtan⁡θ=0 \sqrt{10} - \sqrt{5} \sec \theta \tan \theta = 0
Solve for θ\theta:
10=5sec⁡θtan⁡θ \sqrt{10} = \sqrt{5} \sec \theta \tan \theta
sec⁡θtan⁡θ=105=2 \sec \theta \tan \theta = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}
This equation may not have a simple algebraic solution, so we will evaluate the function at the endpoints and check for any critical points numerically or graphically if needed.

STEP 5

Evaluate the function y y at the endpoints of the interval [0,π3][0, \frac{\pi}{3}].
1. At θ=0\theta = 0:
y(0)=10⋅0−5⋅sec⁡(0)=−5 y(0) = \sqrt{10} \cdot 0 - \sqrt{5} \cdot \sec(0) = -\sqrt{5}
2. At θ=π3\theta = \frac{\pi}{3}:
y(π3)=10⋅π3−5⋅sec⁡(π3) y\left(\frac{\pi}{3}\right) = \sqrt{10} \cdot \frac{\pi}{3} - \sqrt{5} \cdot \sec\left(\frac{\pi}{3}\right)
Since sec⁡(π3)=2\sec\left(\frac{\pi}{3}\right) = 2, we have:
y(π3)=10⋅π3−5⋅2 y\left(\frac{\pi}{3}\right) = \sqrt{10} \cdot \frac{\pi}{3} - \sqrt{5} \cdot 2
y(π3)=10π3−25 y\left(\frac{\pi}{3}\right) = \frac{\sqrt{10} \pi}{3} - 2\sqrt{5}

STEP 6

Determine the minimum and maximum values of y y .
From the evaluations:
- At θ=0\theta = 0, y=−5 y = -\sqrt{5} - At θ=π3\theta = \frac{\pi}{3}, y=10π3−25 y = \frac{\sqrt{10} \pi}{3} - 2\sqrt{5}
The minimum value is already given as −5-\sqrt{5}.
To find the maximum, compare the values:
10π3−25 \frac{\sqrt{10} \pi}{3} - 2\sqrt{5}
This is the maximum value of y y on the interval, assuming no critical points within the interval yield a larger value.
The maximum value is 10π3−25 \boxed{\frac{\sqrt{10} \pi}{3} - 2\sqrt{5}} .

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