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Math  /  Calculus

QuestionWe wish to find the length of the shortest ladder that will reach from the ground over the fence to the wall of the building. [A] First, find a formula for the length of the ladder in terms of θ\theta. (Hint: split the ladder into 2 parts.) Type theta for θ\theta. L(θ)=15sin⁡(θ)+5cos⁡(θ)L(\theta)=\frac{15}{\sin (\theta)}+\frac{5}{\cos (\theta)} [B] Now, find the derivative, L′(θ)L^{\prime}(\theta). Type theta for θ\theta. L′(θ)=−15cos⁡(θ)sin⁡2(θ)+5sin⁡(θ)cos⁡2(θ)L^{\prime}(\theta)=-\frac{15 \cos (\theta)}{\sin ^{2}(\theta)}+\frac{5 \sin (\theta)}{\cos ^{2}(\theta)} [C] Once you find the value of θ\theta that makes L′(θ)=0L^{\prime}(\theta)=0, substitute that into your original function to find the length of the shortest ladder. (Give your answer accurate to 5 decimal places.) L(θmin⁡)≈1 feet L\left(\theta_{\min }\right) \approx 1 \text { feet } □\square Enter an integer or decimal number [more.-1] Submit Question

Studdy Solution

STEP 1

What is this asking? We need to find the shortest ladder that can lean over a fence and reach a building wall. Watch out! Don't forget that the ladder's length depends on the angle it makes with the ground.
Also, remember that we're looking for the *shortest* ladder, so we'll need to use calculus to find the minimum length.

STEP 2

1. Set up the length function
2. Find the critical points
3. Calculate the shortest ladder length

STEP 3

Alright, so we've got this ladder situation!
We're given the function for the ladder's length, L(θ)L(\theta), in terms of the angle θ\theta.
It's given by L(θ)=15sin⁡(θ)+5cos⁡(θ)L(\theta) = \frac{15}{\sin(\theta)} + \frac{5}{\cos(\theta)}.
This formula comes from splitting the ladder into two parts: one part goes from the ground to the top of the fence, and the other part goes from the top of the fence to the building.

STEP 4

To find the shortest ladder, we need to find the minimum value of L(θ)L(\theta).
We do this by finding where the derivative, L′(θ)L'(\theta), is equal to zero.
This tells us where the slope of the length function is zero, meaning we might have a minimum (or maximum).
We're given L′(θ)=−15cos⁡(θ)sin⁡2(θ)+5sin⁡(θ)cos⁡2(θ)L'(\theta) = -\frac{15\cos(\theta)}{\sin^2(\theta)} + \frac{5\sin(\theta)}{\cos^2(\theta)}.

STEP 5

Now, we set L′(θ)L'(\theta) equal to zero and solve for θ\theta: −15cos⁡(θ)sin⁡2(θ)+5sin⁡(θ)cos⁡2(θ)=0-\frac{15\cos(\theta)}{\sin^2(\theta)} + \frac{5\sin(\theta)}{\cos^2(\theta)} = 0 5sin⁡(θ)cos⁡2(θ)=15cos⁡(θ)sin⁡2(θ) \frac{5\sin(\theta)}{\cos^2(\theta)} = \frac{15\cos(\theta)}{\sin^2(\theta)} **Multiply** both sides by sin⁡2(θ)\sin^2(\theta) and cos⁡2(θ)\cos^2(\theta) to get: 5sin⁡3(θ)=15cos⁡3(θ) 5\sin^3(\theta) = 15\cos^3(\theta) sin⁡3(θ)cos⁡3(θ)=155 \frac{\sin^3(\theta)}{\cos^3(\theta)} = \frac{15}{5} tan⁡3(θ)=3 \tan^3(\theta) = 3 tan⁡(θ)=33 \tan(\theta) = \sqrt[3]{3} θ=arctan⁡(33) \theta = \arctan(\sqrt[3]{3})

STEP 6

Using a calculator, we find that θ≈0.907571211\theta \approx \bold{0.907571211} radians.
This is our **critical angle**!

STEP 7

Now, we **plug** this value of θ\theta back into our original L(θ)L(\theta) function: L(θ)=15sin⁡(θ)+5cos⁡(θ) L(\theta) = \frac{15}{\sin(\theta)} + \frac{5}{\cos(\theta)} L(0.907571211)=15sin⁡(0.907571211)+5cos⁡(0.907571211) L(0.907571211) = \frac{15}{\sin(0.907571211)} + \frac{5}{\cos(0.907571211)} L(0.907571211)≈150.788601945+50.615412215 L(0.907571211) \approx \frac{15}{0.788601945} + \frac{5}{0.615412215} L(0.907571211)≈19.01923789+8.1240384 L(0.907571211) \approx 19.01923789 + 8.1240384 L(0.907571211)≈27.14327629 L(0.907571211) \approx \bold{27.14327629}

STEP 8

The shortest ladder length is approximately **27.14328** feet (rounded to 5 decimal places).

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