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Math  /  Calculus

QuestionUsing Local Linearity Linearization For each differential equation, let y=f(x)y=f(x) be the particular solution to the differential equation with the given aitial condition. 1.) dydx=(5−y)sin⁡x\frac{d y}{d x}=(5-y) \sin x and f(π2)=2f\left(\frac{\pi}{2}\right)=2. a. Write an equation for the line tangent to the graph of ff at the point (π2,2)\left(\frac{\pi}{2}, 2\right) b. Use the tangent line to approximate f(1.5)f(1.5). 2.) dydx=−4xy\frac{d y}{d x}=-\frac{4 x}{y} and f(1)=3f(1)=3. a. Write an equation for the line tangent to the graph of ff at the point (1,3)(1,3) b. Use the tangent line to approximate f(1.1)f(1.1).
Answer the questions for each function listed. 3.) f(x)=2cos⁡x+1f(x)=2 \cos x+1 is concave down on [0,π2]\left[0, \frac{\pi}{2}\right]. 4.) f(x)=e2xx+1f(x)=\frac{e^{2 x}}{x+1} is concave up on x>−1x>-1. a. What is the estimate for f(3)f(3) using the local c. What is the estimate for f(0.1)f(0.1) using the linear approximation for ff at x=π2x=\frac{\pi}{2} ? Give an local linear approximation for ff at x=0x=0 ? exact answer. b. Is it an underestimate or overestimate? d. Is it an underestimate or overestimate? Explain. Explain.

Studdy Solution

STEP 1

1. We are using local linearity (linearization) to approximate function values.
2. The tangent line at a point provides a linear approximation of the function near that point.
3. We need to find the tangent line equations and use them for approximation.
4. We will determine whether the approximation is an underestimate or overestimate based on concavity.

STEP 2

1. Solve Problem 1: Find the tangent line and approximate f(1.5) f(1.5) .
2. Solve Problem 2: Find the tangent line and approximate f(1.1) f(1.1) .
3. Analyze Problem 3: Determine if the approximation is an underestimate or overestimate.
4. Analyze Problem 4: Determine if the approximation is an underestimate or overestimate.

STEP 3

To find the tangent line for the differential equation dydx=(5−y)sin⁡x\frac{d y}{d x}=(5-y) \sin x at the point (π2,2)\left(\frac{\pi}{2}, 2\right), we first calculate the derivative at that point.
dydx=(5−2)sin⁡(π2)=3⋅1=3\frac{d y}{d x} = (5 - 2) \sin\left(\frac{\pi}{2}\right) = 3 \cdot 1 = 3
The equation of the tangent line is given by:
y−2=3(x−π2)y - 2 = 3\left(x - \frac{\pi}{2}\right)

STEP 4

To approximate f(1.5) f(1.5) using the tangent line:
Substitute x=1.5 x = 1.5 into the tangent line equation:
y−2=3(1.5−π2)y - 2 = 3\left(1.5 - \frac{\pi}{2}\right)
Calculate the value:
y=2+3(1.5−π2)y = 2 + 3\left(1.5 - \frac{\pi}{2}\right)

STEP 5

For the differential equation dydx=−4xy\frac{d y}{d x}=-\frac{4 x}{y} at the point (1,3)(1, 3), calculate the derivative:
dydx=−4⋅13=−43\frac{d y}{d x} = -\frac{4 \cdot 1}{3} = -\frac{4}{3}
The equation of the tangent line is:
y−3=−43(x−1)y - 3 = -\frac{4}{3}(x - 1)

STEP 6

To approximate f(1.1) f(1.1) using the tangent line:
Substitute x=1.1 x = 1.1 into the tangent line equation:
y−3=−43(1.1−1)y - 3 = -\frac{4}{3}(1.1 - 1)
Calculate the value:
y=3−43(0.1)y = 3 - \frac{4}{3}(0.1)

STEP 7

For f(x)=2cos⁡x+1 f(x) = 2 \cos x + 1 , the function is concave down on [0,π2][0, \frac{\pi}{2}].
Since the function is concave down, the tangent line approximation at any point in this interval will be an overestimate.

STEP 8

For f(x)=e2xx+1 f(x) = \frac{e^{2x}}{x+1} , the function is concave up on x>−1 x > -1 .
Since the function is concave up, the tangent line approximation will be an underestimate.
The solutions for the approximations and concavity analysis are:
1. f(1.5)≈2+3(1.5−π2) f(1.5) \approx 2 + 3(1.5 - \frac{\pi}{2})
2. f(1.1)≈3−43(0.1) f(1.1) \approx 3 - \frac{4}{3}(0.1)
3. Overestimate on [0,π2][0, \frac{\pi}{2}]
4. Underestimate on x>−1 x > -1

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