- f(x) = x + 4 (0,2) 12. f(x) = tan x y = x (0,0) DA Comparing Ay and dy In Exercises 13–18, use the information to find and compare Ay and dy. x-Value x = 1 x = -2 Function 13. y = 0.5×3 14. y = 6 – 2×2 15. y = x4 + 1 16, y = 2 – ** 17. y = x – 2r3 18. y = 7×2 – 5x Differential of x Ax = dx = 0.1 Ax = dx = 0.1 Ax = dx = 0.01 Ax = dx = 0.01 Ax = dx = 0.001 Ax = dx = 0.001 x = -4
I need help with Question 16. Thank You.
The correct answer and explanation is:
Question 16:
The function given is y=2−x2y = 2 – x^2, and you need to compare Δy\Delta y and dydy at x=−4x = -4, where Δx=dx=0.01\Delta x = dx = 0.01.
Step 1: Define the Changes
- Change in yy (Δy\Delta y): Δy=y(x+Δx)−y(x)\Delta y = y(x + \Delta x) – y(x).
- Differential (dydy): The differential is an approximation of Δy\Delta y and is calculated as: dy=dydx⋅dxdy = \frac{dy}{dx} \cdot dx where dydx\frac{dy}{dx} is the derivative of yy with respect to xx.
Step 2: Derivative of y=2−x2y = 2 – x^2
The derivative is: dydx=−2x\frac{dy}{dx} = -2x
Step 3: Compute Δy\Delta y
- Original yy:
At x=−4x = -4: y(−4)=2−(−4)2=2−16=−14y(-4) = 2 – (-4)^2 = 2 – 16 = -14 - New xx:
x+Δx=−4+0.01=−3.99x + \Delta x = -4 + 0.01 = -3.99. - New yy: y(−3.99)=2−(−3.99)2=2−(15.9201)=−13.9201y(-3.99) = 2 – (-3.99)^2 = 2 – (15.9201) = -13.9201
- Change in yy: Δy=y(−3.99)−y(−4)=−13.9201−(−14)=0.0799\Delta y = y(-3.99) – y(-4) = -13.9201 – (-14) = 0.0799
Step 4: Compute dydy
dy=dydx⋅dx=(−2x)⋅dxdy = \frac{dy}{dx} \cdot dx = (-2x) \cdot dx
At x=−4x = -4: dydx=−2(−4)=8\frac{dy}{dx} = -2(-4) = 8 dy=8⋅0.01=0.08dy = 8 \cdot 0.01 = 0.08
Step 5: Compare Δy\Delta y and dydy
- Δy=0.0799\Delta y = 0.0799
- dy=0.08dy = 0.08
The values are very close, showing that dydy provides a good linear approximation to Δy\Delta y when Δx\Delta x is small.
Explanation:
The differential dydy is a linear approximation of the actual change in yy, denoted Δy\Delta y, based on the derivative of the function at a given point. Here, Δx=0.01\Delta x = 0.01 is small, so dydy closely approximates Δy\Delta y. This relationship is more accurate as Δx\Delta x becomes smaller because the curve of the function is locally linear over tiny intervals.
For y=2−x2y = 2 – x^2, the derivative dydx=−2x\frac{dy}{dx} = -2x reflects the instantaneous rate of change at any point xx. At x=−4x = -4, the rate of change is positive (dydx=8\frac{dy}{dx} = 8), indicating that yy increases as xx increases near this point. Calculating both Δy\Delta y and dydy confirms the accuracy of this approximation.