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Linear Approximation - Calculus 1

  1. The position of a particle along the x-axis is given by p(t)=sin⁡(πt4)p(t) = \sin(\frac{\pi t}{4}). Use the tangent line at t=1t=1 to approximate the position of the particle at t=1.1t=1.1.
    Answer: 22+2π80\frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{80} This question tests linear approximation using tangent lines, a fundamental application of derivatives. When you need to estimate a function value near a known point, the tangent line provides the best linear approximation. To find the tangent line at t=1t=1, you need both the function value and the derivative at that point. First, find p(1)=sin⁡(π⋅14)=sin⁡(π4)=22p(1) = \sin(\frac{\pi \cdot 1}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}. Next, find the derivative: p′(t)=cos⁡(πt4)⋅π4p'(t) = \cos(\frac{\pi t}{4}) \cdot \frac{\pi}{4}. At t=1t=1: p′(1)=cos⁡(π4)⋅π4=22⋅π4=2π8p'(1) = \cos(\frac{\pi}{4}) \cdot \frac{\pi}{4} = \frac{\sqrt{2}}{2} \cdot \frac{\pi}{4} = \frac{\sqrt{2}\pi}{8}. The tangent line equation is L(t)=p(1)+p′(1)(t−1)=22+2π8(t−1)L(t) = p(1) + p'(1)(t-1) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{8}(t-1). For t=1.1t=1.1: L(1.1)=22+2π8(0.1)=22+2π80L(1.1) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{8}(0.1) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{80}. This confirms answer C is correct. Answer A uses 12\frac{1}{2} instead of 22\frac{\sqrt{2}}{2}, which would come from incorrectly evaluating sin⁡(π4)\sin(\frac{\pi}{4}) as sin⁡(30°)\sin(30°) rather than sin⁡(45°)\sin(45°). Answer B has the wrong denominator in the linear term (40 instead of 80), likely from forgetting to multiply by the change in tt (0.1). Answer D has the correct calculation but wrong sign, suggesting a sign error when computing the tangent line. Remember: linear approximation problems always follow the pattern f(a)+f′(a)(x−a)f(a) + f'(a)(x-a). Double-check your trigonometric values and arithmetic carefully.
  2. Let h(x)=ln⁡(f(x))h(x) = \ln(f(x)), where f(x)f(x) is a differentiable function with f(2)=e3f(2)=e^3 and f′(2)=−2e3f'(2)=-2e^3. Use a linear approximation centered at x=2x=2 to estimate the value of h(1.9)h(1.9).
    Answer: 3.23.2 The linear approximation of h(x)h(x) at a=2a=2 is L(x)=h(2)+h′(2)(x−2)L(x) = h(2) + h'(2)(x-2). First, we find h(2)=ln⁡(f(2))=ln⁡(e3)=3h(2) = \ln(f(2)) = \ln(e^3) = 3. Next, we find the derivative h′(x)h'(x) using the chain rule: h′(x)=1f(x)⋅f′(x)=f′(x)f(x)h'(x) = \frac{1}{f(x)} \cdot f'(x) = \frac{f'(x)}{f(x)}. At x=2x=2, the slope is h′(2)=f′(2)f(2)=−2e3e3=−2h'(2) = \frac{f'(2)}{f(2)} = \frac{-2e^3}{e^3} = -2. Now we can approximate h(1.9)h(1.9): L(1.9)=h(2)+h′(2)(1.9−2)=3+(−2)(−0.1)=3+0.2=3.2L(1.9) = h(2) + h'(2)(1.9 - 2) = 3 + (-2)(-0.1) = 3 + 0.2 = 3.2.
  3. The linear approximation for a differentiable function f(x)f(x) at x=ax=a is given by the line L(x)=−12x+3L(x) = -\frac{1}{2}x + 3. If it is also known that f(a)=1f(a)=1, what is the value of aa?
    Answer: 44 The tangent line L(x)L(x) at x=ax=a must pass through the point of tangency (a,f(a))(a, f(a)). This means that L(a)=f(a)L(a) = f(a). We are given the equation for L(x)L(x) and the value of f(a)f(a). Substituting x=ax=a into the equation for L(x)L(x) gives L(a)=−12a+3L(a) = -\frac{1}{2}a + 3. Setting this equal to f(a)=1f(a)=1, we get the equation 1=−12a+31 = -\frac{1}{2}a + 3. To solve for aa, we subtract 3 from both sides: 1−3=−12a1 - 3 = -\frac{1}{2}a, which simplifies to −2=−12a-2 = -\frac{1}{2}a. Multiplying both sides by −2-2 gives a=4a=4.
  4. Let L(x)L(x) be the linearization of the function f(x)=e2xf(x) = e^{2x} at x=0x=0. When using L(0.1)L(0.1) to approximate the value of f(0.1)f(0.1), is the approximation an overestimate or an underestimate, and why?
    Answer: An underestimate, because the graph of f(x)f(x) is concave up at x=0x=0. The relationship between a linear approximation and the actual function value is determined by the function's concavity at the point of tangency. To find the concavity of f(x)=e2xf(x) = e^{2x} at x=0x=0, we examine the second derivative. The first derivative is f′(x)=2e2xf'(x) = 2e^{2x}. The second derivative is f′′(x)=4e2xf''(x) = 4e^{2x}. At x=0x=0, f′′(0)=4e0=4f''(0) = 4e^{0} = 4. Since f′′(0)>0f''(0) > 0, the graph of f(x)f(x) is concave up at x=0x=0. For a concave up function, the tangent line lies below the curve, which means the linear approximation will be an underestimate of the actual function value.
  5. Let L(x)L(x) be the linear approximation of the function f(x)=xcos⁡(x)f(x) = x \cos(x) at x=π2x=\frac{\pi}{2}. Which of the following is the value of L(π2−0.1)L(\frac{\pi}{2} - 0.1)?
    Answer: 0.05π0.05\pi The linear approximation of f(x)f(x) at a=π2a=\frac{\pi}{2} is L(x)=f(π2)+f′(π2)(x−π2)L(x) = f(\frac{\pi}{2}) + f'(\frac{\pi}{2})(x-\frac{\pi}{2}). First, find f(π2)=π2cos⁡(π2)=π2⋅0=0f(\frac{\pi}{2}) = \frac{\pi}{2}\cos(\frac{\pi}{2}) = \frac{\pi}{2} \cdot 0 = 0. Next, find the derivative f′(x)f'(x) using the product rule: f′(x)=1⋅cos⁡(x)+x⋅(−sin⁡(x))=cos⁡(x)−xsin⁡(x)f'(x) = 1 \cdot \cos(x) + x \cdot (-\sin(x)) = \cos(x) - x\sin(x). Evaluate the derivative at x=π2x=\frac{\pi}{2}: f′(π2)=cos⁡(π2)−π2sin⁡(π2)=0−π2(1)=−π2f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) - \frac{\pi}{2}\sin(\frac{\pi}{2}) = 0 - \frac{\pi}{2}(1) = -\frac{\pi}{2}. Now, assemble the approximation for x=π2−0.1x = \frac{\pi}{2} - 0.1: L(π2−0.1)=0+(−π2)((π2−0.1)−π2)=−π2(−0.1)=0.05πL(\frac{\pi}{2} - 0.1) = 0 + (-\frac{\pi}{2}) ( (\frac{\pi}{2} - 0.1) - \frac{\pi}{2}) = -\frac{\pi}{2}(-0.1) = 0.05\pi.
  6. The side length of a cube is measured to be 1010 cm with a possible measurement error of ±0.03\pm 0.03 cm. Use differentials to estimate the maximum possible relative error in calculating the volume of the cube.
    Answer: ±0.009\pm 0.009 The volume of a cube is V=s3V=s^3. The relative error in the volume is ΔVV\frac{\Delta V}{V}, which can be approximated by the differential ratio dVV\frac{dV}{V}. The differential dVdV is given by dV=V′(s)ds=3s2dsdV = V'(s)ds = 3s^2 ds. Therefore, the relative error is dVV=3s2dss3=3dss\frac{dV}{V} = \frac{3s^2 ds}{s^3} = \frac{3 ds}{s}. We are given the side length s=10s=10 cm and the error in the side length ds=±0.03ds = \pm 0.03 cm. Plugging these values in, the relative error is 3(±0.03)10=±0.0910=±0.009\frac{3 (\pm 0.03)}{10} = \frac{\pm 0.09}{10} = \pm 0.009. The maximum possible relative error is the magnitude of this value, 0.0090.009.
  7. The volume of a spherical balloon is increasing. A linear approximation is used to estimate the increase in volume as the radius increases from r=5r=5 cm to r=5.1r=5.1 cm. Which of the following is the best estimate for the increase in volume? (The volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3.)
    Answer: 10π10\pi cm3^3 The increase in volume ΔV\Delta V can be approximated by the differential dVdV. The formula is dV=V′(r)drdV = V'(r)dr. First, we find the derivative of the volume function V(r)=43πr3V(r) = \frac{4}{3}\pi r^3, which is V′(r)=4πr2V'(r) = 4\pi r^2. We are using a linear approximation centered at r=5r=5 cm, so we evaluate the derivative at this point: V′(5)=4π(52)=100πV'(5) = 4\pi (5^2) = 100\pi. The change in radius is dr=5.1−5=0.1dr = 5.1 - 5 = 0.1 cm. The approximate increase in volume is dV=V′(5)dr=(100π)(0.1)=10πdV = V'(5)dr = (100\pi)(0.1) = 10\pi cm3^3.
  8. The function g(x)g(x) has a local linear approximation at x=3x=3 given by L(x)=5−2(x−3)L(x) = 5 - 2(x-3). Let h(x)=(g(x))2h(x) = (g(x))^2. Which of the following is the local linear approximation for h(x)h(x) at x=3x=3?
    Answer: 25−20(x−3)25 - 20(x-3) When you encounter a problem asking for the linear approximation of a composite function like h(x)=(g(x))2h(x) = (g(x))^2, you need to use the chain rule along with the given linear approximation. From the given information, L(x)=5−2(x−3)L(x) = 5 - 2(x-3) tells us that g(3)=5g(3) = 5 and g′(3)=−2g'(3) = -2. To find the linear approximation of h(x)=(g(x))2h(x) = (g(x))^2 at x=3x = 3, you need h(3)h(3) and h′(3)h'(3). First, h(3)=(g(3))2=52=25h(3) = (g(3))^2 = 5^2 = 25. Next, using the chain rule: h′(x)=2g(x)⋅g′(x)h'(x) = 2g(x) \cdot g'(x), so h′(3)=2g(3)⋅g′(3)=2(5)(−2)=−20h'(3) = 2g(3) \cdot g'(3) = 2(5)(-2) = -20. Therefore, the linear approximation is h(3)+h′(3)(x−3)=25−20(x−3)h(3) + h'(3)(x-3) = 25 - 20(x-3). Looking at the wrong answers: Choice A gives 10−4(x−3)10 - 4(x-3), which incorrectly uses h(3)=2g(3)=10h(3) = 2g(3) = 10 instead of (g(3))2=25(g(3))^2 = 25, and also miscalculates the derivative. Choice B gives 25−4(x−3)25 - 4(x-3), which correctly finds h(3)=25h(3) = 25 but incorrectly computes h′(3)=2g′(3)=−4h'(3) = 2g'(3) = -4, forgetting to multiply by g(3)g(3) in the chain rule. Choice D gives 5−20(x−3)5 - 20(x-3), which correctly finds the derivative but uses h(3)=g(3)=5h(3) = g(3) = 5 instead of squaring it. Study tip: When finding linear approximations of composite functions, always apply the chain rule carefully to find the derivative, and don't forget that the constant term comes from evaluating the original function, not its derivative.
  9. Let f(x)=x5+2x+1f(x) = x^5 + 2x + 1 and let f−1f^{-1} be the inverse function of ff. Use a linear approximation to estimate the value of f−1(3.8)f^{-1}(3.8).
    Answer: 3435\frac{34}{35} We need to find a linear approximation for f−1(x)f^{-1}(x) near x=3.8x=3.8. First, we find a convenient point for our approximation. By inspection, f(1)=15+2(1)+1=4f(1) = 1^5 + 2(1) + 1 = 4. This means f−1(4)=1f^{-1}(4)=1. We will center our approximation at b=4b=4. The linearization of f−1f^{-1} at b=4b=4 is L(x)=f−1(4)+(f−1)′(4)(x−4)L(x) = f^{-1}(4) + (f^{-1})'(4)(x-4). We need the derivative of the inverse function, (f−1)′(4)=1f′(f−1(4))=1f′(1)(f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))} = \frac{1}{f'(1)}. The derivative of f(x)f(x) is f′(x)=5x4+2f'(x) = 5x^4 + 2. So, f′(1)=5(1)4+2=7f'(1) = 5(1)^4 + 2 = 7. Therefore, (f−1)′(4)=17(f^{-1})'(4) = \frac{1}{7}. The approximation is L(3.8)=1+17(3.8−4)=1+17(−0.2)=1−0.27=1−270=1−135=3435L(3.8) = 1 + \frac{1}{7}(3.8 - 4) = 1 + \frac{1}{7}(-0.2) = 1 - \frac{0.2}{7} = 1 - \frac{2}{70} = 1 - \frac{1}{35} = \frac{34}{35}.
  10. A curve is defined by the equation x3+y3=9xyx^3 + y^3 = 9xy. Use the tangent line to the curve at the point (4,2)(4,2) to approximate the y-coordinate of a point on the curve with x-coordinate 4.14.1.
    Answer: 2.1252.125 First, we find the slope of the tangent line at (4,2)(4,2) using implicit differentiation on x3+y3=9xyx^3 + y^3 = 9xy. Differentiating with respect to xx gives 3x2+3y2dydx=9y+9xdydx3x^2 + 3y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}. Solving for dydx\frac{dy}{dx} yields dydx=9y−3x23y2−9x=3y−x2y2−3x\frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x} = \frac{3y - x^2}{y^2 - 3x}. At the point (4,2)(4,2), the slope is m=3(2)−4222−3(4)=6−164−12=−10−8=54m = \frac{3(2) - 4^2}{2^2 - 3(4)} = \frac{6-16}{4-12} = \frac{-10}{-8} = \frac{5}{4}. The equation of the tangent line (the linearization) at (4,2)(4,2) is L(x)=y0+m(x−x0)=2+54(x−4)L(x) = y_0 + m(x-x_0) = 2 + \frac{5}{4}(x-4). To approximate the y-coordinate at x=4.1x=4.1, we evaluate L(4.1)=2+54(4.1−4)=2+54(0.1)=2+0.54=2+0.125=2.125L(4.1) = 2 + \frac{5}{4}(4.1-4) = 2 + \frac{5}{4}(0.1) = 2 + \frac{0.5}{4} = 2 + 0.125 = 2.125.

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