Multivariable Calculus Quiz: Tangent Planes And Linear Approximation
2 questions · exam conditions
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Tangent Planes And Linear ApproximationQuestion 1 of 2

The function g(x,y)=xy+yxg(x,y) = \frac{x}{y} + \frac{y}{x} is approximated linearly near (2,1)(2,1) to estimate g(1.98,1.01)g(1.98, 1.01). What is the error in this linear approximation compared to the actual value?

The error is approximately 0.0010.001
The error is approximately −0.001-0.001
The error is approximately 0.010.01
The error is approximately −0.01-0.01
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Tangent Planes And Linear Approximation

Practice Tangent Planes And Linear Approximation in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Tangent Planes And Linear Approximation, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The function g(x,y)=xy+yxg(x,y) = \frac{x}{y} + \frac{y}{x} is approximated linearly near (2,1)(2,1) to estimate g(1.98,1.01)g(1.98, 1.01). What is the error in this linear approximation compared to the actual value?

  1. The error is approximately 0.0010.001 (correct answer)
  2. The error is approximately −0.001-0.001
  3. The error is approximately 0.010.01
  4. The error is approximately −0.01-0.01
Explanation: First, compute the linear approximation. g(2,1)=21+12=2.5g(2,1) = \frac{2}{1} + \frac{1}{2} = 2.5. The partial derivatives are gx=1y−yx2g_x = \frac{1}{y} - \frac{y}{x^2} and gy=−xy2+1xg_y = -\frac{x}{y^2} + \frac{1}{x}. At (2,1)(2,1): gx(2,1)=1−14=0.75g_x(2,1) = 1 - \frac{1}{4} = 0.75 and gy(2,1)=−2+0.5=−1.5g_y(2,1) = -2 + 0.5 = -1.5. The linear approximation is L(1.98,1.01)=2.5+0.75(−0.02)+(−1.5)(0.01)=2.5−0.015−0.015=2.47L(1.98,1.01) = 2.5 + 0.75(-0.02) + (-1.5)(0.01) = 2.5 - 0.015 - 0.015 = 2.47. Now compute the actual value: g(1.98,1.01)=1.981.01+1.011.98≈1.9604+0.5101=2.4705g(1.98,1.01) = \frac{1.98}{1.01} + \frac{1.01}{1.98} \approx 1.9604 + 0.5101 = 2.4705. The error is 2.4705−2.47=0.0005≈0.0012.4705 - 2.47 = 0.0005 \approx 0.001.

Question 2

Consider the function w=f(x,y)=ln⁡(x2+y2)w = f(x,y) = \ln(x^2 + y^2). If the linear approximation at point (3,4)(3,4) is used to estimate f(3.05,3.95)f(3.05, 3.95), what is the approximated value?

  1. ln⁡(25)+0.01\ln(25) + 0.01
  2. ln⁡(25)−0.01\ln(25) - 0.01
  3. ln⁡(25)+0.004\ln(25) + 0.004
  4. ln⁡(25)−0.004\ln(25) - 0.004 (correct answer)
Explanation: The linear approximation is L(x,y)=f(3,4)+fx(3,4)(x−3)+fy(3,4)(y−4)L(x,y) = f(3,4) + f_x(3,4)(x-3) + f_y(3,4)(y-4). First, f(3,4)=ln⁡(9+16)=ln⁡(25)f(3,4) = \ln(9+16) = \ln(25). Next, fx=2xx2+y2f_x = \frac{2x}{x^2+y^2}, so fx(3,4)=625f_x(3,4) = \frac{6}{25}. Similarly, fy=2yx2+y2f_y = \frac{2y}{x^2+y^2}, so fy(3,4)=825f_y(3,4) = \frac{8}{25}. For the point (3.05,3.95)(3.05, 3.95), Δx=0.05\Delta x = 0.05 and Δy=−0.05\Delta y = -0.05. Therefore, L(3.05,3.95)=ln⁡(25)+625(0.05)+825(−0.05)=ln⁡(25)+6⋅0.05−8⋅0.0525=ln⁡(25)+−0.125=ln⁡(25)−0.004L(3.05,3.95) = \ln(25) + \frac{6}{25}(0.05) + \frac{8}{25}(-0.05) = \ln(25) + \frac{6 \cdot 0.05 - 8 \cdot 0.05}{25} = \ln(25) + \frac{-0.1}{25} = \ln(25) - 0.004.