Multivariable Calculus Quiz: Level Curves And Gradient Orthogonality
2 questions · exam conditions
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Level Curves And Gradient OrthogonalityQuestion 1 of 2

For the function F(x,y)=ex+ysin⁡(xy)F(x,y) = e^{x+y} \sin(xy), suppose the level curve F(x,y)=kF(x,y) = k passes through the origin and has a well-defined tangent line there. What additional information about the level curves can be determined?

The level curve through the origin is the line y=0y = 0, and nearby level curves have slopes approaching 11
The level curve through the origin is the line x=0x = 0, and nearby level curves have slopes approaching −1-1
The level curve through the origin consists of both coordinate axes, and the tangent concept is undefined at the origin
The level curve through the origin is the line y=xy = x, and nearby level curves are approximately parallel to this line
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Level Curves And Gradient Orthogonality

Practice Level Curves And Gradient Orthogonality 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 Level Curves And Gradient Orthogonality, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

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

For the function F(x,y)=ex+ysin⁡(xy)F(x,y) = e^{x+y} \sin(xy), suppose the level curve F(x,y)=kF(x,y) = k passes through the origin and has a well-defined tangent line there. What additional information about the level curves can be determined?

  1. The level curve through the origin is the line y=0y = 0, and nearby level curves have slopes approaching 11
  2. The level curve through the origin is the line x=0x = 0, and nearby level curves have slopes approaching −1-1
  3. The level curve through the origin consists of both coordinate axes, and the tangent concept is undefined at the origin (correct answer)
  4. The level curve through the origin is the line y=xy = x, and nearby level curves are approximately parallel to this line
Explanation: First, F(0,0)=e0sin⁡(0)=1⋅0=0F(0,0) = e^0 \sin(0) = 1 \cdot 0 = 0. So the level curve through the origin is F(x,y)=0F(x,y) = 0, which means ex+ysin⁡(xy)=0e^{x+y} \sin(xy) = 0. Since ex+y>0e^{x+y} > 0 always, we need sin⁡(xy)=0\sin(xy) = 0, so xy=nπxy = n\pi for integer nn. Near the origin, this gives xy=0xy = 0, which consists of both the xx-axis (y=0y = 0) and yy-axis (x=0x = 0). At the origin, these two curves intersect, so there's no unique tangent line. The gradient ∇F=(ex+y(sin⁡(xy)+ycos⁡(xy)),ex+y(sin⁡(xy)+xcos⁡(xy)))\nabla F = (e^{x+y}(\sin(xy) + y\cos(xy)), e^{x+y}(\sin(xy) + x\cos(xy))) equals (0,0)(0,0) at the origin, confirming the tangent is undefined.

Question 2

Consider the surface z=f(x,y)=x3−3xy2+2yz = f(x,y) = x^3 - 3xy^2 + 2y. If you project the curve where this surface intersects the plane z=5z = 5 onto the xyxy-plane, what is the slope of the resulting level curve at the point where x=2x = 2 and y=1y = 1?

  1. The point (2,1)(2,1) is not on the level curve f(x,y)=5f(x,y) = 5, so the slope is undefined (correct answer)
  2. 34\frac{3}{4} because this is the ratio of the partial derivatives at (2,1)(2,1)
  3. −34-\frac{3}{4} because the level curve slope is the negative reciprocal of the gradient direction
  4. −43-\frac{4}{3} because this is −fxfy-\frac{f_x}{f_y} evaluated at (2,1)(2,1)
Explanation: First check if (2,1)(2,1) is on the level curve f(x,y)=5f(x,y) = 5: f(2,1)=23−3(2)(1)2+2(1)=8−6+2=4≠5f(2,1) = 2^3 - 3(2)(1)^2 + 2(1) = 8 - 6 + 2 = 4 \neq 5. Since the point is not on the specified level curve, the question about the slope at this point is meaningless. The other choices assume the point is on the curve and apply various (some incorrect) formulas for level curve slopes.