Calculus 1 Quiz: Implicit Relation Behaviors
20 questions · exam conditions
0:00
Implicit Relation BehaviorsQuestion 1 of 20

For the curve defined by the equation ax2+by2=cax^2 + by^2 = c, where a, b, and c are positive constants, the second derivative d2y/dx2d^2y/dx^2 is proportional to which of the following?

y−1y^{-1}
y−2y^{-2}
xy−2x y^{-2}
y−3y^{-3}
← Back to quizzes

Calculus 1 Quiz

Calculus 1 Quiz: Implicit Relation Behaviors

Practice Implicit Relation Behaviors in Calculus 1 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 Implicit Relation Behaviors, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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 curve defined by the equation ax2+by2=cax^2 + by^2 = c, where a, b, and c are positive constants, the second derivative d2y/dx2d^2y/dx^2 is proportional to which of the following?

  1. y−1y^{-1}
  2. y−2y^{-2}
  3. xy−2x y^{-2}
  4. y−3y^{-3} (correct answer)
Explanation: When you encounter implicit differentiation problems involving conic sections like ellipses, you need to find the second derivative by differentiating twice and then analyzing the resulting expression. Starting with ax2+by2=cax^2 + by^2 = c, use implicit differentiation to find the first derivative. Differentiating both sides: 2ax+2bydydx=02ax + 2by\frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx}: dydx=−axby\frac{dy}{dx} = -\frac{ax}{by}. To find the second derivative, differentiate dydx=−axby\frac{dy}{dx} = -\frac{ax}{by} using the quotient rule: d2ydx2=−ab⋅y⋅1−x⋅dydxy2\frac{d^2y}{dx^2} = -\frac{a}{b} \cdot \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}. Substituting dydx=−axby\frac{dy}{dx} = -\frac{ax}{by}: d2ydx2=−ab⋅y+ax2byy2=−aby3(by+ax2)\frac{d^2y}{dx^2} = -\frac{a}{b} \cdot \frac{y + \frac{ax^2}{by}}{y^2} = -\frac{a}{by^3}(by + ax^2). Since ax2+by2=cax^2 + by^2 = c, we have by+ax2=c−by2+by=c−by(y−1)by + ax^2 = c - by^2 + by = c - by(y-1). However, more simply, the expression shows that d2ydx2\frac{d^2y}{dx^2} is proportional to y−3y^{-3}, making D correct. Choice A (y−1y^{-1}) would result from a linear relationship. Choice B (y−2y^{-2}) might come from incorrectly applying the quotient rule or stopping at the denominator term. Choice C (xy−2xy^{-2}) incorrectly suggests the second derivative depends on both xx and yy in this specific way. Remember: for conic sections, the second derivative typically involves yy raised to the −3-3 power due to the quotient rule applied to rational expressions containing yy in both numerator and denominator.

Question 2

Consider the curve defined by the equation x2+xy+y2=7x^2 + xy + y^2 = 7. At what positive y-coordinate is the tangent line to this curve horizontal?

  1. y=7/3y = \sqrt{7/3}
  2. y=27/3y = 2\sqrt{7/3} (correct answer)
  3. y=7y = \sqrt{7}
  4. y=27y = 2\sqrt{7}
Explanation: To find where the tangent line is horizontal, we first find the derivative dy/dxdy/dx using implicit differentiation: 2x+(1⋅y+x⋅dy/dx)+2y⋅dy/dx=02x + (1 \cdot y + x \cdot dy/dx) + 2y \cdot dy/dx = 0. Grouping the dy/dxdy/dx terms gives dy/dx(x+2y)=−2x−ydy/dx(x + 2y) = -2x - y, so dy/dx=−(2x+y)/(x+2y)dy/dx = -(2x+y)/(x+2y). A horizontal tangent occurs when dy/dx=0dy/dx = 0, which means the numerator must be zero: 2x+y=02x+y=0, or y=−2xy = -2x. We substitute this relationship back into the original equation: x2+x(−2x)+(−2x)2=7x^2 + x(-2x) + (-2x)^2 = 7, which simplifies to x2−2x2+4x2=7x^2 - 2x^2 + 4x^2 = 7, or 3x2=73x^2 = 7. This gives x=±7/3x = \pm\sqrt{7/3}. The corresponding y-coordinates are found using y=−2xy=-2x. If x=7/3x = \sqrt{7/3}, y=−27/3y = -2\sqrt{7/3}. If x=−7/3x = -\sqrt{7/3}, y=27/3y = 2\sqrt{7/3}. The positive y-coordinate is 27/32\sqrt{7/3}.

Question 3

What is the sum of the y-coordinates of all points on the circle x2+y2−4x+2y=4x^2 + y^2 - 4x + 2y = 4 where the tangent line is vertical?

  1. -2 (correct answer)
  2. -1
  3. 0
  4. 2
Explanation: First, find dy/dxdy/dx by implicit differentiation: 2x+2y(dy/dx)−4+2(dy/dx)=02x + 2y(dy/dx) - 4 + 2(dy/dx) = 0. Collecting terms: dy/dx(2y+2)=4−2xdy/dx(2y+2) = 4-2x, so dy/dx=(4−2x)/(2y+2)=(2−x)/(y+1)dy/dx = (4-2x)/(2y+2) = (2-x)/(y+1). A tangent line is vertical when the denominator is zero and the numerator is non-zero. So, y+1=0  ⟹  y=−1y+1=0 \implies y=-1. To find the corresponding x-coordinates, substitute y=−1y=-1 into the original equation: x2+(−1)2−4x+2(−1)=4x^2 + (-1)^2 - 4x + 2(-1) = 4, which gives x2+1−4x−2=4x^2 + 1 - 4x - 2 = 4, or x2−4x−5=0x^2 - 4x - 5 = 0. Factoring gives (x−5)(x+1)=0(x-5)(x+1) = 0, so x=5x=5 or x=−1x=-1. The two points with vertical tangents are (5,−1)(5, -1) and (−1,−1)(-1, -1). The sum of the y-coordinates of these points is −1+(−1)=−2-1 + (-1) = -2. Note that the center of the circle is (2,−1)(2, -1), and vertical tangents occur at the endpoints of the horizontal diameter, which are at y=−1y=-1.

Question 4

Consider the curve defined by sin⁡(x)+cos⁡(y)=sin⁡(x)cos⁡(y)\sin(x) + \cos(y) = \sin(x)\cos(y). What is the slope of the tangent line at the point (0,π/2)(0, \pi/2)?

  1. -1
  2. 0
  3. 1 (correct answer)
  4. Undefined
Explanation: First, verify the point is on the curve: sin⁡(0)+cos⁡(π/2)=0+0=0\sin(0) + \cos(\pi/2) = 0+0=0 and sin⁡(0)cos⁡(π/2)=0⋅0=0\sin(0)\cos(\pi/2) = 0\cdot0=0. The point is valid. Differentiate implicitly with respect to x: cos⁡(x)−sin⁡(y)(dy/dx)=cos⁡(x)cos⁡(y)+sin⁡(x)(−sin⁡(y)(dy/dx))\cos(x) - \sin(y)(dy/dx) = \cos(x)\cos(y) + \sin(x)(-\sin(y)(dy/dx)). Rearrange to solve for dy/dxdy/dx: (dy/dx)(sin⁡(x)sin⁡(y)−sin⁡(y))=cos⁡(x)cos⁡(y)−cos⁡(x)(dy/dx)(\sin(x)\sin(y) - \sin(y)) = \cos(x)\cos(y) - \cos(x). So, dy/dx=(cos⁡(x)(cos⁡(y)−1))/(sin⁡(y)(sin⁡(x)−1))dy/dx = (\cos(x)(\cos(y)-1))/(\sin(y)(\sin(x)-1)). Evaluate at (0,π/2)(0, \pi/2): dy/dx=(cos⁡(0)(cos⁡(π/2)−1))/(sin⁡(π/2)(sin⁡(0)−1))=(1(0−1))/(1(0−1))=(−1)/(−1)=1dy/dx = (\cos(0)(\cos(\pi/2)-1))/(\sin(\pi/2)(\sin(0)-1)) = (1(0-1))/(1(0-1)) = (-1)/(-1) = 1.

Question 5

The curve defined by y5−y=x3−xy^5 - y = x^3 - x is symmetric with respect to the origin. If the slope of the tangent line at a point (a,b)(a,b) on the curve is mm, what is the slope of the tangent line at the point (−a,−b)(-a, -b)?

  1. mm (correct answer)
  2. −m-m
  3. 1/m1/m
  4. −1/m-1/m
Explanation: We first find the derivative dy/dxdy/dx in terms of x and y. Differentiating the equation implicitly gives: 5y4(dy/dx)−dy/dx=3x2−15y^4(dy/dx) - dy/dx = 3x^2 - 1. Factoring gives dy/dx(5y4−1)=3x2−1dy/dx(5y^4-1) = 3x^2-1, so dy/dx=(3x2−1)/(5y4−1)dy/dx = (3x^2-1)/(5y^4-1). The slope at (a,b)(a,b) is m=(3a2−1)/(5b4−1)m = (3a^2-1)/(5b^4-1). Now we evaluate the slope at the point (−a,−b)(-a,-b): dy/dx∣(−a,−b)=(3(−a)2−1)/(5(−b)4−1)=(3a2−1)/(5b4−1)dy/dx|_{(-a,-b)} = (3(-a)^2-1)/(5(-b)^4-1) = (3a^2-1)/(5b^4-1). This is the same expression as the slope at (a,b)(a,b). Therefore, the slope at (−a,−b)(-a, -b) is also mm. This is a property of curves where both x and y terms have odd or even degrees in a specific combination that leads to this symmetry.

Question 6

The curve known as the folium of Descartes is given by x3+y3=3xyx^3 + y^3 = 3xy. This curve has a vertical tangent at a point (x0,y0)(x_0, y_0) in the first quadrant where x0≠0x_0 \neq 0. What is the value of x0x_0?

  1. 21/32^{1/3}
  2. 22/32^{2/3} (correct answer)
  3. 24/32^{4/3}
  4. 22
Explanation: First, find dydx\frac{dy}{dx} using implicit differentiation: 3x2+3y2dydx=3y+3xdydx3x^2 + 3y^2\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}. Rearranging gives dydx(y2−x)=y−x2\frac{dy}{dx}(y^2 - x) = y - x^2, so dydx=y−x2y2−x\frac{dy}{dx} = \frac{y-x^2}{y^2-x}. A vertical tangent occurs where the denominator is zero and the numerator is non-zero. Set y2−x=0y^2 - x = 0, so x=y2x = y^2. Substitute this into the original equation: (y2)3+y3=3(y2)y(y^2)^3 + y^3 = 3(y^2)y, which simplifies to y6+y3=3y3y^6 + y^3 = 3y^3. This gives y6=2y3y^6 = 2y^3, or y3(y3−2)=0y^3(y^3 - 2) = 0. The solutions are y=0y=0 (which gives x=0x=0) and y3=2y^3 = 2, so y=21/3y = 2^{1/3}. The corresponding xx-coordinate is x0=y2=(21/3)2=22/3x_0 = y^2 = (2^{1/3})^2 = 2^{2/3}. We must check that the numerator y−x2y-x^2 is non-zero at this point. x2=(22/3)2=24/3x^2 = (2^{2/3})^2 = 2^{4/3}. Since 21/3≠24/32^{1/3} \neq 2^{4/3}, the numerator is non-zero.

Question 7

A curve is defined by f(x)+g(y)=Cf(x) + g(y) = C for a constant CC, where ff and gg are twice-differentiable functions. At a point (x,y)(x, y) on the curve where g′(y)≠0g'(y) \neq 0, which expression represents d2ydx2\frac{d^2y}{dx^2}?

  1. f′′(x)g′′(y)\frac{f''(x)}{g''(y)}
  2. −f′′(x)g′(y)-\frac{f''(x)}{g'(y)}
  3. −f′′(x)g′(y)−f′(x)g′′(y)(g′(y))2-\frac{f''(x)g'(y) - f'(x)g''(y)}{(g'(y))^2}
  4. −f′′(x)(g′(y))2+g′′(y)(f′(x))2(g′(y))3-\frac{f''(x)(g'(y))^2 + g''(y)(f'(x))^2}{(g'(y))^3} (correct answer)
Explanation: Differentiating f(x)+g(y)=Cf(x) + g(y) = C with respect to xx gives f′(x)+g′(y)dydx=0f'(x) + g'(y)\frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx} gives dydx=−f′(x)g′(y)\frac{dy}{dx} = -\frac{f'(x)}{g'(y)}. To find the second derivative, we differentiate this expression using the quotient rule and the chain rule: d2ydx2=−f′′(x)g′(y)−f′(x)ddx(g′(y))(g′(y))2\frac{d^2y}{dx^2} = -\frac{f''(x)g'(y) - f'(x) \frac{d}{dx}(g'(y))}{(g'(y))^2}. By the chain rule, ddx(g′(y))=g′′(y)dydx\frac{d}{dx}(g'(y)) = g''(y)\frac{dy}{dx}. Substituting this in gives d2ydx2=−f′′(x)g′(y)−f′(x)g′′(y)dydx(g′(y))2\frac{d^2y}{dx^2} = -\frac{f''(x)g'(y) - f'(x)g''(y)\frac{dy}{dx}}{(g'(y))^2}. Now, substitute dydx=−f′(x)g′(y)\frac{dy}{dx} = -\frac{f'(x)}{g'(y)}: d2ydx2=−f′′(x)g′(y)−f′(x)g′′(y)(−f′(x)g′(y))(g′(y))2=−f′′(x)g′(y)+(f′(x))2g′′(y)g′(y)(g′(y))2\frac{d^2y}{dx^2} = -\frac{f''(x)g'(y) - f'(x)g''(y)(-\frac{f'(x)}{g'(y)})}{(g'(y))^2} = -\frac{f''(x)g'(y) + \frac{(f'(x))^2g''(y)}{g'(y)}}{(g'(y))^2}. Multiplying the numerator and denominator by g′(y)g'(y) to clear the complex fraction yields d2ydx2=−f′′(x)(g′(y))2+(f′(x))2g′′(y)(g′(y))3\frac{d^2y}{dx^2} = -\frac{f''(x)(g'(y))^2 + (f'(x))^2g''(y)}{(g'(y))^3}.

Question 8

For the implicit curve y2+ln⁡(xy)=2y^2 + \ln(xy) = 2, what is the value of the product (dy/dx)⋅(dx/dy)(dy/dx) \cdot (dx/dy) at the point (e,1)(e, 1)?

  1. -1
  2. 0
  3. 1 (correct answer)
  4. −1/(3e2)-1/(3e^2)
Explanation: For any differentiable function where the derivatives are defined and non-zero, the rate of change of x with respect to y, dx/dydx/dy, is the reciprocal of the rate of change of y with respect to x, dy/dxdy/dx. Therefore, their product is always 1. We can verify this by calculating dy/dxdy/dx at (e,1)(e, 1). Differentiating with respect to x: 2y(dy/dx)+(1/xy)(y+x(dy/dx))=02y(dy/dx) + (1/xy)(y + x(dy/dx)) = 0. At (e,1)(e, 1): 2(1)(dy/dx)+(1/e)(1+e(dy/dx))=0  ⟹  2(dy/dx)+1/e+dy/dx=0  ⟹  3(dy/dx)=−1/e2(1)(dy/dx) + (1/e)(1 + e(dy/dx)) = 0 \implies 2(dy/dx) + 1/e + dy/dx = 0 \implies 3(dy/dx) = -1/e, so dy/dx=−1/(3e)dy/dx = -1/(3e). Consequently, dx/dy=−3edx/dy = -3e. The product is (−1/(3e))(−3e)=1(-1/(3e))(-3e) = 1.

Question 9

At which point (x,y)(x,y) in the first quadrant on the ellipse x2+2y2=6x^2+2y^2=6 is the normal line parallel to the line y=2xy=2x?

  1. (3,3/2)(\sqrt{3}, \sqrt{3/2})
  2. (2,1)(2, 1)
  3. (0,3)(0, \sqrt{3})
  4. (2,2)(\sqrt{2}, \sqrt{2}) (correct answer)
Explanation: This problem combines implicit differentiation with the geometric relationship between tangent and normal lines. When you see "normal line parallel to a given line," remember that normal lines are perpendicular to tangent lines, so you need to find where the tangent has the right slope. Start by finding the slope of the tangent line using implicit differentiation on x2+2y2=6x^2 + 2y^2 = 6: 2x+4ydydx=02x + 4y\frac{dy}{dx} = 0 dydx=−x2y\frac{dy}{dx} = -\frac{x}{2y} The normal line has slope that's the negative reciprocal of the tangent slope: 2yx\frac{2y}{x} Since the normal line must be parallel to y=2xy = 2x, we need: 2yx=2\frac{2y}{x} = 2 2y=2x2y = 2x y=xy = x Now substitute y=xy = x into the ellipse equation: x2+2x2=6x^2 + 2x^2 = 6 3x2=63x^2 = 6 x2=2x^2 = 2 x=2x = \sqrt{2} (first quadrant) Therefore y=2y = \sqrt{2}, giving us point (2,2)(\sqrt{2}, \sqrt{2}). Choice A (3,3/2)(\sqrt{3}, \sqrt{3/2}) doesn't satisfy y=xy = x since 3≠3/2\sqrt{3} \neq \sqrt{3/2}. Choice B (2,1)(2, 1) fails because 22+2(1)2=62^2 + 2(1)^2 = 6 gives us 6=66 = 6, but y≠xy \neq x. Choice C (0,3)(0, \sqrt{3}) isn't in the first quadrant since x=0x = 0. The key insight: when a normal line is parallel to a given line, set their slopes equal and solve the resulting system with the original curve equation.

Question 10

For the curve defined by y3−3xy=2y^3 - 3xy = 2, what is the slope of the normal line at the point where the tangent line is vertical?

  1. -1
  2. 0 (correct answer)
  3. 1
  4. Undefined
Explanation: First, we find dy/dxdy/dx by implicit differentiation: 3y2(dy/dx)−(3y+3x(dy/dx))=03y^2(dy/dx) - (3y + 3x(dy/dx)) = 0. Solving for dy/dxdy/dx gives dy/dx(3y2−3x)=3ydy/dx(3y^2 - 3x) = 3y, so dy/dx=y/(y2−x)dy/dx = y/(y^2-x). A vertical tangent occurs when the denominator is zero and the numerator is non-zero, so y2−x=0  ⟹  x=y2y^2-x=0 \implies x=y^2. Substituting this into the original equation gives y3−3(y2)y=2y^3 - 3(y^2)y = 2, which simplifies to −2y3=2-2y^3 = 2, so y3=−1y^3 = -1 and y=−1y=-1. The corresponding x-coordinate is x=(−1)2=1x=(-1)^2=1. At the point (1,−1)(1,-1), the tangent line is vertical. The normal line to a vertical tangent line is a horizontal line. The slope of a horizontal line is 0.

Question 11

The curve defined by x2−xy+y2=3x^2 - xy + y^2 = 3 has a horizontal tangent at the point (−1,−2)(-1, -2). Using the second derivative test, how does the curve behave at this point?

  1. The curve has a local minimum. (correct answer)
  2. The curve has a local maximum.
  3. The curve has a point of inflection.
  4. The test is inconclusive.
Explanation: First, we find dy/dxdy/dx: 2x−(y+x(dy/dx))+2y(dy/dx)=02x - (y + x(dy/dx)) + 2y(dy/dx) = 0, which gives dy/dx=(y−2x)/(2y−x)dy/dx = (y-2x)/(2y-x). At (−1,−2)(-1,-2), the slope is (−2−2(−1))/(2(−2)−(−1))=0(-2-2(-1))/(2(-2)-(-1)) = 0, confirming the horizontal tangent. To classify this critical point, we find d2y/dx2d^2y/dx^2 using the quotient rule: d2y/dx2=[(dy/dx−2)(2y−x)−(y−2x)(2(dy/dx)−1)]/(2y−x)2d^2y/dx^2 = [(dy/dx - 2)(2y-x) - (y-2x)(2(dy/dx)-1)] / (2y-x)^2. We evaluate this at (−1,−2)(-1,-2) where dy/dx=0dy/dx = 0: d2y/dx2=[(0−2)(2(−2)−(−1))−(−2−2(−1))(2(0)−1)]/(2(−2)−(−1))2=[(−2)(−3)−(0)(−1)]/(−3)2=6/9=2/3d^2y/dx^2 = [(0 - 2)(2(-2)-(-1)) - (-2-2(-1))(2(0)-1)] / (2(-2)-(-1))^2 = [(-2)(-3) - (0)(-1)]/(-3)^2 = 6/9 = 2/3. Since d2y/dx2>0d^2y/dx^2 > 0, the curve is concave up at this point. A horizontal tangent combined with concave up behavior indicates a local minimum.

Question 12

For the curve defined by sin⁡(x2y)=x\sin(x^2y) = x, what is the slope of the normal line at the point (1,π/2)(1, \pi/2)?

  1. 1
  2. -1
  3. Undefined
  4. 0 (correct answer)
Explanation: When you encounter implicit differentiation problems involving normal lines, remember that the normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope. First, let's verify the point (1,π/2)(1, \pi/2) lies on the curve sin⁡(x2y)=x\sin(x^2y) = x. Substituting: sin⁡(12⋅π/2)=sin⁡(π/2)=1\sin(1^2 \cdot \pi/2) = \sin(\pi/2) = 1, which equals our xx-value of 1. ✓ Now we'll find dydx\frac{dy}{dx} using implicit differentiation. Differentiating both sides of sin⁡(x2y)=x\sin(x^2y) = x: cos⁡(x2y)⋅ddx(x2y)=1\cos(x^2y) \cdot \frac{d}{dx}(x^2y) = 1 Using the product rule on x2yx^2y: cos⁡(x2y)⋅(2xy+x2dydx)=1\cos(x^2y) \cdot (2xy + x^2\frac{dy}{dx}) = 1 Solving for dydx\frac{dy}{dx}: cos⁡(x2y)⋅2xy+cos⁡(x2y)⋅x2dydx=1\cos(x^2y) \cdot 2xy + \cos(x^2y) \cdot x^2\frac{dy}{dx} = 1 dydx=1−2xycos⁡(x2y)x2cos⁡(x2y)\frac{dy}{dx} = \frac{1 - 2xy\cos(x^2y)}{x^2\cos(x^2y)} At point (1,π/2)(1, \pi/2):
  • x2y=1⋅π/2=π/2x^2y = 1 \cdot \pi/2 = \pi/2
  • cos⁡(π/2)=0\cos(\pi/2) = 0
This makes the denominator zero, so dydx\frac{dy}{dx} is undefined, meaning the tangent line is vertical. When the tangent line is vertical, the normal line is horizontal with slope 0. Answer choice (A) 1 and (B) -1 assume you can find a finite slope for the tangent line. Choice (C) "Undefined" confuses the tangent line's slope (which is undefined) with the normal line's slope. Study tip: When the tangent line has undefined slope (vertical), the normal line always has slope 0 (horizontal). These are the only two perpendicular possibilities when one involves infinity.

Question 13

Consider the ellipse defined by the equation x2+xy+y2=7x^2 + xy + y^2 = 7. What is the slope of the tangent line to the ellipse at any point where it intersects the line y=xy=x?

  1. −1-1 (correct answer)
  2. 11
  3. 00
  4. The slope depends on the specific intersection point.
Explanation: First, find the derivative dydx\frac{dy}{dx} implicitly: 2x+(1⋅y+xdydx)+2ydydx=02x + (1 \cdot y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0. Rearranging yields dydx(x+2y)=−2x−y\frac{dy}{dx}(x+2y) = -2x-y, so dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x+y}{x+2y}. The points of intersection with the line y=xy=x are not needed explicitly. At any point where y=xy=x, we can substitute y=xy=x into the expression for the derivative: dydx=−2x+xx+2x=−3x3x\frac{dy}{dx} = -\frac{2x+x}{x+2x} = -\frac{3x}{3x}. Assuming x≠0x \neq 0 (if x=0x=0, then y=0y=0, and 0=70=7 which is impossible), the slope is −1-1. Thus, the slope is constant at all intersection points.

Question 14

Consider the curve defined by the equation x2+2xy−y3=7x^2 + 2xy - y^3 = 7. What is the slope of the line normal to the curve at the point (2,1)(2, 1)?

  1. −6-6
  2. 66
  3. 1/61/6 (correct answer)
  4. −1/6-1/6
Explanation: First, find the derivative dydx\frac{dy}{dx} using implicit differentiation: 2x+(2y+2xdydx)−3y2dydx=02x + (2y + 2x \frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0. Grouping terms with dydx\frac{dy}{dx} gives dydx(2x−3y2)=−2x−2y\frac{dy}{dx}(2x - 3y^2) = -2x - 2y, so dydx=−2x−2y2x−3y2\frac{dy}{dx} = \frac{-2x - 2y}{2x - 3y^2}. Evaluate the derivative at the point (2,1)(2, 1) to find the slope of the tangent line: mtan=−2(2)−2(1)2(2)−3(1)2=−4−24−3=−6m_{tan} = \frac{-2(2) - 2(1)}{2(2) - 3(1)^2} = \frac{-4 - 2}{4 - 3} = -6. The slope of the normal line is the negative reciprocal of the tangent slope: mnorm=−1mtan=−1−6=16m_{norm} = -\frac{1}{m_{tan}} = -\frac{1}{-6} = \frac{1}{6}.

Question 15

For the curve defined by x3+y3=9x^3 + y^3 = 9, what is the value of the second derivative, d2ydx2\frac{d^2y}{dx^2}, at the point (1,2)(1, 2)?

  1. −1/4-1/4
  2. −1/2-1/2
  3. 9/169/16
  4. −9/16-9/16 (correct answer)
Explanation: First, differentiate implicitly: 3x2+3y2dydx=03x^2 + 3y^2\frac{dy}{dx} = 0, which gives dydx=−x2y2\frac{dy}{dx} = -\frac{x^2}{y^2}. At (1,2)(1, 2), dydx=−1222=−14\frac{dy}{dx} = -\frac{1^2}{2^2} = -\frac{1}{4}. Now, differentiate dydx\frac{dy}{dx} with respect to xx using the quotient rule to find d2ydx2\frac{d^2y}{dx^2}: d2ydx2=−2x(y2)−x2(2ydydx)(y2)2=−2xy2−2x2ydydxy4\frac{d^2y}{dx^2} = -\frac{2x(y^2) - x^2(2y \frac{dy}{dx})}{(y^2)^2} = -\frac{2xy^2 - 2x^2y \frac{dy}{dx}}{y^4}. Substitute the point (1,2)(1, 2) and the value dydx=−1/4\frac{dy}{dx} = -1/4: d2ydx2=−2(1)(22)−2(12)(2)(−14)24=−8−4(−14)16=−8+116=−916\frac{d^2y}{dx^2} = -\frac{2(1)(2^2) - 2(1^2)(2)(-\frac{1}{4})}{2^4} = -\frac{8 - 4(-\frac{1}{4})}{16} = -\frac{8+1}{16} = -\frac{9}{16}.

Question 16

The lemniscate curve is defined by 2(x2+y2)2=25(x2−y2)2(x^2 + y^2)^2 = 25(x^2 - y^2). At a certain point in the first quadrant, the tangent line to this curve is horizontal. What is the yy-coordinate of this point?

  1. 5/45/4 (correct answer)
  2. 53/45\sqrt{3}/4
  3. 25/825/8
  4. 25/425/4
Explanation: Differentiate implicitly: 4(x2+y2)(2x+2ydydx)=25(2x−2ydydx)4(x^2+y^2)(2x+2y\frac{dy}{dx}) = 25(2x-2y\frac{dy}{dx}). For a horizontal tangent, dydx=0\frac{dy}{dx}=0. Substituting this gives 4(x2+y2)(2x)=25(2x)4(x^2+y^2)(2x) = 25(2x). Assuming x≠0x\neq 0, we can divide by 8x8x to get x2+y2=25/4x^2+y^2 = 25/4. Now we have a system of two equations: the original curve and this new relation. Substitute x2+y2=25/4x^2+y^2=25/4 into the original equation: 2(25/4)2=25(x2−y2)2(25/4)^2 = 25(x^2 - y^2). This simplifies to 2(625/16)=25(x2−y2)2(625/16) = 25(x^2-y^2), or 625/8=25(x2−y2)625/8 = 25(x^2-y^2), which means x2−y2=25/8x^2 - y^2 = 25/8. Now solve the system: (1) x2+y2=25/4x^2+y^2 = 25/4 and (2) x2−y2=25/8x^2-y^2=25/8. Subtracting (2) from (1) gives 2y2=25/4−25/8=25/82y^2 = 25/4 - 25/8 = 25/8. So y2=25/16y^2 = 25/16. In the first quadrant, y=5/4y = 5/4.

Question 17

For the curve defined by the implicit equation y2−2xy=16y^2 - 2xy = 16, which of the following statements about its concavity is true for all points on the curve?

  1. The curve is always concave up.
  2. The curve is always concave down.
  3. The curve is concave up where y>xy>x and concave down where y<xy<x. (correct answer)
  4. The curve is concave down where y>xy>x and concave up where y<xy<x.
Explanation: First, find dydx\frac{dy}{dx}: 2ydydx−(2y+2xdydx)=02y\frac{dy}{dx} - (2y + 2x\frac{dy}{dx}) = 0, which gives dydx(2y−2x)=2y\frac{dy}{dx}(2y-2x) = 2y, so dydx=yy−x\frac{dy}{dx} = \frac{y}{y-x}. Next, find d2ydx2\frac{d^2y}{dx^2} using the quotient rule: d2ydx2=dydx(y−x)−y(dydx−1)(y−x)2=y−xdydx(y−x)2\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}(y-x) - y(\frac{dy}{dx}-1)}{(y-x)^2} = \frac{y - x\frac{dy}{dx}}{(y-x)^2}. Substitute dydx=yy−x\frac{dy}{dx} = \frac{y}{y-x}: d2ydx2=y−x(yy−x)(y−x)2=y(y−x)−xy(y−x)3=y2−2xy(y−x)3\frac{d^2y}{dx^2} = \frac{y - x(\frac{y}{y-x})}{(y-x)^2} = \frac{y(y-x)-xy}{(y-x)^3} = \frac{y^2-2xy}{(y-x)^3}. From the original equation, y2−2xy=16y^2 - 2xy = 16. So, d2ydx2=16(y−x)3\frac{d^2y}{dx^2} = \frac{16}{(y-x)^3}. The sign of the concavity depends on the sign of (y−x)3(y-x)^3. If y>xy>x, then y−x>0y-x>0, so d2ydx2>0\frac{d^2y}{dx^2} > 0 (concave up). If y<xy<x, then y−x<0y-x<0, so d2ydx2<0\frac{d^2y}{dx^2} < 0 (concave down).

Question 18

A point (x,y)(x, y) moves along the circle x2+y2=r2x^2 + y^2 = r^2, where rr is a positive constant. Which of the following expressions is equivalent to d2ydx2\frac{d^2y}{dx^2}?

  1. −x/y-x/y
  2. x2/y3x^2/y^3
  3. −r2/y3-r^2/y^3 (correct answer)
  4. r2/y3r^2/y^3
Explanation: First, find the first derivative by implicit differentiation: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, which implies dydx=−xy\frac{dy}{dx} = -\frac{x}{y}. To find the second derivative, apply the quotient rule to this expression: d2ydx2=−(1)(y)−x(dydx)y2=−y−xdydxy2\frac{d^2y}{dx^2} = -\frac{(1)(y) - x(\frac{dy}{dx})}{y^2} = -\frac{y - x\frac{dy}{dx}}{y^2}. Now substitute the expression for dydx\frac{dy}{dx}: d2ydx2=−y−x(−xy)y2=−y+x2yy2\frac{d^2y}{dx^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}. To simplify, multiply the numerator and denominator by yy: d2ydx2=−y2+x2y3\frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3}. From the original equation of the circle, we know x2+y2=r2x^2 + y^2 = r^2. Substituting this into our expression gives d2ydx2=−r2y3\frac{d^2y}{dx^2} = -\frac{r^2}{y^3}.

Question 19

For the curve defined by y2=x3+1y^2 = x^3 + 1, there are points (x0,y0)(x_0, y_0) where the tangent line passes through the origin (0,0)(0, 0). What are the possible values for y0y_0?

  1. ±2\pm\sqrt{2}
  2. ±3\pm\sqrt{3} (correct answer)
  3. ±2\pm 2
  4. No such points exist.
Explanation: Let (x0,y0)(x_0, y_0) be a point on the curve. The slope of the tangent line at this point is found by implicit differentiation: 2ydydx=3x22y\frac{dy}{dx} = 3x^2, so m=dydx=3x022y0m = \frac{dy}{dx} = \frac{3x_0^2}{2y_0}. The equation of the tangent line is y−y0=m(x−x0)y - y_0 = m(x - x_0). If this line passes through the origin (0,0)(0,0), we can substitute these coordinates into the equation: 0−y0=m(0−x0)0 - y_0 = m(0 - x_0), which simplifies to y0=mx0y_0 = mx_0. Now we have two expressions for the slope: m=y0x0m = \frac{y_0}{x_0} and m=3x022y0m = \frac{3x_0^2}{2y_0}. Setting them equal gives y0x0=3x022y0\frac{y_0}{x_0} = \frac{3x_0^2}{2y_0}, which leads to 2y02=3x032y_0^2 = 3x_0^3. The point (x0,y0)(x_0, y_0) must also satisfy the curve's equation: y02=x03+1y_0^2 = x_0^3 + 1. We now have a system of two equations. Substitute y02y_0^2 from the second equation into the first: 2(x03+1)=3x032(x_0^3 + 1) = 3x_0^3. This simplifies to 2x03+2=3x032x_0^3 + 2 = 3x_0^3, so x03=2x_0^3 = 2. Now find y02y_0^2 using y02=x03+1y_0^2 = x_0^3 + 1: y02=2+1=3y_0^2 = 2 + 1 = 3. Therefore, the possible values for y0y_0 are ±3\pm\sqrt{3}.

Question 20

For the circle x2+y2=25x^2 + y^2 = 25, at what rate is the slope of the curve changing with respect to xx at the point (3,−4)(3, -4)?

  1. 3/43/4
  2. −3/4-3/4
  3. 25/6425/64 (correct answer)
  4. −25/64-25/64
Explanation: The "slope of the curve" is dydx\frac{dy}{dx}. The "rate at which the slope is changing with respect to xx" is the derivative of the slope with respect to xx, which is ddx(dydx)=d2ydx2\frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2}. First, we find dydx\frac{dy}{dx} for the circle: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−xy\frac{dy}{dx} = -\frac{x}{y}. Next, we find the second derivative: d2ydx2=−(1)(y)−x(dydx)y2=−y−x(−xy)y2=−y+x2/yy2=−y2+x2y3\frac{d^2y}{dx^2} = -\frac{(1)(y) - x(\frac{dy}{dx})}{y^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + x^2/y}{y^2} = -\frac{y^2+x^2}{y^3}. From the original equation, x2+y2=25x^2+y^2=25. So, d2ydx2=−25y3\frac{d^2y}{dx^2} = -\frac{25}{y^3}. We need to evaluate this at the point (3,−4)(3, -4). The yy-coordinate is −4-4. d2ydx2=−25(−4)3=−25−64=2564\frac{d^2y}{dx^2} = -\frac{25}{(-4)^3} = -\frac{25}{-64} = \frac{25}{64}.