Implicit Relation Behaviors - Calculus 1

1. A point $(x, y)$ moves along the circle $x^2 + y^2 = r^2$, where $r$ is a positive constant. Which of the following expressions is equivalent to $\frac{d^2y}{dx^2}$?
   Answer: $-r^2/y^3$ First, find the first derivative by implicit differentiation: $2x + 2y\frac{dy}{dx} = 0$, which implies $\frac{dy}{dx} = -\frac{x}{y}$. To find the second derivative, apply the quotient rule to this expression: $\frac{d^2y}{dx^2} = -\frac{(1)(y) - x(\frac{dy}{dx})}{y^2} = -\frac{y - x\frac{dy}{dx}}{y^2}$. Now substitute the expression for $\frac{dy}{dx}$: $\frac{d^2y}{dx^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$. To simplify, multiply the numerator and denominator by $y$: $\frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3}$. From the original equation of the circle, we know $x^2 + y^2 = r^2$. Substituting this into our expression gives $\frac{d^2y}{dx^2} = -\frac{r^2}{y^3}$.
2. The curve defined by $x^2 - xy + y^2 = 3$ has a horizontal tangent at the point $(-1, -2)$. Using the second derivative test, how does the curve behave at this point?
   Answer: The curve has a local minimum. First, we find $dy/dx$: $2x - (y + x(dy/dx)) + 2y(dy/dx) = 0$, which gives $dy/dx = (y-2x)/(2y-x)$. At $(-1,-2)$, the slope is $(-2-2(-1))/(2(-2)-(-1)) = 0$, confirming the horizontal tangent. To classify this critical point, we find $d^2y/dx^2$ using the quotient rule: $d^2y/dx^2 = [(dy/dx - 2)(2y-x) - (y-2x)(2(dy/dx)-1)] / (2y-x)^2$. We evaluate this at $(-1,-2)$ where $dy/dx = 0$: $d^2y/dx^2 = [(0 - 2)(2(-2)-(-1)) - (-2-2(-1))(2(0)-1)] / (2(-2)-(-1))^2 = [(-2)(-3) - (0)(-1)]/(-3)^2 = 6/9 = 2/3$. Since $d^2y/dx^2 > 0$, the curve is concave up at this point. A horizontal tangent combined with concave up behavior indicates a local minimum.
3. A curve is defined by $f(x) + g(y) = C$ for a constant $C$, where $f$ and $g$ are twice-differentiable functions. At a point $(x, y)$ on the curve where $g'(y) \neq 0$, which expression represents $\frac{d^2y}{dx^2}$?
   Answer: $-\frac{f''(x)(g'(y))^2 + g''(y)(f'(x))^2}{(g'(y))^3}$ Differentiating $f(x) + g(y) = C$ with respect to $x$ gives $f'(x) + g'(y)\frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx} = -\frac{f'(x)}{g'(y)}$. To find the second derivative, we differentiate this expression using the quotient rule and the chain rule: $\frac{d^2y}{dx^2} = -\frac{f''(x)g'(y) - f'(x) \frac{d}{dx}(g'(y))}{(g'(y))^2}$. By the chain rule, $\frac{d}{dx}(g'(y)) = g''(y)\frac{dy}{dx}$. Substituting this in gives $\frac{d^2y}{dx^2} = -\frac{f''(x)g'(y) - f'(x)g''(y)\frac{dy}{dx}}{(g'(y))^2}$. Now, substitute $\frac{dy}{dx} = -\frac{f'(x)}{g'(y)}$: $\frac{d^2y}{dx^2} = -\frac{f''(x)g'(y) - f'(x)g''(y)(-\frac{f'(x)}{g'(y)})}{(g'(y))^2} = -\frac{f''(x)g'(y) + \frac{(f'(x))^2g''(y)}{g'(y)}}{(g'(y))^2}$. Multiplying the numerator and denominator by $g'(y)$ to clear the complex fraction yields $\frac{d^2y}{dx^2} = -\frac{f''(x)(g'(y))^2 + (f'(x))^2g''(y)}{(g'(y))^3}$.
4. Find the equation of the tangent line to the ellipse $x^2 + 4y^2 = 8$ at its point in the second quadrant where $y=1$.
   Answer: $y = \frac{1}{2}x + 2$ First, find the $x$-coordinate of the point. Substitute $y=1$ into the ellipse equation: $x^2 + 4(1)^2 = 8$, so $x^2 + 4 = 8$, which means $x^2=4$. The possible $x$-values are $2$ and $-2$. Since the point is in the second quadrant, $x$ must be negative, so the point of tangency is $(-2, 1)$. Next, find the slope by implicit differentiation: $2x + 8y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y}$. At $(-2, 1)$, the slope is $m = -\frac{-2}{4(1)} = \frac{2}{4} = \frac{1}{2}$. Using the point-slope form, the tangent line equation is $y - 1 = \frac{1}{2}(x - (-2))$, which simplifies to $y - 1 = \frac{1}{2}(x+2)$, or $y = \frac{1}{2}x + 1 + 1$, which is $y = \frac{1}{2}x + 2$.
5. For the curve defined by $\sin(x^2y) = x$, what is the slope of the normal line at the point $(1, \pi/2)$?
   Answer: 0 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 $(1, \pi/2)$ lies on the curve $\sin(x^2y) = x$. Substituting: $\sin(1^2 \cdot \pi/2) = \sin(\pi/2) = 1$, which equals our $x$-value of 1. ✓ Now we'll find $\frac{dy}{dx}$ using implicit differentiation. Differentiating both sides of $\sin(x^2y) = x$: $\cos(x^2y) \cdot \frac{d}{dx}(x^2y) = 1$ Using the product rule on $x^2y$: $\cos(x^2y) \cdot (2xy + x^2\frac{dy}{dx}) = 1$ Solving for $\frac{dy}{dx}$: $\cos(x^2y) \cdot 2xy + \cos(x^2y) \cdot x^2\frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1 - 2xy\cos(x^2y)}{x^2\cos(x^2y)}$ At point $(1, \pi/2)$: - $x^2y = 1 \cdot \pi/2 = \pi/2$ - $\cos(\pi/2) = 0$ This makes the denominator zero, so $\frac{dy}{dx}$ 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.
6. The curve defined by $x^2 - xy + y^2 = 3$ has two points where the tangent line is horizontal. What is the positive $y$-coordinate of one of these points?
   Answer: $2$ First, find $\frac{dy}{dx}$ by implicit differentiation: $2x - (1 \cdot y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0$. This simplifies to $\frac{dy}{dx}(2y - x) = y - 2x$, so $\frac{dy}{dx} = \frac{y - 2x}{2y - x}$. A horizontal tangent occurs when the slope is zero, which means the numerator must be zero (and the denominator non-zero). Set $y - 2x = 0$, which implies $y = 2x$. Substitute this into the original equation: $x^2 - x(2x) + (2x)^2 = 3$. This gives $x^2 - 2x^2 + 4x^2 = 3$, or $3x^2 = 3$, so $x = \pm 1$. The corresponding $y$-coordinates are found using $y=2x$. If $x=1$, $y=2$. If $x=-1$, $y=-2$. The positive $y$-coordinate is $2$.
7. For the curve defined by $y^2 = x^3 + 1$, there are points $(x_0, y_0)$ where the tangent line passes through the origin $(0, 0)$. What are the possible values for $y_0$?
   Answer: $\pm\sqrt{3}$ Let $(x_0, y_0)$ be a point on the curve. The slope of the tangent line at this point is found by implicit differentiation: $2y\frac{dy}{dx} = 3x^2$, so $m = \frac{dy}{dx} = \frac{3x_0^2}{2y_0}$. The equation of the tangent line is $y - y_0 = m(x - x_0)$. If this line passes through the origin $(0,0)$, we can substitute these coordinates into the equation: $0 - y_0 = m(0 - x_0)$, which simplifies to $y_0 = mx_0$. Now we have two expressions for the slope: $m = \frac{y_0}{x_0}$ and $m = \frac{3x_0^2}{2y_0}$. Setting them equal gives $\frac{y_0}{x_0} = \frac{3x_0^2}{2y_0}$, which leads to $2y_0^2 = 3x_0^3$. The point $(x_0, y_0)$ must also satisfy the curve's equation: $y_0^2 = x_0^3 + 1$. We now have a system of two equations. Substitute $y_0^2$ from the second equation into the first: $2(x_0^3 + 1) = 3x_0^3$. This simplifies to $2x_0^3 + 2 = 3x_0^3$, so $x_0^3 = 2$. Now find $y_0^2$ using $y_0^2 = x_0^3 + 1$: $y_0^2 = 2 + 1 = 3$. Therefore, the possible values for $y_0$ are $\pm\sqrt{3}$.
8. Consider the curve defined by the equation $x^2 + 2xy - y^3 = 7$. What is the slope of the line normal to the curve at the point $(2, 1)$?
   Answer: $1/6$ First, find the derivative $\frac{dy}{dx}$ using implicit differentiation: $2x + (2y + 2x \frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0$. Grouping terms with $\frac{dy}{dx}$ gives $\frac{dy}{dx}(2x - 3y^2) = -2x - 2y$, so $\frac{dy}{dx} = \frac{-2x - 2y}{2x - 3y^2}$. Evaluate the derivative at the point $(2, 1)$ to find the slope of the tangent line: $m_{tan} = \frac{-2(2) - 2(1)}{2(2) - 3(1)^2} = \frac{-4 - 2}{4 - 3} = -6$. The slope of the normal line is the negative reciprocal of the tangent slope: $m_{norm} = -\frac{1}{m_{tan}} = -\frac{1}{-6} = \frac{1}{6}$.
9. For the curve defined by $y^3 - 3xy = 2$, what is the slope of the normal line at the point where the tangent line is vertical?
   Answer: 0 First, we find $dy/dx$ by implicit differentiation: $3y^2(dy/dx) - (3y + 3x(dy/dx)) = 0$. Solving for $dy/dx$ gives $dy/dx(3y^2 - 3x) = 3y$, so $dy/dx = y/(y^2-x)$. A vertical tangent occurs when the denominator is zero and the numerator is non-zero, so $y^2-x=0 \implies x=y^2$. Substituting this into the original equation gives $y^3 - 3(y^2)y = 2$, which simplifies to $-2y^3 = 2$, so $y^3 = -1$ and $y=-1$. The corresponding x-coordinate is $x=(-1)^2=1$. At the point $(1,-1)$, 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.
10. For the curve defined by the equation $ax^2 + by^2 = c$, where a, b, and c are positive constants, the second derivative $d^2y/dx^2$ is proportional to which of the following?
    Answer: $y^{-3}$ 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 $ax^2 + by^2 = c$, use implicit differentiation to find the first derivative. Differentiating both sides: $2ax + 2by\frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{ax}{by}$. To find the second derivative, differentiate $\frac{dy}{dx} = -\frac{ax}{by}$ using the quotient rule: $\frac{d^2y}{dx^2} = -\frac{a}{b} \cdot \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}$. Substituting $\frac{dy}{dx} = -\frac{ax}{by}$: $\frac{d^2y}{dx^2} = -\frac{a}{b} \cdot \frac{y + \frac{ax^2}{by}}{y^2} = -\frac{a}{by^3}(by + ax^2)$. Since $ax^2 + by^2 = c$, we have $by + ax^2 = c - by^2 + by = c - by(y-1)$. However, more simply, the expression shows that $\frac{d^2y}{dx^2}$ is proportional to $y^{-3}$, making D correct. Choice A ($y^{-1}$) would result from a linear relationship. Choice B ($y^{-2}$) might come from incorrectly applying the quotient rule or stopping at the denominator term. Choice C ($xy^{-2}$) incorrectly suggests the second derivative depends on both $x$ and $y$ in this specific way. Remember: for conic sections, the second derivative typically involves $y$ raised to the $-3$ power due to the quotient rule applied to rational expressions containing $y$ in both numerator and denominator.
11. Consider the folium of Descartes given by $x^3 + y^3 = 6xy$. Find the slope of the tangent line at the point in the first quadrant where $x=y$.
    Answer: -1 First, find the point on the curve where $x=y$. Substitute $y=x$ into the equation: $x^3 + x^3 = 6x(x)$, which simplifies to $2x^3 = 6x^2$. This gives $2x^3 - 6x^2 = 0$, or $2x^2(x-3) = 0$. The solutions are $x=0$ and $x=3$. The point in the first quadrant is $(3,3)$ (since $(0,0)$ is the origin). Now, differentiate the equation implicitly: $3x^2 + 3y^2(dy/dx) = 6y + 6x(dy/dx)$. We could solve for $dy/dx$ and then plug in the point, or plug in the point directly: $3(3)^2 + 3(3)^2(dy/dx) = 6(3) + 6(3)(dy/dx)$. This simplifies to $27 + 27(dy/dx) = 18 + 18(dy/dx)$. Rearranging the terms, we get $9(dy/dx) = -9$, so $dy/dx = -1$.
12. The lemniscate curve is defined by $2(x^2 + y^2)^2 = 25(x^2 - y^2)$. At a certain point in the first quadrant, the tangent line to this curve is horizontal. What is the $y$-coordinate of this point?
    Answer: $5/4$ Differentiate implicitly: $4(x^2+y^2)(2x+2y\frac{dy}{dx}) = 25(2x-2y\frac{dy}{dx})$. For a horizontal tangent, $\frac{dy}{dx}=0$. Substituting this gives $4(x^2+y^2)(2x) = 25(2x)$. Assuming $x\neq 0$, we can divide by $8x$ to get $x^2+y^2 = 25/4$. Now we have a system of two equations: the original curve and this new relation. Substitute $x^2+y^2=25/4$ into the original equation: $2(25/4)^2 = 25(x^2 - y^2)$. This simplifies to $2(625/16) = 25(x^2-y^2)$, or $625/8 = 25(x^2-y^2)$, which means $x^2 - y^2 = 25/8$. Now solve the system: (1) $x^2+y^2 = 25/4$ and (2) $x^2-y^2=25/8$. Subtracting (2) from (1) gives $2y^2 = 25/4 - 25/8 = 25/8$. So $y^2 = 25/16$. In the first quadrant, $y = 5/4$.
13. For the curve defined by $x^3 + y^3 = 9$, what is the value of the second derivative, $\frac{d^2y}{dx^2}$, at the point $(1, 2)$?
    Answer: $-9/16$ First, differentiate implicitly: $3x^2 + 3y^2\frac{dy}{dx} = 0$, which gives $\frac{dy}{dx} = -\frac{x^2}{y^2}$. At $(1, 2)$, $\frac{dy}{dx} = -\frac{1^2}{2^2} = -\frac{1}{4}$. Now, differentiate $\frac{dy}{dx}$ with respect to $x$ using the quotient rule to find $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = -\frac{2x(y^2) - x^2(2y \frac{dy}{dx})}{(y^2)^2} = -\frac{2xy^2 - 2x^2y \frac{dy}{dx}}{y^4}$. Substitute the point $(1, 2)$ and the value $\frac{dy}{dx} = -1/4$: $\frac{d^2y}{dx^2} = -\frac{2(1)(2^2) - 2(1^2)(2)(-\frac{1}{4})}{2^4} = -\frac{8 - 4(-\frac{1}{4})}{16} = -\frac{8+1}{16} = -\frac{9}{16}$.
14. For the curve $x^3 + y^3 = 9$, what is the value of $d^2y/dx^2$ at the point $(2, 1)$?
    Answer: -36 First, find $dy/dx$ by implicit differentiation: $3x^2 + 3y^2(dy/dx) = 0$, which gives $dy/dx = -x^2/y^2$. At $(2,1)$, $dy/dx = -2^2/1^2 = -4$. Now, we find the second derivative, $d^2y/dx^2$, by differentiating $dy/dx$ using the quotient rule: $d^2y/dx^2 = -[(2x)(y^2) - (x^2)(2y \cdot dy/dx)] / (y^2)^2$. Substitute the point $(2,1)$ and $dy/dx = -4$ into this expression: $d^2y/dx^2 = -[(2(2))(1^2) - (2^2)(2(1)(-4))] / (1^2)^2 = -[4 - (4)(-8)] / 1 = -[4+32] = -36$. Alternatively, one can show that $d^2y/dx^2 = -18x/y^5$, which at $(2,1)$ is $-18(2)/1^5 = -36$.
15. For the circle $x^2 + y^2 = 25$, at what rate is the slope of the curve changing with respect to $x$ at the point $(3, -4)$?
    Answer: $25/64$ The "slope of the curve" is $\frac{dy}{dx}$. The "rate at which the slope is changing with respect to $x$" is the derivative of the slope with respect to $x$, which is $\frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2}$. First, we find $\frac{dy}{dx}$ for the circle: $2x + 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$. Next, we find the second derivative: $\frac{d^2y}{dx^2} = -\frac{(1)(y) - x(\frac{dy}{dx})}{y^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + x^2/y}{y^2} = -\frac{y^2+x^2}{y^3}$. From the original equation, $x^2+y^2=25$. So, $\frac{d^2y}{dx^2} = -\frac{25}{y^3}$. We need to evaluate this at the point $(3, -4)$. The $y$-coordinate is $-4$. $\frac{d^2y}{dx^2} = -\frac{25}{(-4)^3} = -\frac{25}{-64} = \frac{25}{64}$.
16. Consider the ellipse defined by the equation $x^2 + xy + y^2 = 7$. What is the slope of the tangent line to the ellipse at any point where it intersects the line $y=x$?
    Answer: $-1$ First, find the derivative $\frac{dy}{dx}$ implicitly: $2x + (1 \cdot y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0$. Rearranging yields $\frac{dy}{dx}(x+2y) = -2x-y$, so $\frac{dy}{dx} = -\frac{2x+y}{x+2y}$. The points of intersection with the line $y=x$ are not needed explicitly. At any point where $y=x$, we can substitute $y=x$ into the expression for the derivative: $\frac{dy}{dx} = -\frac{2x+x}{x+2x} = -\frac{3x}{3x}$. Assuming $x \neq 0$ (if $x=0$, then $y=0$, and $0=7$ which is impossible), the slope is $-1$. Thus, the slope is constant at all intersection points.
17. Consider the curve defined by $x \cos(y) = 1$. What is the equation of the normal line to the curve at the point where it intersects the line $x=2$?
    Answer: $y - \frac{\pi}{3} = -2\sqrt{3}(x-2)$ First, find the point of intersection. If $x=2$, then $2\cos(y)=1$, so $\cos(y) = 1/2$. A solution in $[0, \pi]$ is $y = \pi/3$. The point is $(2, \pi/3)$. Next, find the slope of the tangent line, $dy/dx$, using implicit differentiation: $1\cdot\cos(y) + x(-\sin(y)dy/dx) = 0$, so $dy/dx = \cos(y)/(x\sin(y))$. At $(2, \pi/3)$, the slope is $m_t = \cos(\pi/3) / (2\sin(\pi/3)) = (1/2) / (2\sqrt{3}/2) = 1/(2\sqrt{3})$. The slope of the normal line is the negative reciprocal, $m_n = -2\sqrt{3}$. Using the point-slope form, the equation of the normal line is $y - \pi/3 = -2\sqrt{3}(x-2)$.
18. Consider the curve defined by the equation $x^2 + xy + y^2 = 7$. At what positive y-coordinate is the tangent line to this curve horizontal?
    Answer: $y = 2\sqrt{7/3}$ To find where the tangent line is horizontal, we first find the derivative $dy/dx$ using implicit differentiation: $2x + (1 \cdot y + x \cdot dy/dx) + 2y \cdot dy/dx = 0$. Grouping the $dy/dx$ terms gives $dy/dx(x + 2y) = -2x - y$, so $dy/dx = -(2x+y)/(x+2y)$. A horizontal tangent occurs when $dy/dx = 0$, which means the numerator must be zero: $2x+y=0$, or $y = -2x$. We substitute this relationship back into the original equation: $x^2 + x(-2x) + (-2x)^2 = 7$, which simplifies to $x^2 - 2x^2 + 4x^2 = 7$, or $3x^2 = 7$. This gives $x = \pm\sqrt{7/3}$. The corresponding y-coordinates are found using $y=-2x$. If $x = \sqrt{7/3}$, $y = -2\sqrt{7/3}$. If $x = -\sqrt{7/3}$, $y = 2\sqrt{7/3}$. The positive y-coordinate is $2\sqrt{7/3}$.
19. The curve known as the folium of Descartes is given by $x^3 + y^3 = 3xy$. This curve has a vertical tangent at a point $(x_0, y_0)$ in the first quadrant where $x_0 \neq 0$. What is the value of $x_0$?
    Answer: $2^{2/3}$ First, find $\frac{dy}{dx}$ using implicit differentiation: $3x^2 + 3y^2\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$. Rearranging gives $\frac{dy}{dx}(y^2 - x) = y - x^2$, so $\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$. A vertical tangent occurs where the denominator is zero and the numerator is non-zero. Set $y^2 - x = 0$, so $x = y^2$. Substitute this into the original equation: $(y^2)^3 + y^3 = 3(y^2)y$, which simplifies to $y^6 + y^3 = 3y^3$. This gives $y^6 = 2y^3$, or $y^3(y^3 - 2) = 0$. The solutions are $y=0$ (which gives $x=0$) and $y^3 = 2$, so $y = 2^{1/3}$. The corresponding $x$-coordinate is $x_0 = y^2 = (2^{1/3})^2 = 2^{2/3}$. We must check that the numerator $y-x^2$ is non-zero at this point. $x^2 = (2^{2/3})^2 = 2^{4/3}$. Since $2^{1/3} \neq 2^{4/3}$, the numerator is non-zero.
20. The slope of the tangent to the curve $2x^2 + 3y^2 = 5$ at a point $(x_0, y_0)$ in the first quadrant is $-1$. What is the concavity of the curve at this point?
    Answer: Concave down First, find $dy/dx$: $4x + 6y(dy/dx) = 0$, so $dy/dx = -2x/(3y)$. Given that $dy/dx = -1$, we have $-2x_0/(3y_0) = -1$, which implies $2x_0 = 3y_0$. To find the second derivative, we use $d^2y/dx^2 = -10/(9y^3)$. Since the point is in the first quadrant, $y_0 > 0$, so $d^2y/dx^2 < 0$. Therefore, the curve is concave down at this point.

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