Tangent AND Normal flash cards

1. 1.For a curve $y=f(x)$, write the equations of the tangent and normal at point $(x_1,y_1)$.

   Tangent: $y-y_1=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}(x-x_1)$
   Normal: $y-y_1=-\dfrac{1}{\left(\frac{dy}{dx}\right)_{(x_1,y_1)}}(x-x_1)$

   Hint: Normal slope is negative reciprocal of tangent slope

2. 2.What are the lengths of the tangent, normal, subtangent, and subnormal at a point $(x_1,y_1)$ on $y=f(x)$, in terms of $m=\dfrac{dy}{dx}$?

   Tangent length $=\left|\dfrac{y_1\sqrt{1+m^2}}{m}\right|$; Normal length $=|y_1\sqrt{1+m^2}|$; Subtangent $=\left|\dfrac{y_1}{m}\right|$; Subnormal $=|y_1 m|$

   Hint: Subtangent = projection on x-axis of tangent segment

3. 3.If the tangent to a curve at $(x_1,y_1)$ makes equal intercepts with the axes, what slope condition does this impose, and how is it used to find such points on a curve like $y=f(x)$?

   Equal intercepts of the same sign on both axes gives slope $m=-1$; intercepts equal in magnitude but opposite in sign gives slope $m=1$. Set $\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=-1$ (or $=1$) and solve together with the curve equation to find the point

   Hint: Equal intercepts (same sign) $\Rightarrow m=-1$; opposite sign $\Rightarrow m=1$

4. 4.State the condition for two curves $y=f(x)$ and $y=g(x)$ to intersect orthogonally at a common point.

   If $m_1=f'(x_1)$ and $m_2=g'(x_1)$ are the slopes of the two curves at the point of intersection, orthogonality requires $m_1 m_2=-1$

   Hint: Product of slopes = $-1$

5. 5.For the curve $y=f(x)$, how do you find points where the tangent is parallel to a given line $y=mx+c$, and where it is perpendicular to it?

   Parallel: solve $\dfrac{dy}{dx}=m$ using the curve equation to get the point(s). Perpendicular: solve $\dfrac{dy}{dx}=-\dfrac{1}{m}$ using the curve equation to get the point(s)

   Hint: Match or negative-reciprocal the given slope

6. 6.For the curve $y=\ln x$, find the length of the subtangent and subnormal at the point where $x=e$.

   At $x=e$, $y=1$ and $\dfrac{dy}{dx}=\dfrac1x=\dfrac1e=m$. Subtangent $=\left|\dfrac{y}{m}\right|=e$; subnormal $=|my|=\dfrac1e$. So subtangent $=\boxed{e}$ and subnormal $=\boxed{1/e}$.

   Hint: Evaluate the slope at $x=e$ first, then apply the standard length formulas.

7. 7.Prove that the curve $y=ae^{x/a}$ has a constant subtangent, and find its value.

   Differentiating, $\dfrac{dy}{dx}=e^{x/a}=\dfrac{y}{a}=m$. Subtangent $=\dfrac{y}{m}=\dfrac{y}{y/a}=a$, independent of the point of contact — the curve's defining property is constant subtangent equal to $a$.

   Hint: Express the slope in terms of $y$ itself before forming $y/m$.

8. 8.For the parabola $y^2=4ax$, show that the subnormal is constant at every point, and find its value.

   Differentiating implicitly, $2y\dfrac{dy}{dx}=4a\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}$. Subnormal $=\left|y\cdot\dfrac{dy}{dx}\right|=\left|y\cdot\dfrac{2a}{y}\right|=2a$, the same at every point — a classic invariant of the parabola: subnormal $=2a$.

   Hint: Differentiate implicitly and simplify $y\cdot(dy/dx)$.

9. 9.For the rectangular hyperbola $xy=c^2$, prove that the point of contact bisects the segment of the tangent intercepted between the coordinate axes.

   At $(ct,c/t)$, differentiating $xy=c^2$ gives $\dfrac{dy}{dx}=-\dfrac{y}{x}=-\dfrac1{t^2}$, and the tangent works out to $\dfrac{x}{t}+ty=2c$, meeting the axes at $(2ct,0)$ and $(0,2c/t)$. The midpoint of these intercepts is $\left(ct,\dfrac{c}{t}\right)$ — exactly the point of tangency, proving the point of contact bisects the intercepted segment.

   Hint: Find where the tangent meets each axis, then locate the midpoint.

10. 10.For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, find the length of the subnormal at the point $(a\cos\theta,b\sin\theta)$.

    Implicit differentiation gives $\dfrac{2x}{a^2}+\dfrac{2y}{b^2}\dfrac{dy}{dx}=0\Rightarrow \dfrac{dy}{dx}=-\dfrac{b^2x}{a^2y}$, which at the given point is $-\dfrac{b\cos\theta}{a\sin\theta}$. Subnormal $=\left|y\cdot\dfrac{dy}{dx}\right|=\left|b\sin\theta\cdot\left(-\dfrac{b\cos\theta}{a\sin\theta}\right)\right|=\boxed{\dfrac{b^2\cos\theta}{a}}$.

    Hint: Use implicit differentiation, then apply subnormal $=|y\,dy/dx|$.

Open the interactive deck for the other 75 cards, with self-grading so the ones you keep missing come back.