We need to write a complete article of at least 900 words about: "2.Now, 2 tangent lines and the derivative homework answer key". So presumably the article should provide answer key for homework problems in section 2.On top of that, 2 tangent lines and the derivative. Worth adding: we need to produce educational content, with intro, subheadings, bold, italics, lists, etc. Even so, must be at least 900 words. Here's the thing — no meta opening sentences like "Okay, here's a comprehensive article". Must start directly with main content. Should not mention that it's an article. Should be SEO-friendly, with main keyword likely "tangent lines and the derivative homework answer key". Use semantic keywords. Provide answer key with explanations. Provide steps, scientific explanation, FAQ, conclusion. Use markdown formatting Most people skip this — try not to..
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We need to produce at least 900 words, maybe around 1000-1100.
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Let's craft article:
Title: "2.2 Tangent Lines and the Derivative Homework Answer Key".
2.2 Tangent Lines and the Derivative – Homework Answer Key
Overview
Section 2.2 of most introductory calculus texts introduces tangent lines as the geometric counterpart of the derivative. The homework that follows typically asks students to:
- Identify the point of tangency on a given curve.
- Compute the derivative (f'(x)) using limit definitions or differentiation rules.
- Write the equation of the tangent line in point‑slope form.
- Interpret the result in the context of rates of change or motion.
The answer key below supplies complete solutions for the most common problem set, complete with step‑by‑step reasoning, common pitfalls, and a short FAQ for quick reference Worth keeping that in mind..
1. Problem 1 – Limit‑Definition Tangent Line
Problem.
Find the equation of the tangent line to the curve
[ y = \frac{1}{x} ]
at the point where (x = 2).
Answer Key.
- Compute the derivative using the limit definition
[ f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} ]
Here (f(x)=x^{-1}) and (a=2).
[ \begin{aligned} f'(2) &= \lim_{h\to0}\frac{(2+h)^{-1}-2^{-1}}{h} \ &= \lim_{h\to0}\frac{\frac{1}{2+h}-\frac{1}{2}}{h} \ &= \lim_{h\to0}\frac{2-(2+h)}{h(2+h)2} \ &= \lim_{h\to0}\frac{-h}{2h(2+h)} \ &= \lim_{h\to0}\frac{-1}{2(2+h)} = -\frac{1}{4}. \end{aligned} ]
Thus (f'(2) = -\tfrac14).
- Find the point ((x_0,y_0)).
[ y_0 = f(2) = \frac{1}{2}=0.5. ]
- Write the tangent line using point‑slope form (y-y_0=m(x-x_0)):
[ y-0.5 = -\frac14,(x-2) \quad\Longrightarrow\quad \boxed{,y = -\frac14x + 1,}. ]
2. Problem 2 – Power Rule Application
Problem.
Given (f(x)=3x^{4}-5x^{2}+7), determine the tangent line at (x=1).
Answer Key.
- Differentiate with the power rule ( \frac{d}{dx}[x^{n}] = nx^{,n-1}):
[ f'(x)=12x^{3}-10x. ]
- Evaluate at (x=1)
[ f'(1)=12(1)^{3}-10(1)=12-10=2. ]
- Find the point on the curve
[ f(1)=3(1)^{4}-5(1)^{2}+7=3-5+7=5. ]
- Tangent line
[ y-5 = 2(x-1) ;\Longrightarrow; \boxed{,y = 2x+3,}. ]
3. Problem 3 – Implicit Differentiation
Problem.
For the curve defined implicitly by
[ x^{2}+xy+y^{2}=7, ]
find the slope of the tangent line at the point ((1,2)).
Answer Key.
- Differentiate both sides with respect to (x) (treat (y) as a function of (x)):
[ 2x + (x,y' + y) + 2y,y' = 0. ]
- Collect terms containing (y'):
[ (x+2y),y' + (2x + y) = 0 ;\Longrightarrow; y' = -\frac{2x + y}{x+2y}. ]
- Plug in ((x,y)=(1,2)):
[ y' = -\frac{2(1)+2}{1+2(2)} = -\frac{4}{5}. ]
Thus the slope at ((1,2)) is (-\tfrac45).
- Equation of the tangent line (optional):
[ y-2 = -\frac45,(x-1) ;\Longrightarrow; \boxed{,5y = -4x + 14,}. ]
4. Problem 4 – Related Rates (Derivative in Motion)
Problem.
A particle moves along the graph (y = \sqrt{x}). At the instant when (x=9), the particle’s horizontal speed is (\dfrac{dx}{dt}=3) units/s. Find the vertical speed (\dfrac{dy}{dt}) at that instant.
Answer Key.
- Differentiate the relation (y = x^{1/2}) implicitly with respect to (t):
[ \frac{dy}{dt}= \frac12 x^{-1/2}\frac{dx}{dt}. ]
- Insert the known values ((x=9), (\frac{dx}{dt}=3)):
[ \frac{dy}{dt}= \frac12 (9)^{-1/2}\cdot 3 = \frac12\cdot\frac{1}{3}\cdot 3 = \frac12. ]
Therefore (\boxed{\dfrac{dy}{dt}=0.5\text{ units/s}}) Practical, not theoretical..
5. Problem 5 – Tangent Line to a Trigonometric Function
Problem.
Find the tangent line to (f(x)=\sin x) at (x=\frac{\pi}{6}).
Answer Key.
- Derivative of (\sin x)
[ f'(x)=\cos x. ]
- Slope at (x=\frac{\pi}{6})
[ m = \cos!\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}. ]
- Point on the curve
[ \bigl(x_0,y_0\bigr)=\Bigl(\frac{\pi}{6},\sin!\frac{\pi}{6}\Bigr)=\Bigl(\frac{\pi}{6},\frac12\Bigr). ]
- Tangent line
[ y-\frac12 = \frac{\sqrt{3}}{2}!\left(x-\frac{\pi}{6}\right) ]
or, after simplifying,
[ \boxed{,y = \frac{\sqrt{3}}{2}x + \Bigl(\frac12-\frac{\sqrt{3},\pi}{12}\Bigr),}. ]
6. Problem 6 – Using the Difference Quotient Directly
Problem.
For (f(x)=\sqrt{x+4}), compute the derivative at (x=5) by evaluating the limit
[ \lim_{h\to0}\frac{f(5+h)-f(5)}{h}. ]
Answer Key.
- Set up the expression
[ \frac{\sqrt{5+h+4}-\sqrt{9}}{h} = \frac{\sqrt{h+9}-3}{h}. ]
- Rationalize the numerator
[ \frac{\sqrt{h+9}-3}{h}\cdot\frac{\sqrt{h+9}+3}{\sqrt{h+9}+3} = \frac{(h+9)-9}{h\bigl(\sqrt{h+9}+3\bigr)} = \frac{h}{h\bigl(\sqrt{h+9}+3\bigr)} = \frac{1}{\sqrt{h+9}+3}. ]
- Take the limit as (h\to0):
[ \lim_{h\to0}\frac{1}{\sqrt{h+9}+3}= \frac{1}{\sqrt{9}+3}= \frac{1}{3+3}= \frac16. ]
Hence (f'(5)=\boxed{\dfrac16}) Simple as that..
7. Problem 7 – Piecewise Function
Problem.
Let
[ f(x)= \begin{cases} x^{2}+1, & x\le 1,\[4pt] 2x+3, & x>1. \end{cases} ]
Determine whether a tangent line exists at (x=1).
Answer Key.
- Check continuity at (x=1):
[ \lim_{x\to1^-}f(x)=1^{2}+1=2,\qquad \lim_{x\to1^+}f(x)=2(1)+3=5. ]
Since the left‑ and right‑hand limits differ, (f) is discontinuous at (x=1); therefore a tangent line does not exist there And that's really what it comes down to..
Key point: a tangent line can only be defined at points where the function is continuous and the derivative from both sides agrees The details matter here..
8. Problem 8 – Higher‑Order Tangent Approximation
Problem.
Find the linear approximation (first‑order Taylor polynomial) of (g(x)=e^{2x}) near (x=0) and use it to estimate (e^{0.1}).
Answer Key.
- Derivative of (g):
[ g'(x)=2e^{2x}. ]
- Evaluate at (x=0):
[ g(0)=e^{0}=1,\qquad g'(0)=2e^{0}=2. ]
- Linear approximation
[ L(x)=g(0)+g'(0)x = 1+2x. ]
- Estimate (e^{0.1}=g(0.05)) (since (g(x)=e^{2x}) and (2\cdot0.05=0.1)):
[ L(0.05)=1+2(0.05)=1.10. ]
Thus the approximation is (\boxed{e^{0.1}\approx1.10}). (The true value is (1.105170\ldots), confirming the linear model’s accuracy.
Frequently Asked Questions
| Question | Brief Answer |
|---|---|
| *Why does the limit definition give the same result as the power rule?, denominator (\neq0)). Which means | |
| *Do implicit curves always have a tangent line? And * | The power rule is derived from the limit definition. |
| *Can a tangent line exist at a cusp?Points where the denominator vanishes correspond to vertical tangents or singularities. | |
| *How do I know which differentiation rule to apply? | |
| *What is the geometric meaning of a negative derivative?e.Which means when the function is a simple power, the limit simplifies to the rule (nx^{n-1}). Think about it: * | Only where the implicit derivative (dy/dx) is defined (i. At a cusp the left‑hand and right‑hand slopes are different (or undefined), so the derivative—and thus a unique tangent line—does not exist. * |
Key Takeaways
- Tangent lines are the geometric manifestation of the derivative; the slope of the line equals (f'(a)) at the point ((a,f(a))).
- The limit definition (\displaystyle f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}) works for any differentiable function, but differentiation rules (power, product, quotient, chain) dramatically speed up computation.
- Implicit differentiation allows us to find slopes for curves not solved for (y). Remember to treat (y) as a function of (x) and multiply by (dy/dx) whenever you differentiate a term containing (y).
- Related‑rates problems are simply applications of the chain rule in a physical context; keep track of units and the variable being differentiated with respect to time.
- A tangent line fails to exist at points of discontinuity, cusps, or where the derivative is infinite (vertical tangents).
Final Thoughts
Mastering the connection between tangent lines and derivatives is a cornerstone of calculus. The answer key presented above not only supplies the final equations but also walks through each logical step, reinforcing the underlying concepts. By practicing these problems repeatedly—first using the limit definition, then transitioning to shortcut rules—students develop intuition for how a curve behaves locally.
When you encounter a new homework problem in Section 2.2, follow this four‑step checklist:
- Identify the point of tangency (plug the given (x)‑value into the original function).
- Differentiate the function using the most efficient rule.
- Evaluate the derivative at the point to obtain the slope.
- Write the tangent line in point‑slope form and simplify.
If the function is piecewise, implicit, or involves a physical rate, adapt steps 2–3 accordingly (use implicit differentiation or the chain rule) Most people skip this — try not to..
With these strategies in hand, the tangent‑line and derivative homework becomes a series of systematic, manageable tasks rather than a mysterious hurdle. Keep the answer key as a reference, but strive to internalize each reasoning step—because true mastery comes from understanding the process, not just copying the final answer. Happy differentiating!
Extending the Toolkit: Advanced Scenarios
When the function you are differentiating is not presented in a simple algebraic form, the same systematic approach still applies—only the execution changes slightly.
1. Piecewise Functions
A piecewise definition forces you to treat each interval separately It's one of those things that adds up..
- Step 1: Determine which piece contains the point of tangency.
- Step 2: Differentiate the applicable piece using the appropriate rule (power, product, etc.).
- Step 3: Verify continuity of the derivative at the boundary. If the left‑hand and right‑hand derivatives differ, the curve has a corner there and a tangent line does not exist in the classical sense.
Example:
[
g(x)=\begin{cases}
x^{2}, & x\le 1\[4pt]
3x-2, & x>1
\end{cases}
]
At (x=1) the function value is (g(1)=1). The left‑hand derivative is (g'(1^-)=2). The right‑hand derivative is (g'(1^+)=3). Because the slopes differ, the graph has a corner at ((1,1)); thus no single tangent line exists there Took long enough..
2. Implicit Curves with Higher‑Order Terms
When the relationship involves both (x) and (y) raised to powers greater than one, differentiate implicitly and solve for (\dfrac{dy}{dx}).
- Differentiate each term, remembering to multiply every occurrence of (y) by (\dfrac{dy}{dx}).
- Collect the (\dfrac{dy}{dx}) terms on one side and isolate them.
Example:
[x^{2}+xy+y^{2}=7
]
Differentiating implicitly:
[
2x + y + x\frac{dy}{dx}+2y\frac{dy}{dx}=0
]
Factor (\frac{dy}{dx}):
[\frac{dy}{dx}(x+2y)=-(2x+y)
]
[
\frac{dy}{dx}= -\frac{2x+y}{x+2y}
]
Evaluating at ((1,2)) yields (\displaystyle \frac{dy}{dx}= -\frac{4}{5}), which is the slope of the tangent line at that point It's one of those things that adds up..
3. Parametric Curves
When a curve is described by a pair of functions (x(t)) and (y(t)), the slope of the tangent line is obtained via the chain rule:
[
\frac{dy}{dx}= \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}
]
provided (\dfrac{dx}{dt}\neq0) That's the part that actually makes a difference..
Example:
[
x(t)=t^{2},\qquad y(t)=t^{3}-3t
] [
\frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=3t^{2}-3
]
At (t=1), (\displaystyle \frac{dy}{dx}= \frac{0}{2}=0). The point on the curve is ((1, -2)); the tangent line is therefore (y+2=0(x-1)), i.e., (y=-2).
4. Numerical Approximation When Closed Form Is Unavailable
For functions defined only by data points or by a computer routine, you can approximate the derivative using finite differences:
[
f'(a)\approx\frac{f(a+h)-f(a)}{h}
]
Choose a small (h) (e.g., (h=10^{-5})) to obtain a slope that can then be plugged into the point‑slope formula. This method is especially handy in applied settings such as physics simulations or economics where the analytical derivative is cumbersome Small thing, real impact..
The Bigger Picture: Why Tangent Lines Matter Beyond the mechanics of finding an equation, tangent lines provide a local linear approximation of a nonlinear function. In many scientific and engineering contexts, this approximation is the foundation for:
- Linearization of differential equations near an equilibrium point.
- Newton’s method for finding roots, where each iteration uses the tangent line to guide the next guess.
- Error estimation in numerical integration and differentiation. Understanding that the derivative is not just an abstract number but the slope of a line that best mimics the curve at a point equips you to translate calculus into tangible problem‑solving tools.
Final Synthesis
The journey from the definition of a derivative to the practical computation of tangent lines illustrates a central theme of calculus: local behavior determines global structure. By mastering the following workflow, you can tackle any problem that asks for a tangent line in Section 2.2:
- Locate the point of tangency.
- Select the appropriate differentiation technique (power rule, product rule, implicit differentiation, parametric differentiation, or numerical approximation). 3. Compute the derivative and evaluate it at the target point to obtain the slope.
- Construct the tangent line using point‑slope or slope‑inter
...intercept form.
4. Construct the tangent line using point-slope or slope-intercept form. Given the point ((x_0, y_0)) and slope (m = f'(x_0)), the equation is:
[
y - y_0 = m(x - x_0)
]
Alternatively, solve for (y) to express it as (y = mx + b), where (b = y_0 - mx_0) And that's really what it comes down to. Which is the point..
Conclusion
Mastering tangent lines transcends mere algebraic manipulation; it unlocks the ability to approximate complex systems with simple linear models. Whether analyzing motion in physics, optimizing functions in economics, or designing algorithms in engineering, the tangent line serves as a bridge between instantaneous rates of change and tangible predictions. By systematically applying differentiation techniques and leveraging the geometric interpretation of the derivative, you transform abstract calculus into a powerful lens for understanding and shaping the world. This foundational skill not only solves textbook problems but also equips you to dissect real-world phenomena where curves and their linear approximations reveal deeper truths about continuity, growth, and equilibrium.
