Interpretation — instantaneous rate of change, slope of tangent: Exercises

4.1.11 · D4 · Maths › Calculus I — Limits & Derivatives › Interpretation — instantaneous rate of change, slope of tang

Shuru karne se pehle, ek picture — woh ek move jo saare L2 problems share karte hain — secant tangent mein slide hoti hai.

Figure — Interpretation — instantaneous rate of change, slope of tangent

L1 — Recognition

Goal: pehchano kaun sa object kaun sa hai. Heavy algebra nahi.

Q1. Ek car ki 2-ghante ki trip mein average speed $60$ km/h thi. Distance vs. time ke graph par, kiski slope $60$ hogi: $t = 1$ h par tangent ki, ya start aur end points ko join karne wali secant ki? Ek line mein explain karo.

Recall Solution Q1

Secant jo start $(0, d_{0})$ ko end $(2, d_{0} + 120)$ se join karti hai. Average speed $= \frac{total distance}{total time} = \frac{Δ d}{Δ t}$ , jo ki bilkul do real points ke beech rise over run hai — yeh secant slope ki definition hai (dekho Average rate of change). $t = 1$ par tangent instantaneous speed hai, jo generally alag number hoti hai.

Q2. $f (x) = x^{2}$ ke liye, $x = a$ par difference quotient likho bina simplify kiye. Phir batao ki fixed $h \neq = 0$ ke liye yeh geometrically kya represent karta hai.

Recall Solution Q2

$\frac{f ( a + h ) - f ( a )}{h} = \frac{( a + h ) ^{2} - a ^{2}}{h}$ Fixed $h \neq = 0$ ke liye yeh $(a, a^{2})$ aur $(a + h, (a + h)^{2})$ se guzarne wali secant line ka slope hai — interval $[a, a + h]$ par change ka average rate. Yeh derivative tabhi banta hai jab limit $h \to 0$ lene ke baad.

Q3. True ya false: " $f^{'} (a)$ tangent line ka slope hai $(a, f (a))$ par aur yeh woh rate bhi hai jis par $f$ us point par per unit $x$ change karta hai." Justify karo.

Recall Solution Q3

True. Dono $\frac{change in output}{change in input}$ hain us instant par. Tangent curve ki best straight-line copy hai $a$ ke paas; uska slope = rise/run = rate. Geometry aur rate ek hi number ke do costumes hain.

L2 — Application

Goal: four-step machine chalao (build → expand → $h$ cancel karo → limit) end to end.

Q4. Limit definition use karke (koi shortcut rules nahi), $f (x) = x^{2} - 4 x + 1$ ke liye $f^{'} (2)$ nikalo.

Recall Solution Q4

Step 1 (build): $\frac{f ( 2 + h ) - f ( 2 )}{h}$ . Yahan $f (2) = 4 - 8 + 1 = - 3$ . Step 2 (expand): $f (2 + h) = (2 + h)^{2} - 4 (2 + h) + 1 = 4 + 4 h + h^{2} - 8 - 4 h + 1 = h^{2} - 3$ . Toh numerator $= (h^{2} - 3) - (- 3) = h^{2}$ . Step 3 ( $h$ cancel karo): $\frac{h ^{2}}{h} = h$ for $h \neq = 0$ . Limit lene se pehle illegal $0/0$ ko hata do. Step 4 (limit): $h \to 0 lim h = 0$ . Toh $f^{'} (2) = 0$ . Matlab: $x = 2$ par tangent horizontal hai — $x = 2$ is parabola ka vertex hai.

Q5. $g (x) = x^{2}$ ki $x = 3$ par tangent line ki equation nikalo, parent note se given hai ki $g^{'} (3) = 6$ .

Recall Solution Q5

Point: $(3, g (3)) = (3, 9)$ . Slope: $g^{'} (3) = 6$ . Point–slope form (ek line ko ek point + ek slope chahiye): $y - 9 = 6 (x - 3) ⟹ y = 6 x - 9 .$

Q6. Ek ball ki height $s (t) = 5 t^{2}$ metres hai. Limit definition use karke, $t = 3$ s par instantaneous velocity nikalo.

Recall Solution Q6

$s (3) = 45$ . $s (3 + h) = 5 (3 + h)^{2} = 5 (9 + 6 h + h^{2}) = 45 + 30 h + 5 h^{2}$ . $\frac{s ( 3 + h ) - s ( 3 )}{h} = \frac{30 h + 5 h ^{2}}{h} = 30 + 5 h (h \neq = 0) .$ $h \to 0 lim (30 + 5 h) = 30$ . Toh $v (3) = 30 m/s$ (dekho Velocity and acceleration).

Q7. Definition se $f (x) = \frac{1}{x}$ ke liye general $a \neq = 0$ par $f^{'} (a)$ nikalo.

Recall Solution Q7

Step 1: $\frac{\frac{1}{a + h} - \frac{1}{a}}{h}$ . Step 2 (common denominator): $\frac{1}{a + h} - \frac{1}{a} = \frac{a - ( a + h )}{a ( a + h )} = \frac{- h}{a ( a + h )}$ . Step 3 ( $h$ se divide karo): $\frac{- h}{a ( a + h )} \cdot \frac{1}{h} = \frac{- 1}{a ( a + h )}$ for $h \neq = 0$ . Step 4 (limit): $h \to 0 lim \frac{- 1}{a ( a + h )} = \frac{- 1}{a ^{2}}$ . Toh $f^{'} (a) = - \frac{1}{a ^{2}}$ — hamesha negative, isliye $1/ x$ hamesha girta hai (for $a \neq = 0$ ).

L3 — Analysis

Goal: existence, sign, aur behaviour ke baare mein reason karo — sirf compute mat karo.

Q8. $f (x) = ∣ x - 1∣$ ke liye, decide karo ki $f^{'} (1)$ exist karta hai ya nahi. Left aur right computations dikhao.

Recall Solution Q8

$a = 1$ par difference quotient: $\frac{∣ ( 1 + h ) - 1∣ - 0}{h} = \frac{∣ h ∣}{h}$ .

Right se ( $h > 0$ ): $∣ h ∣ = h$ , ratio $= + 1$ .
Left se ( $h < 0$ ): $∣ h ∣ = - h$ , ratio $= - 1$ .

Right limit $+ 1 \neq = - 1$ left limit, toh two-sided limit exist nahi karta $\Rightarrow$ $f^{'} (1)$ exist nahi karta. Geometrically $(1, 0)$ par ek corner hai: koi single tangent slope nahi (dekho Differentiability and continuity). Neeche corner figure dekho.

Q9. Parabola $f (x) = x^{2}$ ka secant slope $[a, a + h]$ par $2 a + h$ hai (yeh given maan lo). Is expression ka use karke explain karo, kyun $a$ par tangent slope $2 a$ hona chahiye — aur $h$ term kya role play karta hai.

Recall Solution Q9

Secant slope $= 2 a + h$ do costs se bana hai: ek true part $2 a$ (fixed, interval par depend nahi karta) aur ek error part $h$ (interval ki width). Jab hum doosra point andar slide karte hain, $h \to 0$ , error vanish ho jaata hai aur sirf true part bachta hai: $f^{'} (a) = lim_{h \to 0} (2 a + h) = 2 a .$ Toh tangent slope secant slope ka "interval-independent skeleton" hai bilkul. Yeh Power Rule ka result hai $\frac{d}{d x} x^{2} = 2 x$ , first principles se dekha gaya.

Q10. $f (x) = x^{3}$ ke liye, origin par tangent line $y = 0$ hai. Kya yeh sach hai ki "ek tangent line curve ko bilkul ek point par touch karti hai"? Origin par investigate karo.

Recall Solution Q10

$0$ par slope: difference quotient $\frac{( 0 + h ) ^{3} - 0}{h} = \frac{h ^{3}}{h} = h^{2} \to 0$ , toh $f^{'} (0) = 0$ aur tangent $y = 0$ hai ( $x$ -axis). Lekin $y = 0$ meets $y = x^{3}$ jahan bhi $x^{3} = 0$ — sirf origin par yahan, phir bhi tangent curve ko rest karne ki jagah cross karke guzarti hai. Generally "ek baar touch karti hai" false hai: tangent ko locally slope match karne se define kiya jaata hai (derivative), intersections count karne se nahi. (Ek point par $y = x^{3} - x$ ki tangent curve ko kahin aur bhi hit kar sakti hai.)

L4 — Synthesis

Goal: slope, tangent line, aur rate ideas ko ek problem mein combine karo.

Q11. $f (x) = x^{2} - 4 x + 1$ ke liye, har woh point nikalo jahan tangent line horizontal ho, aur confirm karo ki tera answer Q4 se match karta hai.

Recall Solution Q11

Pehle general slope nikalo. Difference quotient: $\frac{( x + h ) ^{2} - 4 ( x + h ) + 1 - ( x ^{2} - 4 x + 1 )}{h} = \frac{2 x h + h ^{2} - 4 h}{h} = 2 x - 4 + h .$ Limit: $f^{'} (x) = 2 x - 4$ . Horizontal tangent matlab slope $0$ : $2 x - 4 = 0 \Rightarrow x = 2$ . Point: $f (2) = - 3$ , toh horizontal tangent $(2, - 3)$ par hai, tangent line $y = - 3$ . Yeh Q4 se match karta hai ( $f^{'} (2) = 0$ ).

Q12. Upar feka hua ek ball ka height $s (t) = 20 t - 5 t^{2}$ metres hai. (a) Definition se velocity function $v (t) = s^{'} (t)$ nikalo. (b) Kis time par ball momentarily at rest hai (flight ka top)? (c) Maximum height kya hai?

Recall Solution Q12

(a) $s (t + h) = 20 (t + h) - 5 (t + h)^{2} = 20 t + 20 h - 5 t^{2} - 10 t h - 5 h^{2}$ . $\frac{s ( t + h ) - s ( t )}{h} = \frac{20 h - 10 t h - 5 h ^{2}}{h} = 20 - 10 t - 5 h .$ Limit: $v (t) = s^{'} (t) = 20 - 10 t$ m/s. (b) At rest $\Rightarrow v = 0$ : $20 - 10 t = 0 \Rightarrow t = 2$ s. (c) $s (2) = 40 - 20 = 20$ m. Toh $v (t) = 20 - 10 t, t = 2 s, h_{m a x} = 20 m$ . Note karo ki $v > 0$ (rising) for $t < 2$ , $v < 0$ (falling) for $t > 2$ — derivative ka sign motion ki direction report karta hai (dekho Velocity and acceleration).

L5 — Mastery

Goal: ek general statement prove karo aur ek case design karo.

Q13. Limit definition se prove karo ki kisi bhi constants $m, c$ ke liye line $f (x) = m x + c$ ka $f^{'} (a) = m$ hota hai har $a$ ke liye. Interpret karo: lines ke liye "average vs. instantaneous" ke baare mein yeh kya kehta hai?

Recall Solution Q13

$\frac{f ( a + h ) - f ( a )}{h} = \frac{[ m ( a + h ) + c ] - [ ma + c ]}{h} = \frac{mh}{h} = m (h \neq = 0) .$ Quotient $m$ ke equal hai kisi bhi limit se pehle — isme koi $h$ bacha hi nahi. Toh $h \to 0 lim m = m$ , jisse $f^{'} (a) = m$ milta hai sab $a$ ke liye. Interpretation: ek straight line ke liye secant slope pehle se hi $m$ hota hai interval se regardless, toh average rate = instantaneous rate = $m$ har jagah. Yahi wajah hai ki L1 trap (dono ko confuse karna) sirf lines ke liye kabhi kaam karta hai.

Q14. Ek aisa function design karo jo $x = 0$ par continuous ho lekin wahan differentiable nahi ho, $∣ x ∣$ se alag, aur prove karo ki woh derivative test mein fail karta hai. (Hint: ek "vertical tangent" bhi two-sided derivative ko khatam kar deti hai.)

Recall Solution Q14

Lo $f (x) = x^{1/3}$ (cube root). Yeh har jagah continuous hai (koi jump nahi). $0$ par difference quotient: $\frac{f ( 0 + h ) - f ( 0 )}{h} = \frac{h ^{1/3} - 0}{h} = \frac{h ^{1/3}}{h} = h^{- 2/3} = \frac{1}{h ^{2/3}} .$ Jab $h \to 0$ , $h^{2/3} \to 0^{+}$ (yeh square-type power hai, hamesha $\geq 0$ ), toh $\frac{1}{h ^{2/3}} \to + \infty$ . Limit dono sides se infinite hai, toh $f^{'} (0)$ exist nahi karta — tangent vertical hai (infinite slope). Yeh differentiability fail hone ka ek doosra, alag tarika hai (Q8 mein corner vs. yahan vertical tangent). Dono continuity rakhte hain; dono finite-slope limit tod dete hain (dekho Differentiability and continuity).

Recall Ek-nazar answer key

Q4 $f^{'} (2) = 0$ · Q5 $y = 6 x - 9$ · Q6 $v (3) = 30$ m/s · Q7 $f^{'} (a) = - 1/ a^{2}$ · Q8 DNE (corner) · Q9 slope $= 2 a$ · Q10 "touches once" false · Q11 $(2, - 3)$ · Q12 $v (t) = 20 - 10 t$ , $t = 2$ s, $h_{m a x} = 20$ m · Q13 $f^{'} (a) = m$ · Q14 $f^{'} (0)$ DNE (vertical tangent).

Connections

Average rate of change — secant slopes jo har L2 problem mein limit hote hain.
Limits — formal definition — " $h \to 0$ " engine jo poore mein use hota hai.
Derivative as a function — Q11/Q12 mein $f^{'} (x)$ sabhi $x$ ke liye nikala, sirf ek point ke liye nahi.
Power Rule — woh shortcut jo Q9 $x^{2}$ ke liye scratch se prove karta hai.
Differentiability and continuity — Q8, Q14: corner aur vertical-tangent failures.
Velocity and acceleration — Q6, Q12: rate-of-change physics.