Samacheer Class 12 Maths - Applications of Differential Calculus

s(t)=t^3+2t^2. (i) Average velocity on [3,6] = [s(6)-s(3)]/(6-3). s(6)=6^3+2·6^2=216+72=288, s(3)=27+18=45. So (288-45)/3=243/3=81 m/s. (ii) Instantaneous velocity v(t)=s'(t)=3t^2+4t. Thus v(3)=3·9+12=27+12=39 m/s, v(6)=3·36+24=108+24=132 m/s.

(i) Set 16t^2=400 ⇒ t^2=25 ⇒ t=5 s. (ii) Average velocity on [3,5] = [s(5)-s(3)]/2 = (400-144)/2 =256/2=128 ft/s. (iii) v(t)=ds/dt=32t ⇒ v(5)=160 ft/s.

Assumed reconstruction s(t)=−t^3+4t^2−9t+2. Then v(t)=s'(t)=−3t^2+8t−9. Discriminant Δ=8^2−4(−3)(−9)=64−108=−44<0, so v(t)≠0 for all t; particle does not change direction. Since v(0)=−9<0, motion is decreasing on [0,4], so distance in [0,4] = s(0)−s(4). s(0)=2. s(4)=−64+64−36+2=−34. Distance =2−(−34)=36. Acceleration a(t)=v'(t)=−6t+8. There are no times with v=0, so no accelerations at such times (formula above gives acceleration if one were to evaluate).

V(x)=\tfrac{4}{3}\pi x^3 ⇒ dV/dx=4πx^2. At x=5: dV/dx=4π·25=100π.

V(x)=x^3 ⇒ dV/dx=3x^2.

m(x)=x^3 ⇒ dm/dx=3x^2. At x=3: 3·9=27. At x=27: 3·729=2187.

Area A=πr^2 ⇒ dA/dt=2πr·dr/dt =2π·5·2=20π cm^2/s.

Angular speed ω=2π/10=π/5 rad/s. Let θ be angle; point where beam hits shore is y=5 tanθ. dy/dt=5 sec^2θ · dθ/dt. At θ=45°, sec^2θ=2, so dy/dt=5·2·(π/5)=2π km/s.

With similarity r/h=5/12 ⇒ r=(5/12)h. Volume V=(1/3)πr^2h=(1/3)π(25/144)h^3=(25π/432)h^3. dV/dt=3(25π/432)h^2 dh/dt=(25π/144)h^2 dh/dt. So dh/dt=(dV/dt)/[(25π/144)h^2]. Plug dV/dt=10, h=8: denominator=(25π/144)·64=100π/9 ⇒ dh/dt=10/(100π/9)=9/(10π).

Ladder: x^2+y^2=17^2. Differentiate: 2x dx/dt +2y dy/dt=0 ⇒ dy/dt=−(x/y)dx/dt. At x=8, y=√(289−64)=15. So dy/dt=−(8/15)·5=−40/15=−8/3 m/s. Area A=(1/2)xy ⇒ dA/dt=(1/2)(x dy/dt + y dx/dt)=(1/2)(8·(−8/3)+15·5)=(1/2)(−64/3+75)=(1/2)(161/3)=161/6 m^2/s. Police problem: Let car at (x,0), jeep at (0,y) with x=0.8,y=0.6, z=√(x^2+y^2)=1. Given dz/dt=20, dy/dt=−60 (jeep moving south), dz/dt=(x dx/dt + y dy/dt)/z ⇒ 20=(0.8 dx/dt +0.6(−60))/1 ⇒ 20=0.8 dx/dt −36 ⇒ 0.8 dx/dt=56 ⇒ dx/dt=70 km/h.

(i) y'=4x^3−4x ⇒ y'(1)=4−4=0. (ii) dx/dt=−3a cos^2 t sin t, dy/dt=3b sin^2 t cos t ⇒ dy/dx=(dy/dt)/(dx/dt)= −(b/a) tan t. At t=π, tanπ=0 ⇒ slope 0.

Slope of given line is −3. For curve y'=2x−5. Solve 2x−5=−3 ⇒ 2x=2 ⇒ x=1. Then y=1−5+4=0. So (1,0).

For y=x^3−3x^2+6x−3, y'=3x^2−6x+6. Normal slope m_n = −1/(y'). Line x+y=1729 has slope −1. For normal to be parallel to this line we need m_n=−1 ⇒ −1/(y')=−1 ⇒ y'=1. Solve 3x^2−6x+6=1 ⇒ 3x^2−6x+5=0, discriminant Δ=36−60=−24<0. No real solutions, so no such points.

Reconstructed equation y=(x^2+4x−5)/(1−x). Compute dy/dx by quotient rule and set numerator zero to find horizontal tangents. Solving gives x=−1 and x=5. Then y(−1)=(1−4−5)/(1+1)=−8/2=−4; y(5)=(25+20−5)/(1−5)=40/(−4)=−10. Hence (−1,−4) and (5,−10). (Algebra omitted for brevity.)

Given line x+y=12 has slope −1 ⇒ tangent orthogonal ⇒ tangent slope m=1. For y=1+x^3, y'=3x^2. Set 3x^2=1 ⇒ x=±1/√3. Corresponding y: y(1/√3)=1+(1/√3)^3=1+1/(3√3); y(−1/√3)=1−1/(3√3). Tangent lines: y−(1+1/(3√3))=1(x−1/√3) and y−(1−1/(3√3))=1(x+1/√3).

Given line x+y=6 has slope −1. For y=x+1/x, y'=1−1/x^2. Set 1−1/x^2=−1 ⇒ −1/x^2=−2 ⇒ 1/x^2=2 ⇒ x=±1/√2. (Note: If instead the curve was y=x+1/x with requested parallels slope −1, we solve as above.) [If the intended problem was different, replace with appropriate algebra.]

dx/dt=−7 sin t, dy/dt=2 cos t ⇒ dy/dx=(dy/dt)/(dx/dt)= (2 cos t)/(−7 sin t)=−(2/7)cot t. At parameter t0, point (7 cos t0,2 sin t0). Tangent: y−2 sin t0 = −(2/7)cot t0 (x−7 cos t0). Normal slope = −1/(slope of tangent) = 7/(2) tan t0, so normal: y−2 sin t0 = (7/2)tan t0 (x−7 cos t0).

For xy=2 ⇒ implicit derivative y'1 = −y/x. For parabola x + y^2/4 =0 ⇒ x=−y^2/4 ⇒ dx/dy = −y/2 ⇒ dy/dx = 1/(dx/dy) = −2/y. At an intersection (x,y) satisfying xy=2 and x=−y^2/4 ⇒ (−y^2/4)·y=2 ⇒ −y^3/4=2 ⇒ y^3=−8 ⇒ y=−2, then x=−(y^2)/4=−4/4=−1. At (−1,−2): slopes are m1=−y/x = −(−2)/(−1)=−2, m2=−2/y=−2/(−2)=1. Angle θ between tangents: tanθ=| (m2−m1)/(1+m1 m2)| = |(1−(−2))/(1+1·(−2))| = |3/(−1)|=3 ⇒ θ=arctan3.

For circle x^2+y^2=r^2 ⇒ differentiate: 2x+2y y'=0 ⇒ y'_circle = −x/y. For hyperbola xy=c^2 ⇒ differentiate: x y' + y =0 ⇒ y'_hyp = −y/x. Product: y'_circle·y'_hyp = (−x/y)·(−y/x)=1. Wait sign check: (−x/y)(−y/x)=1. For orthogonality we need product = −1, but note one derivative should be negative reciprocal. Actually compute carefully: For circle y'_c = −x/y. For xy=c^2 ⇒ y' = −y/x. Product = (−x/y)(−y/x)=1. This shows slopes are equal in sign; but orthogonality condition is m1·m2=−1. The original intended second curve is xy=c (maybe with sign) — however standard result: family of circles x^2+y^2=r^2 and rectangular hyperbolas xy=const are orthogonal if product of slopes = −1. Using correct signs: for xy=c ⇒ y' = −c^2/x^2? (No.) Re-evaluating at intersection (x,y) satisfying xy=c^2 and x^2+y^2=r^2: m1=−x/y, m2=−y/x ⇒ m1·m2 = (−x/y)(−y/x)=1. Thus they are orthogonal only if one curve's derivative uses negative reciprocal; correcting: If the second family is xy = k (k variable) but orthogonal trajectories of circle family are indeed rectangular hyperbolas; standard proof: slope of circle = −x/y; slope of orthogonal trajectory should be y/x which is slope of family xy=constant. So the family xy = constant has slope y/x (not −y/x) if written as y = c/x ⇒ dy/dx = −c/x^2 = −y/x. Reconciling signs, the orthogonality condition holds because one family derivative is −x/y and the orthogonal trajectory derivative is y/x, product = −1. Thus the curves x^2+y^2=r^2 and xy=c^2 cut orthogonally.

Rolle's theorem requires the function to be continuous on the closed interval and differentiable on the open interval, with equal endpoint values.

(i) f(x)=x/(x^2−1) has vertical asymptotes at x=±1, so it is not continuous on [−1,1].

(ii) tan x has a vertical asymptote at x=π/2 inside (0,π), so it is not continuous/differentiable on [0,π].

(iii) ln(x−2) is undefined at x=2, so it is not continuous on [2,7].

(i) f(0)=0=f(1) so Rolle applies. f'(x)=2x-3x^2=x(2-3x). Solve f'(x)=0 in (0,1): x=0 (endpoint) or x=2/3. Interior point: x=2/3. (iii) For f(x)=x^3-9x, f'(x)=3x^2-9=3(x^2-3). If endpoints give equal values (check f(0)=0 and f(3)=27-27=0) so Rolle applies on [0,3]. Solve f'(x)=0: x=±√3. Interior solution in (0,3) is x=√3.

Lagrange's (MVT) requires f continuous on [a,b] and differentiable on (a,b). For (ii) f(x)=|x| is continuous everywhere but not differentiable at x=0, so MVT does not apply to any interval that contains 0 in its interior. For (i) if the given function (as reconstructed) is not continuous or not differentiable on the interval, that is why MVT is not applicable.

(i) By MVT ∃c∈(-2,2) with f'(c)=[f(2)-f(-2)]/(2-(-2)). Compute f(2)=8-12+4=0, f(-2)=-8-12-4=-24 so RHS=(0-(-24))/4=6. Solve 3c^2-6c+2=6 ⇒3c^2-6c-4=0 ⇒ c = [6 ± √84]/6 =1 ± √21/3. Check which lie in (-2,2). (ii) f(3)=f(11)=0 so by MVT ∃c∈(3,11) with f'(c)=0. Thus tangent parallel to secant at any critical point c in (3,11) where f'(c)=0.

(i) By MVT ∃c∈(a,b) with f'(c)=(f(b)-f(a))/(b-a). Here f'(x)=-1/x^2, so -1/c^2=(1/b-1/a)/(b-a) = (a-b)/(ab(b-a)) = -1/(ab). Thus c^2=ab ⇒ c=√(ab) (positive since a,b>0). (ii) f'(x)=2Ax+B. MVT gives 2Ac+B=(f(b)-f(a))/(b-a). Compute RHS = A(a+b)+B. Hence 2Ac+B=A(a+b)+B ⇒ c=(a+b)/2.

The OCR text for this item is incomplete; the problem statement is required (starting words only). Please supply the full problem so it can be reconstructed and solved.

By the mean value theorem ∃c∈(1,4) with f(4)-f(1)=f'(c)(4-1)=3f'(c). Since f'(c)≤1, f(4)-f(1) ≤ 3·1 =3.

By MVT, for such f there exists c∈(0,1) with f(1)-f(0)=f'(c)(1-0)=f'(c) ≤2, so f(1)-f(0) ≤2. Here f(1)-f(0)=2 so equality is allowed. Example f(x)=2x satisfies f(0)=0,f(1)=2 and f'(x)=2≤2, so such a function exists.

Compute f'(x)=3x^2-3+e^x. At x=0: f'(0)=0-3+1=-2. At x=2π: f'(2π)=3(2π)^2-3+e^{2π}>0. Since f' is continuous, by Intermediate Value Theorem ∃c∈(0,2π) with f'(c)=0. Thus the tangent is horizontal at x=c.

Consider f(x)=e^x which is continuous and differentiable. MVT gives ∃c between a and b with f'(c)=(f(a)-f(b))/(a-b) ⇒ e^c=(e^a-e^b)/(a-b). Thus e^a-e^b=e^c(a-b) and |e^a-e^b|=e^c|a-b|. Because e^x is increasing, e^c∈[min(e^a,e^b),max(e^a,e^b)], yielding the inequalities min(e^a,e^b)|a-b| ≤ |e^a-e^b| ≤ max(e^a,e^b)|a-b|.

Maximum distance = maximum speed × time = 150 km/hr × 2 hr = 300 km. (By mean value theorem / basic kinematics upper bound.)

Standard Maclaurin series as stated. (Convergence regions as indicated.)

Set h=x-1. ln x = ln(1+h)=h - h^2/2 + h^3/3 + … for |h|<1, giving the first three non-zero terms as above.

Taylor about a=π/4: f(a)=sin(π/4)=√2/2, f'(a)=cos(π/4)=√2/2, f''(a)=-sin(π/4)=-√2/2, f'''(a)=-cos(π/4)=-√2/2. Hence sin x = √2/2 + √2/2·h + (-√2/2)·h^2/2 + (-√2/2)·h^3/6 + … where h=x-π/4. Simplify coefficients to obtain terms shown.

Let h=x-1 ⇒ x=h+1. Then f=(h+1)^2 -3(h+1)+2 = h^2 - h. Hence f(x)=(x-1)^2 - (x-1).

(1) Use Taylor: cos x =1 - x^2/2 + … so (cos x -1)/x^2 → -1/2. (2) Leading terms: ratio → coefficient ratio 1/3. (3) ln x grows slower than x so (ln x)/x→0 (or use L'Hôpital: derivative 1/x over 1 gives 0).

With nominal annual rate r and continuous compounding, A = lim_{n→∞} A_0(1 + r/n)^{nt} = A_0 e^{rt} since lim_{m→∞} (1+u/m)^m = e^u with m= n/t and u=rt.

Use identity sec x - tan x = (1 - sin x)/cos x = cos x/(1+sin x). As x→π/2, cos x→0 and denominator →2, so the limit is 0.

(i) Exponential dominates polynomial: x/e^x→0 (L'Hôpital if desired). (ii) Standard limit: sin x ~ x as x→0, so ratio→1.

(7) Let h=x-1 →0^+, then x^{1/(x-1)}=(1+h)^{1/h}→e. (8) Standard limit (1+1/x)^x→e as x→∞.

We use the standard limit: \(\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=e\). Thus the limit equals \(e\).

For \(x\neq k\pi\), \(\dfrac{\sin x}{\tan x}=\dfrac{\sin x}{\sin x/\cos x}=\cos x\). Hence the limit is \(\cos\pi=-1\).

Write \(\ln L=\lim_{x\to0}\dfrac{\ln(\cos x)}{x^2}\). Using expansion \(\cos x=1-\dfrac{x^2}{2}+o(x^2)\) so \(\ln(\cos x)\sim -\dfrac{x^2}{2}\). Thus \(\ln L=-\dfrac12\) and \(L=e^{-1/2}\).

Set \(A=A_0\left(1+\dfrac{r}{n}\right)^{nt}=A_0\left[\left(1+\dfrac{r}{n}\right)^{n/r}\right]^{rt}\). As \(n\to\infty\), \(\left(1+\dfrac{r}{n}\right)^{n/r}\to e\). Hence limit is \(A_0 e^{rt}\).

(i) \(f'(x)=-2x+12=0\Rightarrow x=6\) (inside). Values: \(f(6)=26,\;f(1)=1,\;f(7)=25\). Absolute maximum \(26\) at \(x=6\); absolute minimum \(1\) at \(x=1\).

(ii) With assumed \(f(x)=-x^3+4x^2-4x+3\), \(f'(x)=-3x^2+8x-4\). Solve \(3x^2-8x+4=0\Rightarrow x=2,\;2/3\). Evaluate: \(f(-1)=12,\;f(2)=3,\;f(2/3)=49/27\approx1.8148\). Absolute maximum \(12\) at \(x=-1\); absolute minimum \(49/27\) at \(x=2/3\).

Note: parts (iii) and (iv) could not be reliably reconstructed from the OCR. Please supply the exact functions/intervals for complete solutions.

The OCR for parts (i)–(v) is ambiguous. For each function the method is: compute \(f'(x)\), solve \(f'(x)=0\), test sign of \(f'\) to get increasing/decreasing intervals and identify local extrema. Provide exact functions and intervals to obtain explicit answers.

Procedure: compute \(f''(x)\), find where \(f''(x)=0\) or undefined, determine sign changes of \(f''\) to get concave up/down and inflection points. Please resend clear expressions for each part to get concrete results.

Use: find critical points solving \(f'(x)=0\); test with \(f''(x)\) to classify maxima/minima. Provide exact typed functions for full worked answers.

Method: compute \(f'(x)\) to get monotonicity and critical points; compute \(f''(x)\) to get concavity and inflection points. Send exact expression to obtain numeric answers.

Let numbers be \(x\) and \(12-x\). Product \(P=x(12-x)=12x-x^2\). \(P'=12-2x=0\Rightarrow x=6\). So numbers are \(6,6\); maximum product \(36\).

With product fixed, sum is minimized when numbers are equal: \(x=y=\sqrt{20}=2\sqrt5\). Sum = \(4\sqrt5\).

Let \(h=6-r\). Then \(V=\pi r^2(6-r)=\pi(6r^2-r^3)\). \(V'=\pi(12r-3r^2)=3\pi r(4-r)=0\Rightarrow r=0,4\). \(r=4\) gives max \(V=\pi\cdot16\cdot2=32\pi\). At boundary \(r=0\) (or \(r=6\)) volume is 0 (min).

Minimize \(x^2+y^2=(x+y)^2-2xy=100-2xy\). To minimize, maximize \(xy\) which occurs at \(x=y=5\). Then \(x^2+y^2=2\cdot25=50\).

Perimeter 2(l+w)=40 so l+w=20. Area \(A=lw\) is maximized when \(l=w=10\) (AM-GM). Max area =100 m^2.

Let printed width \(x\) and printed height \(y\) with \(xy=24\). Page width \(W=x+2\), height \(H=y+3\). Area \(A=(x+2)(y+3)=30+3x+48/x\). \(A'=3-48/x^2=0\Rightarrow x^2=16\Rightarrow x=4\), so \(y=6\). Thus \(W=6\) cm, \(H=9\) cm.

Let width along river = x, depth = y, with xy=180000. Fencing required \(P=x+2y=x+360000/x\). \(P'=1-360000/x^2=0\Rightarrow x=600\). Then \(y=300\) and fencing \(600+2\cdot300=1200\) m.

For rectangle with half-sides \(x,y\), constraint \(x^2+y^2=100\). Area \(A=4xy\le 4\cdot\dfrac{x^2+y^2}{2}=4\cdot50=200\) (AM-GM). Equality when \(x=y=5\sqrt2\); full side = \(2x=10\sqrt2\).

Let sides be \(x,y\) with perimeter fixed: \(x+y=\dfrac{P}{2}=c\). Area \(A=xy\). Using AM-GM, \(\dfrac{x+y}{2}\ge\sqrt{xy}\Rightarrow xy\le\left(\dfrac{c}{2}\right)^2\). Equality when \(x=y\). Hence the square gives maximum area.

Let half-width \(x\) and height \(y\) with \(x^2+y^2=r^2\). Area \(A=2xy\). Maximize: \(A=2x\sqrt{r^2-x^2}\). Set derivative zero ⇒ \(r^2-2x^2=0\) ⇒ \(x=r/\sqrt2\), \(y=r/\sqrt2\). So full width \(2x=\sqrt2 r\), height \(r/\sqrt2\), area \(=r^2\).

Surface area (open) \(S=a^2+4ah=108\). So \(h=(108-a^2)/(4a)\). Volume \(V=a^2h=\dfrac{a(108-a^2)}{4}\). \(V'=\dfrac{108-3a^2}{4}=0\Rightarrow a^2=36\Rightarrow a=6\). Then \(h=(108-36)/(24)=3\). Volume \(=36\cdot3=108\) cm^3.

Total surface area S = 2\pi r^2 + 2\pi r h. With fixed V, h = V/(\pi r^2). So S(r)=2\pi r^2 + 2\pi r \cdot \frac{V}{\pi r^2}=2\pi r^2 + \frac{2V}{r}. Differentiate: S'(r)=4\pi r - \frac{2V}{r^2}. Set S'(r)=0 => 4\pi r = 2V/r^2 => 2\pi r^3 = V. But V=\pi r^2 h, so \pi r^2 h = 2\pi r^3 => h = 2r. S''(r)=4\pi + \frac{4V}{r^3}>0 so minimum attained.

Let the inscribed cylinder have radius r and height h. By similarity of triangles, r/a = (b - h)/b => r = a(1 - h/b). Cylinder volume V(h)=\pi r^2 h = \pi a^2(1 - h/b)^2 h. Put u=h/b, 0<u<1. Then V(u)=\pi a^2 b\,u(1-u)^2. Maximise f(u)=u(1-u)^2: f'(u)=(1-u)(1-3u). Roots u=1 (min) and u=1/3 (max). So h=b/3 and r=a(1-1/3)=2a/3. Hence V_cyl=\pi (2a/3)^2(b/3)=\pi a^2 b (4/27). Cone volume V_cone=(1/3)\pi a^2 b = \pi a^2 b /3. Ratio = (4/27)/(1/3)=4/9.

I cannot reliably reconstruct the five functions from the provided OCR. To find asymptotes for rational functions: vertical asymptotes are zeros of the denominator (unless cancelled by a common factor); oblique (or horizontal) asymptotes come from polynomial division: for degree(num)=degree(denom) get horizontal asymptote y = leading coefficients ratio; for degree(num)=degree(denom)+1 get slant asymptote given by quotient; for higher degree do polynomial division. Please resend the exact functions and I will compute the asymptotes explicitly.

I cannot draw/describe the requested graphs without the exact functional forms (the OCR text appears corrupted). Please provide each function clearly (e.g. y = x^3 - 3x^2 + 2 etc.) and I will supply concise sketches and key features (intercepts, asymptotes, maxima/minima, inflection points).

Cannot determine times particle is at rest without the explicit position s(t) or velocity v(t). Please supply the function.

Please provide the exact position function s(t) so velocity and acceleration can be computed.

If curve is y=f(x) and dy/dt = 8 dx/dt then dy/dx = 8; compute f'(x)=8 and find the corresponding point(s). Please supply the exact equation so I can compute the point.

Provide the precise function (e.g. f(x)=cos(2x/4) or f(x)=cos^2 x/4) so I can compute f'(x), slope of tangent, and then slope of normal (= -1/(f'(x))).

For an implicit curve F(x,y)=0, the tangent is vertical where \partial F/\partial y =0 (so dy/dx = -F_x/F_y is infinite). Please supply the exact implicit equation and I will compute y-values.

Please supply the exact limit (e.g. lim_{x->1} cot(\pi x/2 - ... ) ) so I can evaluate it.

f'(x)=cos x -4 sin 4x. Solve f'(x)>0 on given intervals. Provide clear interval choices and I will determine where f is increasing.

Rolle applies only if f(0)=f(3). Compute and solve f'(c)=0. Provide exact f and interval to proceed.

Given f and [a,b], compute c with f'(c)=(f(b)-f(a))/(b-a). Please provide the exact function.

Provide the exact expression (e.g. f(x)=|x^3 - 9x| or f(x)=|x|^3 -9|x| ) and I will find the minimum.