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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4

Question 1.
Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector \(4 \hat{i}+3 \hat{j}-7 \hat{k}\) and parallel to the vector \(2 \hat{i}-6 \hat{j}+7 \hat{k}\)
Solution:

Question 2.
Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (-2, 3, 4) and parallel to the straight line \(\frac{x-1}{-4}=\frac{y+3}{5}=\frac{8-z}{6}\)
Solution:

Question 3.
Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
Solution:
Given straight line passing through the points (6, 7, 4) and (8, 4, 9).
Direction ratio of the straight line joining these two points 2, -3, -5.
Cartesian equation:

(ii) The straight Line cuts yz-plane
So we get x = 0
2t + 6 = 0 ⇒ 2t = -6
t = -3
-3t + 7 = -3 (-3) + 7 = 9 + 7 = 16
5t + 4 = 5(-3) + 4 = -15 + 4 = -11
The required point (0, 16, -11).

Question 4.
Find the direction cosines of the straight line passing through the points (5, 6, 7) and (7,9,13). Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.
Solution:
Given straight line passing through the points (5, 6, 7) and (7, 9, 13)
∴ d.r.s : 2, 3, 6

Note: Selection of \(\vec{a}\) and \(\vec{b}\) is your choice.

Question 5.
Find the acute angle between the following lines.

(iii) 2x = 3y = -z and 6x = -y = -4z
Solution:

Question 6.
The vertices of ∆ABC are A(7, 2, 1), B(6, 0, 3), and C(4, 2, 4) . Find ∠ABC.
Solution:

Question 7.
If the straight line joining the points (2, 1, 4) and (a – 1, 4, -1) is parallel to the line joining the points (0, 2, b – 1) and (5, 3, -2), find the values of a and b.
Solution:

Question 8.
If the straight lines are perpendicular to each other, find the value of m.
Solution:

Question 9.
Show that the points (2, 3, 4),(-1, 4, 5) and (8, 1, 2) are collinear.
Solution:
Given points are (2, 3, 4), (-1, 4, 5) and (8, 1, 2) Equation of the line joining of the first and second point is
\(\frac{x-2}{-3}=\frac{y-3}{1}=\frac{z-4}{1}\) = m (say)
(-3m + 2, m + 3, m + 4)
On putting m = -2, we get the third point is (8, 1, 2)
∴ Given points are collinear.

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 Additional Problems

Question 1.
Find the vector and cartesian equations of the straight line passing through the point A with position vector \(3 \vec{i}-\vec{j}+4 \vec{k}\) and parallel to the vector \(-5 \vec{i}+7 \vec{j}+3 \vec{k}\).
Solution:
We know that vector equation of the line through the point with position vector \(\vec{a}\) and parallel to \(\vec{v}\) is given by \(\vec{r}=\vec{a}+t \vec{v}\) where t is a scalar.
Here \(\vec{a}=3 \vec{i}-\vec{j}+4 \vec{k}\) and \(\vec{v}=-5 \vec{i}+7 \vec{j}+3 \vec{k}\)
Vector equation of the line is
\(\vec{r}=(3 \vec{i}-\vec{j}+4 \vec{k})+t(-5 \vec{i}+7 \vec{j}+3 \vec{k})\) ………………. (1)
The cartesian equation of the line passing through (xp yx, zx) and parallel to a vector whose d.r.s are l, m, n is

Question 2.
Find the vector and cartesian equations of the straight line passing through the points (- 5, 2, 3) and (4, – 3, 6).
Solution:
Vector equation of the straight line passing through two points with position vectors \(\vec{a}\) and \(\vec{b}\) is given by

Question 3.
Find the angle between the lines.

Solution:
Let the given lines be in the direction of \(\vec{u}\) and \(\vec{v}\)

Question 4.
Find the angle between the following lines
Solution:
Angle between two lines is the same as angle between their parallel vectors.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.
Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x-axis.
Solution:

Question 2.
Find the area of the region bounded by 2x – y + 1 = 0, y = – 1, y = 3 and y – axis.
Solution:
2x – y + 1 = 0

To find further limit put x = 0, we get y = 1

Question 3.
Find the area of the region bounded by the curve 2 + x – x² + y = 0, x – axis, x = – 3 and x = 3.
Solution:

Question 4.
Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.
Solution:
To find point of intersection of the curves
y = 2x + 5 and y = x² – 2x we get (-1, 3) and (5, 15)

Question 5.
Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π.
Solution:
To find the points of intersection,

Question 6.
Find the area of the region bounded by y = tan x, y = cot x and the line x = 0, x = \(\frac{\pi}{2}\), 0
Solution:
To find the points of intersection of these two curves between 0 to \(\frac{\pi}{2}\) is \(\frac{\pi}{4}\)

Question 7.
Find the area of the region bounded by parabola y² = x and the line y = x – 2
Solution:
To find the points of intersection solve the two equations y² = x and y = x – 2

Question 8.
Father of a family wishes to divide his square field bounded by x = 0, x = 4 , y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
To find the points of intersection of the two curves, y² = 4x and x² = 4y are (0, 0) and (4, 4).
Area of the square field = 4 × 4 = 16 sq. units

So, the remaining each of the two parts must be \(\frac{16}{3}\) sq.units.
∴ Yes, It is possible to divide the square field into three equal parts.

Question 9.
The curves = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Solution:
y = (x – 2)² + 1



∴ x = 2 is a minimum point
∴ The point P is (2, 1)
But slope of PQ is 2
∴ Equation of the chord PQ
y – y₁ = m(x – x₁)
y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
On solving the curve and line we get the point Q(4, 5)

Question 10.
Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x.
Solution:
To find points of intersection of x² + y² = 16 and y² = 6x are (2, \(2 \sqrt{3})\)) and (2, –\(2 \sqrt{3})\))
Due to symmetrical property,

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8 Additional Problems

Question 1.
Find the area of the region enclosed by y² = x and y = x – 2.
Solution:
The points of intersection of the parabola y² = x and the line y = x – 2 are (1, -1) and (4, 2)

To compute the region [shown in figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect toy no splitting is necessary.

Question 2.
Find the area bounded by the curve y = x³ and the line y = x.
Solution:
The line y = x lies above the curve y = x³ in the first quadrant and y = x³ lies above the line y = x in the third quadrant. To get the points of intersection, solve the curves y = x³, y = x ⇒ x³ = x. We get x = {0, ± 1}

Question 3.
Find the area of the loop of the curve 3ay² = x (x – a)².
Solution:
Put y = 0; we get x = 0, a
It meets the x – axis at x = 0 and x = a
∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 4.
Find the area between the line y = x + 1 and the curve y = x² – 1.
Solution:
To get the points of intersection of the curves we should solve the equations y = x +1 and y = x² – 1.
we get, x² – 1 = x + 1
x² – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0
x = – 1 or x = 2
∴ The line intersects the curve at x = – 1 and x = 2.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3

Question 1.

Solution:

Question 2.
For any vector \(\vec{a}\), prove that
Solution:

Question 3.
Prove that \([\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}]\) = 0
Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.
If \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) are coplanar vectors, show that \((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}\)
Solution:

Question 7.
If , find the values of l, m, n
Solution:

On solving (3) & (4)

Question 8.
If \(\hat{a}, \hat{b}, \hat{c}\) are three unit vectors such that \(\hat{b} \text { and } \hat{c}\) are non-parallel and \(\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}\), find the angle between \(\hat{a}\) and \(\hat{c}\).
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3 Additional Problems

Question 1.
If and show that they are not equal.
Solution:

Question 2.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.2

Question 1.

Solution:

Question 2.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors .
Solution:
Volume of the parallelepiped = \(\| \vec{a}, \vec{b}, \vec{c}]\)

= -264 + 224 + 760 = 720 cubic units

Question 3.
The volume of the parallelepiped whose coterminus edges are , \(-3 \vec{i}+7 \vec{j}+5 \vec{k}\) is 90 cubic units. Find the value of λ
Solution:
Given, Volume of the parallelepiped = 90 cubic units

Question 4.
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) be the concurrent edges of parallelepiped
Given volume of parallelepiped = 4 cubic units

Question 5.
Find the altitude of a parallelepiped determined by the vectors \(\vec{a}=-2 \hat{i}+5 \hat{j}+3 \hat{k}\), \(\hat{b}=\hat{i}+3 \hat{j}-2 \hat{k}\) and \(\vec{c}=-3 \vec{i}+\vec{j}+4 \vec{k}\) if the base is taken as the parallelogram determined by b and c.
Solution:
Volume = Base Area × Height
\(|[\vec{a}, \vec{b}, \vec{c}]|=|\vec{b} \times \vec{c}|\) × Height

Question 6.
Determine whether the three vectors \(2 \hat{i}+3 \hat{j}+\hat{k}, \hat{i}-2 \hat{j}+2 \hat{k}\) and \(3 \hat{i}+\hat{j}+3 \hat{k}\) are coplanar.
Solution:

Question 7.
Let If c₁ = 1 and c₂ = 2, find c₃ such that \(\vec{a}, \vec{b}\) and \(\vec{c}\) and c are coplanar.
Solution:

Question 8.
If , show that \([\vec{a} \vec{b} \vec{c}]\) depends neither x nor y.
Solution:

Question 9.
If the vectors

are coplanar, prove that c is the geometric mean of a and b.
Solution:

∴ c is the geometric means of ‘a’ and ‘b’.

Question 10.
Let \(\vec{a}, \vec{b}, \vec{c}\) be three non-zero vectors such that \(\vec{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\) show that .
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.2 Additional Problems

Question 1.
If the edges meet a vertex, find the volume of the parallelepiped.
Solution:
Volume of the parallelepiped =
The volume cannot be negative
∴ Volume of parallelepiped = 264 cu. units

Question 2.
If and \(\vec{x} \neq \overrightarrow{0}\) then show that \(\vec{a}, \vec{b}, \vec{c}\) are coplanar.
Solution:

Question 3.
The volume of a parallelepiped whose edges are represented by \(-12 \vec{i}+\lambda k\), \(3 \vec{j}-\vec{k}, 2 \vec{i}+\vec{j}-15 \vec{k}\) is 546. Find the value of λ.
Solution:
Volume of the parallelepiped =
= -12 [-45 + 1] – 0 () + λ [0 – 6] = -12 (-44) -6 λ
= 528 – 6λ = 546 (given)
⇒ -6λ = 546 – 528 = 18
∴ λ = \(\frac{18}{-6}\) = -3

Question 4.
Prove that \(|\vec{a} \vec{b} \vec{c}|\) = abc if and only if \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular.
Solution:

Question 5.
Show that the points (1, 3, 1), (1, 1, -1), (-1, 1, 1), (2, 2, -1) are lying on the same plane.
Solution:

Question 6.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1.
Evaluate the following

Solution:
We know that

(ii)
Solution:

= 0 + 1 + 24 + 4 = 29

Question 2.

Solution:

By using Gamma integral (n = 1; a = α)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6

Question 1.

Solution:

(ii)

Solution:

(iii)

Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:
We know that,

(viii)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6 Additional Problems

Question 1.
Evaluate

Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

Question 2.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.7

Question 1.
In each of the following cases, determine whether the following function is homogeneous or not. If it is so, find the degree.

Solution:

It is not a homogeneous function

∴ It is a homogeneous function with degree 3.


∴ It is homogeneous function of degree 0.

∴ It is not a homogeneous function

Question 2.
Prove that f(x, y) = x³ – 2x² y + 3xy² + y³ is homogeneous; what is the degree? Verify Euler’s Theorem for f.
Solution:
f (x, y) = x³ – 2x²y + 3xy² + y³
f(tx, ty) = t³x³ – 2(t²x²)(ty) + 3(tx)(t²y²) + t³y³
= t³ [x³ – 2x²y + 3xy² + y³]
f(tx, ty) = t³ f(x, y)
‘f’ is a homogeneous function of degree 3. By Euler’s theorem, we have

∴ Euler’s Theorem verified

Question 3.
Prove that g(x, y) = x log (\(\frac{y}{x}\)) is homogeneous; what is the degree? Verify Euler’s Theorem for g.
Solution:


∴ Euler’s Theorem verified

Question 4.

Solution:

Question 5.

Solution:

∴ ‘f’ is a homogeneous function of degree 1. By Euler’s theorem, we have

Question 6.

Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.7 Additional Problems

Question 1.

Solution:

Question 2.

Solution:
R.H.S. is not a homogeneous and hence

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4

Question 1.
Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse 2x² + 7y² = 14 .
Solution:
2x² + 7y² = 14
(÷ by 14) ⇒ \(\frac{x^{2}}{7}+\frac{y^{2}}{2}\) = 1
comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
we get a² = 7 and b² = 2
The equation of tangent to the above ellipse will be of the form
y = mx + \(\sqrt{a^{2} m^{2}+b^{2}}\) ⇒ y = mx + \(\sqrt{7 m^{2}+2}\)
Here the tangents are drawn from the point (5, 2)
⇒ 2 = 5m + \(\sqrt{7 m^{2}+2}\) ⇒ 2 – 5m = \(\sqrt{7 m^{2}+2}\)
Squaring on both sides we get
(2 – 5m)² = 7m² + 2
25m² + 4 – 20m – 7m² – 2 = 0
18m² – 20m + 2 = 0
(÷ by 2) ⇒ 9m² – 10m + 1 = 0
(9m – 1) (m – 1) = 0
‘ m = 1 (or) m = 1/9
When m = 1, the equation of tangent is
y = x + 3 or x – y + 3 = 0
When m = \(\frac{1}{9}\) the equation of tangent is 9
y = \(=\frac{x}{9}+\sqrt{\frac{7}{81}+2}\) (i.e.) y = \(\frac{x}{9}+\frac{13}{9}\)
9y = x + 13 or x – 9y + 13 = 0

Question 2.
Find the equations of tangents to the hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{64}\) = 1 which are parallel to 10x – 3y + 9 = 0.
Solution:
\(\frac{x^{2}}{16}-\frac{y^{2}}{64}\) = 1
Here a² = 16 and b² = 64
The equation of tangents will be of the form y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(i.e.,) y = mx ± \(\sqrt{16 m^{2}-64}\)
Where ‘m’ is the slope of the tangent.
Here we are given that the tangents are parallel to 10x – 3y + 9 = 0
So slope of tangents will be equatl to slope of the given line

Question 3.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12. Also find the coordinates of the point of contact.
Solution:
The given ellipse is x² + 3y² = 12
(÷ by 12) ⇒ \(\frac{x^{2}}{12}+\frac{y^{2}}{4}\) = 1
(ie.,) Here a² = 12 and b² = 4
The given line is x – y + 4 = 0
(ie.,) y = x + 4
Comparing this line with y = mx + c
We get m = 1 and c = 4
The condition for the line y = mx + c
To be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is c² = a²m² + b²
LHS = c² = 4² = 16
RHS: a²m² + b² = 12( 1 )² + 4 = 16
LHS = RHS The given line is a tangent to the ellipse. Also the point of contact is
\(\left(\frac{-a^{2} m}{c}, \frac{b^{2}}{c}\right)=\left[-\left(\frac{12(1)}{4}\right), \frac{4}{4}\right]\) (i.e.,) (-3, 1)

Question 4.
Find the equation of the tangent to the parabola y² = 16x perpendicular to 2x + 2y + 3 =0
Solution:
y² =16x
Comparing this equation with y² = 4ax
we get 4a = 16 ⇒ a = 4
The equation of tangent to the parabola y² = 16x will be of the form

So m = 1
⇒ The equation of tangent will be y = 1(x) + \(\frac{4}{1}\) (i.e.,) y = x + 4
(or) x – y + 4 = 0

Question 5.
Find the equation of the tangent at t = 2 to the parabola y² = 8x. (Hint: use parametric form)
Solution:
y² = 8x .
Comparing this equation with y² = 4ax
we get 4a = 8 ⇒ a = 2
Now, the parametric form for y² = 4ax is x = at², y = 2at
Here a = 2 and t = 2
⇒ x = 2(2)² = 8 and y = 2(2) (2) = 8
So the point is (8, 8)
Now eqution of tangent to y² = 4 ax at (x₁, y₁) is yy₁ = 2a(x + x₁)
Here (x₁, y₁) = (8, 8) and a = 2
So equation of tangent is y(8) = 2(2) (x + 8)
(ie.,) 8y = 4 (x + 8)
(÷ by 4) ⇒ 2y = x + 8 ⇒ x – 2y + 8 = 0
Aliter
The equation of tangent to the parabola y² = 4ax at ‘t’ is
yt = x + at²
Here t = 2 and a = 2
So equation of tangent is
(i.e.,) y(2) = x + 2(2)²
2y = x + 8 ⇒ x – 2y + 8 = 0

Question 6.
Find the equations of the tangent and normal to hyperbola 12x² – 9y² = 108 at θ = \(\frac{\pi}{3}\) .
(Hint: Use parametric form)
Solution:
12x² – 9y² = 108

Normal is a line perpendicular to tangent
So equation of normal will be of the form 3x + 4y + k = 0
The normal is drawn at (6, 6)
⇒ 18 + 24 + k = 0 ⇒ k = – 42
So equation of normal is 3x + 4y – 42 = 0

Question 7.
Prove that the point of intersection of the tangents at ‘t₁‘ and t₂’ on the parabola y² = 4ax is [at₁ t₂, a (t₁ + t₂)].
Solution:
The equation of tangent to y² = 4ax at ‘t’ is given by yt = x + at²
So equation of tangent at ‘t₁‘ is yt₁ = x + at₁²
and equation of tangent at ‘t₂‘ is yt₂ = x + at₂²
To find the point of intersection we have to solve the two equations

Question 8.
If the normal at the point ‘t₁‘ on the parabola y² = 4ax meets the parabola again at the point ‘t₂‘, then prove that t₂ = \(-\left(t_{1}+\frac{2}{t_{1}}\right)\)
Solution:
Equation of normal to y² =4 at’ t’ is y + xt = 2at + at³.
So equation of normal at ‘t₁’ is y + xt₁ = 2at₁ + at₁³
The normal meets the parabola y² = 4ax at ‘t₂’ (ie.,) at (at₂², 2at₂)
⇒ 2at₂ + at₁t₂² = 2at₁ + at₁³
So 2a(t₂ – t₁) = at₁³ – at₁t₂²
⇒ 2a(t₂ – t₁) = at₁(t₁² – t₂²)
⇒ 2(t₂ – t₁) = t₁(t₁ + t₂)(t₁ – t₂)
⇒ 2= -t₁(t₁ + t₂)
⇒ t₁ + t₂ = \(\frac{-2}{t_{1}}\)
⇒ t₂ = \(-t_{1}-\frac{2}{t_{1}}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4 Additional Questions

Question 1.
Find the equations of the tangent and normal to the parabolas : x² + 2x – 4y + 4 = 0 at (0, 1)
Solution:

Question 2.
Find the equations of the tangent and normal to the parabola y² = 8x at t = \(\frac{1}{2}\)
Solution:


∴ Equation of the tangent is 2x – y + 1 = 0 and equation of the normal is 2x + 4y – 9 = 0

Question 3.
Find the equations of the tangents: to the parabola y² = 6x, parallel to 3x – 2y + 5 = 0
Solution:


∴ Equation of the tangent is 3x – 2y + 2 = 0

Question 4.
Find the equations of the tangents: to the parabola 4x² – y² = 64 Which are parallel to 10x – 3y + 9 = 0.
Solution:

Question 5.
Find the equation of the tangent from point (2, -3) to the parabola y² = 4x.
Solution:
y² = 4x
Equation of the tangent to the parabola will be of the form y = \(m x+\frac{1}{m}\)
The tangents pass through (2, -3)


The two tangents drawn from (2, -3) are x + y + 1 = 0, x + 2y + 4 = 0

Question 6.
Find the equation of the tangents from the point (2, -3) to the parabola 2x² – 3y² = 6
Solution:
2x² – 3y² = 6

Question 7.
Prove that the line 5x + 12y = 9 touches the hyperbola x² – 9y² = 9 and find the point of contact.
Solution:

(1) = (2) ⇒ The given line is a tangent to the curve i.e., the given line touches the curve. To find the point of contact we have to solve the line and the curve.
The given line 5x + 12y = 9

Question 8.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12. Find the co-ordinates of the point of contact.
Solution:
The given ellipse is x² + 3y² = 12
Given line is x – y + 4 = 0 ⇒ y = x + 4
Here, m = 1 and c = 4
The condition for the line y = mx + c to be a tangent to the ellipse

RHS: a² m² + b = 12(1) + 4 = 16
LHS: RHS ⇒ the given lines is a tangent to the ellipse.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6

Question 1.
If u(x, y) = x²y + 3xy⁴, x = e^t and y = sin t, find \(\frac{d u}{d t}\) and evaluate it at t = 0.
Solution:

= (2xy + 3y⁴) (e^t) + (x² + 12xy³) (cos t)
= (2e^t sin t + 3 sin⁴ t) e^t + [e^2t + 12e^t sin³ t] cos t
= e^t [2e^t sin t + 3 sin⁴ t + e^t (cos t) + 12 sin³t cos t]
at t = 0

Question 2.

Solution:

Question 3.
If w(x, y, z) = x² + y² + z², x = e^t, y = e^t sin t, and z = e^tcos t, find \(\frac{d w}{d t}\)
Solution:
w(x, y, z) = x² + y² + z² ; x = e^t ; y = e^t sin t, z = e^t cos t

Question 4.

Solution:

(Replace x, y, z value)

Question 5.
If w(x, y) = 6x³ – 3xy + 2y², x = e^s, y = cos s, s ∈ R, find \(\frac{d w}{d s}\), and evaluate at s = 0.
Solution:

Question 6.
If z(x, y) = x tan^-1 (x y), x = t², y = s e^t, s, t ∈ R, Find \(\frac{\partial z}{\partial \mathbf{t}}\) and \(\frac{\partial z}{\partial \mathbf{t}}\) at s = t = 1
Solution:
z (x, y) = x tan^-1 (xy) ; x = t² ; y = se^t

Question 7.
Let U (x, y) = e^x sin y, where x = st², y = s²t, s, t ∈ R. Find them at s = t = 1.
Solution:

(Replace x, y values)

Question 8.
Let z(x, y) = x³ – 3x²y³, where x = se^t, y = se^-t, s, t ∈ R. Find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\)
Solution:

Question 9.
W(x,y, z) = xy + yz + zx, x = u -v, y = uv, z = u + v, u, v e R. Find \(\frac{\partial \boldsymbol{w}}{\partial \boldsymbol{u}}\), \(\frac{\partial \boldsymbol{w}}{\partial \boldsymbol{v}}\) them at \(\left(\frac{1}{2}, 1\right)\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6 Additional Problems

Question 1.
Suppose that z = \(y e^{x^{2}}\) where x = 2t and y = 1 – t then find \(\frac{d z}{d t}\).
Solution:


(Since x = 2t and y = 1 – t)

Question 2.
If w = x + 2y + z² and x = cos t ; y = sin t ; z = t. Find \(\frac{d w}{d t}\).
Solution:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Question 1.
Evaluate the following:

Solution:

Substitute (2) and (3) in (1), we get

(ii)
Solution:

Dividing both Numerator and Denominator by cos² x

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5 Additional Problems

Question 1.

Solution:

Put 2 tan x, ∴ 2sc² x dx = dt

Question 2.

Solution: