1. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(iii)

Lets suppose and let x = 1 and
Then,
Now, we cam say that is approximately equals to dy
Now,
Hence, is approximately equal to 0.8

1. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(ii)

Lets suppose and let x = 49 and
Then,
Now, we can say that is approximately equal to dy
Now,
Hence, is approximately equal to 7.035

1. Using differentials, find the approximate value of each of the following up to 3

places of decimal. (i)

Lets suppose and let x = 25 and
Then,
Now, we can say that is approximate equals to dy
Now,
Hence, is approximately equals to 5.03

27) The line is a tangent to the curve at the point

(A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)

The slope of the given line is 1
given curve equation is
If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve
The slope of tangent =
Now, when y = 2,
Hence, the coordinates are (1,2)
Hence, (A) is the correct answer

26) The slope of the normal to the curveis

(A) 3 (B) 1/3 (C) –3 (D) -1/3

Equation of the given curve is
Slope of tangent =
at x = 0
Now, we know that
Hence, (D) is the correct option

25) Find the equation of the tangent to the curve which is parallel to the line

Parellel to line means the slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get the slope of line m = 2 and c = 5/2
Now, we know that the slope of the tangent at a given point to given curve is given by
Given the equation of curve is
Now, when
,
but y cannot be -ve so we take only positive...

24) Find the equations of the tangent and normal to the hyperbola

at the point

Given equation is
Now ,we know that
slope of tangent =
at point
equation of tangent at point with slope
Now, divide both sides by
Hence, the equation of tangent is
We know that
equation of normal at the point with slope

23) Prove that the curves and xy = k cut at right angles*

Let suppose, Curve and xy = k cut at the right angle
then the slope of their tangent also cut at the right angle
means,
-(i)
Now these values in equation (i)
Hence proved

Find the equations of the tangent and normal to the parabola y ^ 2 = 4 ax at the point (at 2, 2 at).

22) Find the equations of the tangent and normal to the parabola at the point

Equation of the given curve is
Slope of tangent =
at point
Now, the equation of tangent with point and slope is
We know that
Now, the equation of at point with slope -t
Hence, the equations of the tangent and normal to the parabola
at the point are

21) Find the equation of the normals to the curve which are parallel

to the line

Equation of given curve is
Parellel to line means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with our given equation. we get,
Slope of tangent =
We know that
Now, when x = 2,
and
When x = -2 ,
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope
Similarly, the equation of at point...

20) Find the equation of the normal at the point for the curve

Given equation of curve is
Slope of tangent
at point
Now, we know that
equation of normal at point and with slope
Hence, the equation of normal is

19) Find the points on the curve at which the tangents are parallel

to the x-axis.

parellel to x-axis means slope is 0
Given equation of curve is
Slope of tangent =
When x = 1 ,
Hence, the coordinates are (1,2) and (1,-2)

18) For the curve , find all the points at which the tangent passes

through the origin.

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is
Slope of tangent =
Now, equation of tangent is
at (0,0) Y = 0 and X = 0
and we have
Now, when x = 0,
when x = 1 ,
when x= -1 ,
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

17) Find the points on the curve at which the slope of the tangent is equal to the y-coordinate of the point.

Given equation of curve is
Slope of tangent =
it is given that the slope of the tangent is equal to the y-coordinate of the point
We have
So, when x = 0 , y = 0
and when x = 3 ,
Hence, the coordinates are (3,27) and (0,0)

16) Show that the tangents to the curve at the points where x = 2 and x = – 2 are parallel.

Slope of tangent =
When x = 2
When x = -2
Slope is equal when x= 2 and x = - 2
Hence, we can say that both the tangents to curve is parallel

15) Find the equation of the tangent line to the curve which is (b) perpendicular to the line

Perpendicular to line means
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5
Now, we know that the slope of the tangent at a given point to given curve is given by
Given the equation of curve is
Now, when ,
Hence, the coordinates are
Now, the equation of tangent passing through (2,7) and...

15) Find the equation of the tangent line to the curve which is (a) parallel to the line

Parellel to line means slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get slope of line m = 2 and c = 9
Now, we know that the slope of tangent at a given point to given curve is given by
Given equation of curve is
Now, when x = 2 ,
Hence, the coordinates are (2,7)
Now, equation of tangent...

14) Find the equations of the tangent and normal to the given curves at the indicated points:

e)

We know that Slope of the tangent at a point on the given curve is given by
Given the equation of the curve
Now,
and
Now,
Hence slope of the tangent is -1
Now we know that,
Now, the equation of the tangent at the point with slope = -1 is
and
equation of the tangent at
i.e. is
Similarly, the equation of normal at with slope =...

14) Find the equations of the tangent and normal to the given curves at the indicated points

d)

We know that Slope of the tangent at a point on the given curve is given by
Given the equation of the curve
at point (0,0)
Hence slope of tangent is 0
Now we know that,
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = is

14) Find the equations of the tangent and normal to the given curves at the indicated

points:

c)

We know that Slope of the tangent at a point on the given curve is given by
Given the equation of the curve
at point (1,1)
Hence slope of tangent is 3
Now we know that,
Now, equation of tangent at point (1,1) with slope = 3 is
equation of tangent is
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
equation of normal is

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