Find equations of the tangent line and normal line to the curve at the given point?

Question

47. y = sqrt (1+4sinx) , (0,1)

48. x^2 + 4xy + y^2 = 13, (2,1)

please show all steps so i can understand

Answer 1

47.

The curve y = √(1+ 4sinx)

The point of the curve is (0,1)

The slope of the curve m = dy / dx

But y = √(1+ 4sinx)

Diferenciate with respective x

dy / dx = d / dx√(1 + 4sinx)

d / dx(sinx) = cosx

dy / dx = 1 / 2√(1 + 4sinx)(0 + 4cosx)

The slope of the curve at point (0,1) : m = dy / dx = 1 / 2√(1 + 4sin0)(0 + cos0)

m = dy / dx = 1 / 2√(1 + 0)(0+ 1)

m = 1/ 2 √(1)(1)

m = 1 / 2

The slope of the curve m and at the point (x1,y1) then  the equation of the tangent is y - y1 = m(x - x1)

Substitute m = 1 / 2 and x1 = 0 , y1 = 1

Tangent line is y - 1 = 1 / 2(x - 0)

y = x / 2 + 1.

The normal line slope is -1 / m = -1 / 1/2

slope =-2 and point (0,1)

The equation is y - 1 = -2(x - 0)

There fore y = -2x + 1

The equation of the normal line : y = -2x + 1.

 

 

Answer 2

48.

 The curve x² + 4xy + y² = 13

The point of the curve is (2,1)

The slope of the curve m = dy / dx

Diferenciate with respective x to the curve

Recall : Diferenciate formula deriavative of xⁿ  = n x^n-1 and derivative of uv = uv¹ +vu¹

2x + 4[x (dy / dx) + y(1)] + 2y dy / dx = 2x + 4[x (dy / dx) + y] + 2y dy / dx = o

2x + 4xdy /dx + 4xy + 2ydy /dx = 0

2x + 4xdy / dx + 2ydy / dx  + 4xy = 0

2x + 4xy + 2dy / dx (2x + y) = 0

Subtract 2x + 4xy from each side

2dy / dx (2x + y) = -(2x + 4xy)

Take out common term 2x

2dy / dx (2x + y) = -2x (1 + 2y)

Divide each side by 2

dy /dx (2x + y) = -(1 + 2y)

Divide each side by 2x + y

dy /dx = -(1+2y) / (2x + y)

dy / dx at point (2,1)

Substitute x = 2 and y = 1 in the slope (m) dy /dx

The slope m = -(1 + 2(1)) / (2(2) + 1)

Simplify

m = -3 / 5

The slope of the given curve is m = -3 / 5 and at point (2,1)

slope m and point (x1,y1) then

The equation of the tangent line : y - y1 = m( x - x1)

Substitute m = -3 / 5 and x1 = 2 , y1= 1 in the equation of the tangent line y -1 = -3/5(x - 2)

y - 1 = -3/5x +6/5

Add 1 to each side

y = -3/5x + 6 / 5 + 1

Simplify

y =-3/5x + 11 / 5

Multiply each side by 5

There fore 5y = -3x + 11

The equation of the tangent line : y = -3/5x + 11/5 or 3x + 5y = 11

The product slopes of the tangent and normal line is -1.

There fore the normal slope = -1 / tangent slope =-1 / m

Substitute m = -3 / 5

The normal slope = -1 / -3/5 =-5 / -3 = 5 / 3 and at point (2,1)

Substitute slope =5/3, x1 =2 and y1 = 1

The normal line equation y - 1 = 5/3( x - 2)

Simplify

y - 1 = 5/3x - 10/3

Add 1 to each side

y = 5/3x -10/3 +1

Simplify

y = 5/3x -7/3

Multiply each side by 3

3y =5x - 7

There fore the normal line equation : 5x - 3y =7

 

Comments

this is wrong, you have one more x than you should when you derive.

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Please check the updated answer.

Answer 3

 

47.

The curve y = √(1+ 4sinx)

The point of the curve is (0,1)

The slope of the curve m = dy / dx

But y = √(1+ 4sinx)

Diferenciate with respective x

dy / dx = d / dx√(1 + 4sinx)

d / dx(sinx) = cosx

dy / dx = 1 / 2√(1 + 4sinx)(0 + 4cosx)

The slope of the curve at point (0,1) : m = dy / dx = 1 / 2√(1 + 4sin0)(0 + 4cos0)

m = dy / dx =  1/2(0+4)

m = dy / dx = 1/2*4 = 2

m = 2

The slope of the curve at the point (x1,y1) then equation of the tangent is y - y1 = m( x- x1)

Substitute m = 2 and x1 =0 , y1 =1

The equation of tangent line is y - 1 = 2(x - 0)

y - 1 = 2x

The normal line slope is -1/m = -1/2

Slope = -1/2 and point ( 0,1)

The equation is y - 1 =-1/2(x -0)

y-1 =-1/2(x)

2(y - 1) = -x

Therefore x + 2y - 2 = 0

The equation of normal line is x + 2y - 2 =0

 

Answer 4

(48)

Step 1:

The curve equation is and the point is .

Slope of the function is derivative of the function at that point.

Apply derivative on each side with respect to .

.

Find the slope of the tangent line at .

.

Step 2:

Point slope form of line equation is .

Substitute and .

The tangent line equation is .

Step 3:

The normal line is perpendicular to the tangent line.

The equation of normal line at is , Where is slope of the tangent line.

Substitute and .

The normal line equation is .

Solution:

The tangent line equation is .

The normal line equation is .