Question

10 points please help with calculus HW

Answer 1

a. The equation of the tangent at (1,3) is y = -x/3 + 10/3

b. The equation of the normal at (1,3) is y = 3x

c. Find the graph in the attachment

a. How to find the equation of the tangent at (1, 3)?

Since x² + y² = 10, we differentiate the equation with respect to x to find dy/dx which the the equation of the tangent.

So, x² + y² = 10

d(x² + y²)/dx = d10/dx

dx²/dx + dy²/dx = 0

2x + 2ydy/dx = 0

2ydy/dx = -2x

dy/dx = -2x/2y

dy/dx = -x/y

At (1,3), dy/dx = -1/3

Using the equation of a straight line in slope-point form, we have

m = (y - y₁)/(x - x₁) where

• m = gradient of the tangent = dy/dx at (1,3) = -1/3 and
• (x₁, y₁) = (1,3)

So, m = (y - y₁)/(x - x₁)

-1/3 = (y - 3)/(x - 1)

-(x - 1) = 3(y - 3)

-x + 1 = 3y - 9

3y = -x + 1 + 9

3y = -x + 10

3y + x = 10

y = -x/3 + 10/3

So, the equation of the tangent at (1,3) is y = -x/3 + 10/3

b. The equation of the normal at the point (1, 3)

Since the tangent and normal line are perpendicular at the point, for two perpendicular line,

mm' = -1 where

• m = gradient of tangent = -1/3 and
• m' = gradient of normal

So, m' = -1/m

= -1/(-1/3)

= 3

Using the equation of a straight line in slope-point form, we have

m' = (y - y₁)/(x - x₁) where

• m' = gradient of normal at (1, 3)  and (
• x₁, y₁) = (1,3)

So, m = (y - y₁)/(x - x₁)

3 = (y - 3)/(x - 1)

3(x - 1) = (y - 3)

3x - 3 = y - 3

y = 3x - 3 + 3

y = 3x + 0

y = 3x

So, the equation of the normal at (1,3) is y = 3x

c. Find the graph in the attachment

Learn more about equation of tangent and normal here:

brainly.com/question/7252502

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