Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
I am new in differential equation. I just want to solve the following problem, Can any one give me the step wise solution for this question??? Determine the equation of the tangent line for the function f ( x ) = x 2 + 1 at point (3,10).

```
11 years ago

```							Find the slope of the function by differentiation
f '(x) = 2x

Plug in the certain point's values Since this function does not have y we don't plug in y yet
f '(3) = 6 {6 is now the slope of the point 3,10}

Plug both slope and point values into a linear equation
(y - y1) = m(x - x1) {this is the linear equation}
(y - 10) = 6(x - 3)  {Which can be simplified as below}
y = 6x -8

Just as we can find the slope and equation of a tangent line for a function, we can also do the same for a normal line. However, the normal line has two differences from the tangent line.

1. The slope of a normal line is perpendicular to the slope of the tangent line. Or in other words, the negative inverse of the tangent line.

2. The normal line is only defined if x does not = 0.

As a result, to find the slope and equation of the normal line, follow the steps above and convert the slope of the tangent line to the slope of the normal line.
```
11 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 51 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions