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
`        The slope of the tangent to the curve at a point (x,y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10, 9) on it is -3. The equation of the curve isAns. y = (-3/16) (x-2)2 + 3`
one year ago

```							the constant term should be 21 dy/dx=k(x-2)     where k is proportionality constant at (10,9)  dy/dx=-3  hence-3=k(10-2)   implies k=-3/8 dy/dx=(-3/8) *(x-2)dy=(-3/8)*(x-2)*dx integrating we get (-3/8)*((x-2)^2)/2)+c=y now put x=10 and y=9 as the are the point of the curve and we get c=21
```
one year 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