Showing posts with label Tangent Line. Show all posts

Tangent Line - Problem Solving

y = \sqrt{x^{2} - 25} a t t h e p o i n t (13, 12)

S t e p 1 : G e t t h e d e r i v a t i v e o f y = \sqrt{x^{2} - 25}

A p p l y t h e D e r i v a t i v e o f S q u a r e R o o t R u l e

S t e p 2 : G e t t h e s l o p e a t p o i n t (13, 12)

S t e p 3 : U s i n g t h e p o i n t - s l o p e f o r m a t p o i n t (13, 12)

F i n a l A n s w e r : 13 x - 12 y - 25 = 0

Tangent line to the given curves at the given point of tangency

y = x^{3} (x - 1)^{4} a t t h e p o i n t (2, 8)

S t e p 1 : F i n d t h e D e r i v a t i v e b y a p p l y i n g t h e P r o d u c t o f T w o F a c t o r s R u l e

S t e p 2 : U s i n g t h e p o i n t s l o p e f o r m a t p o i n t (2, 8)

F i n a l A n s w e r : 44 x - y - 80 = 0

F i n d t h e e q u a t i o n o f t h e t a n g e n t l i n e a n d n o r m a l l i n e

S t e p 1 : U s e t h e d i f f e r e n t i a t i o n f o r m u l a

: A t p o i n t (1, 2)

Hence, slope of the tangent line at (1, 2) equals - 1

while slope of the normal line at the point is 1.

Step 2: U s i n g t h e p o i n t- s l o p e f o r m o f t h e e q u a t i o n o f t h e l i n e .

E q u a t i o n o f t h e t a n g e n t l i n e :

E q u a t i o n o f t h e n o r m a l l i n e :

Find the slope of the given curve at the indicated point.

y = {(4 x^{2} + 3)}^{2}; (\frac{1}{2}, 16)

S t e p 1 : U s e t h e P o w e r F o r m u l a

\frac{d}{d x} (c u^{n}) = c n u^{n - 1} \frac{d}{d x} (u) t o f i n d f^{'} (x) = \frac{d y}{d x}

f^{'} (x) = \frac{d y}{d x} = 2 (4 x^{2} + 3) (8 x) = 16 x (4 x^{2} + 3)

N o t e : A t p o i n t (\frac{1}{2}, 16), s l o p e o f t a n g e n t l i n e i s g i v e n b e l o w .

F i n a l A n s w e r : T h e S l o p e o f t h e c u r v e a t (\frac{1}{2}, 16) i s 32