Define-a-curve-E-by-the-parametric-equations-x-t-g-1-t-h-1-t-f-1-u-du-y-t-g-2-t-h-2-t-f-2-u-du-where-h-1-g-1-h-2-and-g-2-are-different

Question Number 2284 by Yozzi last updated on 13/Nov/15

D e f i n e a c u r v e E b y t h e p a r a m e t r i c

e q u a t i o n s

x (t) = \int_{g_{1} (t)}^{h_{1} (t)} f_{1} (u) d u

y (t) = \int_{g_{2} (t)}^{h_{2} (t)} f_{2} (u) d u

w h e r e h_{1}, g_{1}, h_{2} a n d g_{2} a r e d i f f e r e n t i a b l e

a n d b o t h f_{1} a n d f_{2} a r e c o n t i n u o u s .

F i n d t h e g r a d i e n t, m, o f t h e t a n g e n t

t o E a t t h e p o i n t N w i t h p a r a m e t e r

v a l u e t (t ⩾ 0) .

L e t h_{1} (t) = t, g_{1} (t) = t^{2}, h_{2} (t) = t^{3}, g_{2} (t) = t^{4}

f_{1} (u) = u^{2} a n d f_{2} (u) = u^{3} . F i n d t h e

v a l u e s o f t w h e n t h e t a n g e n t i s

(i) v e r t i c a l,

(i i) h o r i z o n t a l .

W h a t a r e t h e e q u a t i o n s o f t h e t a n g e n t s

i n (i) a n d (i i) ?

Commented by Yozzi last updated on 14/Nov/15

A c o n s e q u e n c e o f t h e F u n d a m e n t a l

T h e o r e m o f C a l c u l u s i s t h a t, f o r

ϕ b e i n g c o n t i n u o u s, a n d b o t h ψ a n d η

b e i n g d i f f e r e n t i a b l e,

\frac{d}{d x} \int_{η (x)}^{ψ (x)} ϕ (z) d z = ϕ (ψ (x)) ψ^{^{'}} (x) - ϕ (η (x)) η^{^{'}} (x) .

T h e F u n d a m e n t a l T h e o r e m o f C a l c u l u s

s t a t e s t h a t f o r f b e i n g c o n t i n u o u s o v e r

t h e i n t e r v a l a ⩽ t ⩽ b, w i t h o n e g i v e n

a n y x \in [a, b], a n d w e l e t F b e a f u n c t i o n

s u c h t h a t

F (x) = \int_{a}^{x} f (t) d t

t h e n F^{^{'}} (x) = f (x) .

I t m a y i n t e r e s t y o u t o p r o v e t h e r e s u l t

u s i n g t h i s t h e o r e m b e f o r e u s i n g t h e

r e s u l t .

Answered by Yozzi last updated on 15/Nov/15

W e n e e d t o f i n d x^{^{'}} (t) a n d y^{^{'}} (t) .

x^{^{'}} (t) = \frac{d}{d t} \int_{g_{1} (t)}^{h_{1} (t)} f_{1} (u) d u . F r o m c o m m e n t s,

x^{^{'}} (t) = f_{1} (h_{1} (t)) h_{1}^{^{'}} (t) - f_{1} (g_{1} (t)) g_{1}^{^{'}} (t)

S i m i l a r l y, f o r y^{^{'}} (t) = \frac{d}{d t} \int_{g_{2} (t)}^{h_{2} (t)} f_{2} (u) d u,

y^{^{'}} (t) = f_{2} (h_{2} (t)) h_{2}^{^{'}} (t) - f_{2} (g_{2} (t)) g_{2}^{^{'}} (t) .

T h e g r a d i e n t m o f t h e t a n g e n t t o E a t t h e

p o i n t N w i t h p a r a m e t e r v a l u e t (t ⩾ 0)

i s t h e n g i v e n b y

m = \frac{y^{^{'}} (t)}{x^{^{'}} (t)} .

m = \frac{f_{2} (h_{2} (t)) h_{2}^{^{'}} (t) - f_{2} (g_{2} (t)) g_{2}^{^{'}} (t)}{f_{1} (h_{1} (t)) h_{1}^{^{'}} (t) - f_{1} (g_{1} (t)) g_{1}^{^{'}} (t)} .

G i v e n h_{1} (t) = t, g_{1} (t) = t^{2}, h_{2} (t) = t^{3}, g_{2} (t) = t^{4}

\Rightarrow h_{1}^{^{'}} (t) = 1, g_{1}^{^{'}} (t) = 2 t, h_{2}^{^{'}} (t) = 3 t^{2}, g_{2}^{^{'}} (t) = 4 t^{3} .

∵ f_{1} (u) = u^{2}, f_{2} (u) = u^{3}

m = \frac{{(t^{3})}^{3} \times 3 t^{2} - {(t^{4})}^{3} \times 4 t^{3}}{{(t)}^{2} \times 1 - {(t^{2})}^{2} \times 2 t}

= \frac{3 t^{11} - 4 t^{15}}{t^{2} - 2 t^{5}}

= \frac{t^{11} (3 - 4 t^{4})}{t^{2} (1 - 2 t^{3})}

m = \frac{t^{9} (3 - 4 t^{4})}{1 - 2 t^{3}}

(i) I f t h e t a n g e n t t o E a t N i s v e r t i c a l

\Rightarrow m i s u n d e f i n e d f o r s o m e t .

T h i s o c c u r s i f 1 - 2 t^{3} = 0 \Rightarrow t = 2^{- \frac{1}{3}} .

(i i) I f t h e t a n g e n t t o E a t N i s h o r i z o n t a l,

\Rightarrow m = 0 \Rightarrow t^{9} (3 - 4 t^{4}) = 0 \Rightarrow t^{9} = 0 o r 3 - 4 t^{4} = 0 .

t^{9} = 0 \Rightarrow t = 0 .

3 - 4 t^{4} = 0 \Rightarrow t^{4} = 3 / 4 \Rightarrow t = \pm {(\frac{3}{4})}^{1 / 4} .

S i n c e t ⩾ 0 a t N \Rightarrow t = {(\frac{3}{4})}^{1 / 4} .

t = 0 \lor t = {(\frac{3}{4})}^{1 / 4} .

Answered by Yozzi last updated on 15/Nov/15

G i v e n t h e f u n c t i o n s o f h_{1}, g_{1}, h_{2}, g_{2}, f_{1}

a n d f_{2}, w e h a v e

x (t) = \int_{t^{2}}^{t} u^{2} d u = \frac{u^{3}}{3} ∣_{t^{2}}^{t} = \frac{t^{3} - t^{6}}{3}

y (t) = \int_{t^{4}}^{t^{3}} u^{3} d u = \frac{u^{4}}{4} ∣_{t^{4}}^{t^{3}} = \frac{t^{12} - t^{16}}{4} .

I f t h e t a n g e n t i s v e r t i c a l a t t = 2^{- 1 / 3},

w e r e q u i r e a n x - l i n e .

∴ x (2^{- 1 / 3}) = \frac{2^{- 1} - 2^{- 2}}{3} = \frac{1}{12}

\Rightarrow e q u a t i o n o f v e r t i c a l t a n g e n t i s x = \frac{1}{12} .

I f t h e t a n g n e t i s h o r i z o n t a l w e h a v e

a y - l i n e . F o r t = 0, y (0) = 0 .

∴ t h e e q u a t i o n o f t h e t a n g e n t i s y = 0 .

F o r t = {(3 / 4)}^{1 / 4},

y ({(3 / 4)}^{1 / 4}) = \frac{0 . 75^{3} - 0 . 75^{4}}{4} = \frac{\frac{27}{64} - \frac{81}{256}}{4}

y ({(3 / 4)}^{1 / 4}) = \frac{1}{4} \times \frac{108 - 81}{256} = \frac{27}{1024}

T h e e q u a t i o n o f t h e o t h e r h o r i z o n t a l

t a n g e n t i s y = 27 / 1024 .

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