Find-the-equation-of-the-line-tangent-to-the-parametric-curve-given-by-the-equations-x-1-t-3-4-t-2-y-t-5-t-2-2-at-t-1-

Question Number 120452 by john santu last updated on 31/Oct/20

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

c u r v e g i v e n b y t h e e q u a t i o n s {\begin{cases} x = {(1 + t^{3})}^{4} + t^{2} \\ y = t^{5} + t^{2} + 2 \end{cases} a t t = 1

Answered by physicstutes last updated on 31/Oct/20

x = {(1 + t^{3})}^{4} + t^{2}, when t = 1, x = 17

\Rightarrow \frac{d x}{d t} = 12 t^{2} {(1 + t^{3})}^{3} + 2 t

also y = t^{5} + t^{2} + 2 when t = 1, y = 4

\Rightarrow \frac{d y}{d t} = 5 t^{4} + 2 t

\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{5 t^{4} + 2 t}{12 t^{2} {(1 + t^{3})}^{3} + 2 t}

{(\frac{d y}{d x})}_{t = 1} = \frac{5}{98}

y - 4 = \frac{5}{99} (x - 17)

98 (y - 4) = 5 (x - 17) [mistakes corrected]

Commented by bramlexs22 last updated on 31/Oct/20

t y p o i t s h o u l d b e \frac{7}{98}

Commented by peter frank last updated on 31/Oct/20

help please 90208, 90030

Answered by bramlexs22 last updated on 31/Oct/20

g r a d i e n t (m) = \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}

= \frac{5 t^{4} + 2 t}{12 t^{2} {(1 + t^{3})}^{3} + 2} = \frac{7}{98}

f o r t = 1 \to {\begin{cases} x = 17 \\ y = 4 \end{cases}

e q o f l i n e t a n g e n t 7 x - 98 y = 7 (17) - 98 (4)

7 x - 98 y = 119 - 392

\Leftrightarrow 7 x - 98 y + 273 = 0