Question

help please

Find the slope of the tangent line to the given curve at the point corresponding to the specified value of the parameter

Answer 1

\[x=\ln t\]
\[y=1+t^2\]

t= 1

Answer 2

x = ln t y= 1+t^2
x= ln (1) y=1+ (1)^2
x= 0 y= 2

(0,2)

Answer 3

@ganeshie8

Answer 4

\[\frac{ dy }{ dx } 1+t^2\]
____ = dy/dx = 2t^2

\[\frac{ dx }{ dt } \frac{ 1 }{ t }\]

Answer 5

not sure if i did it right..

Answer 6

\[\frac{ dy }{ dx }=\frac{ dy }{ dt }*\frac{ dt }{ dx }\]

Answer 7

oh yea

Answer 8

\[\frac{ dt }{ dx }=e ^{x}\]

Answer 9

\[\frac{ dy }{ dt }=2t\]

Answer 10

t=1, so dy/dt=2 and dt/dx=t=1

Answer 11

so answer is 2

Answer 12

\(\dfrac{dy}{dx} = \dfrac{\dot y}{\dot x} = \dfrac{2t}{\frac{1}{t}} = 2t^2\)

Answer 13

plug in the t value

Answer 14

You can also reason with the geometrical interpretation (pretty much the same as what the others showed above):

The velocity of the point following the curve is: \((x'(t), y'(t))\) (velocity is a vector).


So: \((x'(t),y'(t)) = (\frac1t, 2t)\).
The slope is: \(\frac{\Delta y}{\Delta x} = \frac{2t}{1/t}= 2t^2\). Pretty much the same calculations.