Question

I'm trying to find the line tangent to a parametrized function and I am doing that by eliminating the parameter rather than taking the derivative of both y(t) and x(t) and dividing, but I'm doing something wrong, and I can't figure out what.

Answer 1

\[x(t) = 5t, \ y(t) = \sqrt{t}\]

Answer 2

\[\frac {x}{5} = t\]
\[\sqrt{\frac {x}{5}} = \sqrt{t}\]

Answer 3

\[\frac {\sqrt{x}}{\sqrt{5}} = \sqrt{t} = y\]

Answer 4

\[\frac {dy}{dx} = \frac {1}{\sqrt{5}} \frac {d}{dx} (\sqrt{x})\]

Answer 5

\[\frac {dy}{dx} = \frac {1}{\sqrt{5}}* \frac{1}{2 \sqrt{x}} = \frac {1}{2 \sqrt{5} \sqrt{x}}\]

Answer 6

looks good ?

Answer 7

Yeah, it says that it should be:

Answer 8

Because they chose not to eliminate the parameter, which I haven't taken a look at, but the math should be the same result

Answer 9

lemme post the whole question verbatim just to make sure this isn't a bookkeeping issue

Answer 10

yup !

\(\large \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{dy}{dt} / \frac{dx}{dt} \)

Answer 11

Huh. Guess I'll let me prof. know about this.

Answer 12

about what ?

Answer 13

Lemme go through the parametrized version just to make sure. About this not matching up. IDK why they wouldn't be the same answer.

Answer 14

you wud get the same answer

Answer 15

Oh, derp.

Answer 16

we just need to replace "t" to "x" in the end :)

Answer 17

Yeah, lol...thanks

Answer 18

the second derivative is a bit trickier in parametric form

Answer 19

I'll try to do it myself, as long as I don't make any clumsy errors, I hopefully will be able to do it :3 I'll just leave it in cartesian form. ty.

Answer 20

you will be using parametric form a lot in multivariable calculus,
so it is better to do it in parametric form.. gives some practicce :)

Answer 21

cartesian form is straightforward.. .

Answer 22

good luck !

Answer 23

\[\frac {dy^{2}}{d^{2}x} = \frac {1}{2 \sqrt{5}} \frac {d}{dx}(\sqrt {x}) = - \frac {1}{4 \sqrt{5}x^{3/2}}\]

\[\frac {dy^{2}}{d^{2}x} = \frac {d}{dt} \left( \dfrac {dy}{dx} \right) / \frac {dx}{dt} = -\frac {1}{20 t^{3/2}} / \ 5 = - \frac {1}{100t^{3/2}}\]

Answer 24

(Thanks, btw!)