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2 years ago

5

HELP ME PLZ

1 answer:

Kryger [21]2 years ago

5 0

Explanation If a figure is rotated around a centre point and it still appears exactly as it did before the rotation, it is said to have rotational symmetry.

Examples : A number of shapes like squares, circles, regular hexagon, etc. have rotational symmetry.

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Read 2 more answers

Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e

svet-max [94.6K]

ANSWER

$y = 2x -4$

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

$x = 5 + ln(t)$

$y = {t}^{2} + 5$

From the first equation we make t the subject to get;

$x - 5 = ln(t)$

$t = {e}^{x - 5}$

We put it into the second equation.

$y = { ({e}^{x - 5}) }^{2} + 5$

$y = { ({e}^{2(x - 5)}) } + 5$

We differentiate to get;

$\frac{dy}{dx} = 2 {e}^{2(x - 5)}$

At x=5,

$\frac{dy}{dx} = 2 {e}^{2(5 - 5)}$

$\frac{dy}{dx} = 2 {e}^{0} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

$y-y_1=m(x-x_1)$

$y - 6 = 2(x - 5)$

$y = 2x - 10 + 6$

$y = 2x -4$

Without eliminating the parameter,

$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

$\frac{dy}{dx} = \frac{ 2t}{ \frac{1}{t} }$

$\frac{dy}{dx} = 2 {t}^{2}$

At x=5,

$5 = 5 + ln(t)$

$ln(t) = 0$

$t = {e}^{0} = 1$

This implies that,

$\frac{dy}{dx} = 2 {(1)}^{2} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

$y-y_1=m(x-x_1)$

$y - 6 = 2(x - 5) =$

$y = 2x -4$

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Step-by-step explanation:

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