Write an equation for a line with a slope of<br> 3 and a y-intercept of (0, 2).

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

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

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

$y = 2x -4$

5 0

3 years ago

JL is a diameter of circle k. If tangents to circle I are constructed through points L and K what relationships would exist betw

luda_lava [24]

i. Let t be the line tangent at point J. We know that a tangent line at a point on a circle, is perpendicular to the diameter comprising that certain point. So t is perpendicular to JL
let l be the tangent line through L. Then l is perpendicular to JL ii. So t and l are 2 different lines, both perpendicular to line JL.
2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.
Remark. Draw a picture to check the

3 0

4 years ago

What are the zeros of (x-2)(x^2-9)

WARRIOR [948]

$(x-2)(x^2-9)=0\iff x-2=0\ or\ x^2-9=0\\\\x-2=0\ \ \ \ |add\ 2\ to\ both\ sides\\\\\boxed{x=2}\\\\x^2-9=0\ \ \ \ |add\ 9\ to\ both\ sides\\\\x^2=9\iff x=\pm\sqrt9\to \boxed{x=-3\ or\ x=3}\\\\Answer:\boxed{x=-3\ or\ x=2\ or\ x=3}$

3 0

3 years ago

Determine if each of the following sets is a subspace of Pn, for an appropriate value of n.

snow_tiger [21]

Answer:

1) W₁ is a subspace of Pₙ (R)

2) W₂ is not a subspace of Pₙ (R)

4) W₃ is a subspace of Pₙ (R)

Step-by-step explanation:

Given that;

1.Let W₁ be the set of all polynomials of the form p(t) = at², where a is in R

2.Let W₂ be the set of all polynomials of the form p(t) = t² + a, where a is in R

3.Let W₃ be the set of all polynomials of the form p(t) = at² + at, where a is in R

so

1)

let W₁ = { at² ║ a∈ R }

let ∝ = a₁t² and β = a₂t² ∈W₁

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t²) + c₂(a₂t²)

= c₁a₁t² + c²a₂t²

= (c₁a₁ + c²a₂)t² ∈ W₁

Therefore c₁∝ + c₂β ∈ W₁ for all ∝, β ∈ W₁ and scalars c₁, c₂

Thus, W₁ is a subspace of Pₙ (R)

2)

let W₂ = { t² + a ║ a∈ R }

the zero polynomial 0t² + 0 ∉ W₂

because the coefficient of t² is 0 but not 1

Thus W₂ is not a subspace of Pₙ (R)

3)

let W₃ = { at² + a ║ a∈ R }

let ∝ = a₁t² +a₁t and β = a₂t² + a₂t ∈ W₃

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t² +a₁t) + c₂(a₂t² + a₂t)

= c₁a₁t² +c₁a₁t + c₂a₂t² + c₂a₂t

= (c₁a₁ +c₂a₂)t² + (c₁a₁t + c₂a₂)t ∈ W₃

Therefore c₁∝ + c₂β ∈ W₃ for all ∝, β ∈ W₃ and scalars c₁, c₂

Thus, W₃ is a subspace of Pₙ (R)

8 0

3 years ago

I'm going crazy, someone help me

Sunny_sXe [5.5K]

Answer:

i wonder why

Step-by-step explanation:

4 0

3 years ago

Write an equation for a line with a slope of 3 and a y-intercept of (0, 2).