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mina [271]
3 years ago
12

a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve a

t the point (1, 2). y = Correct: Your answer is correct. (b) At what points does this curve have horizontal tangents? (x, y) = -2 Incorrect: Your answer is incorrect. (smaller y-value) (x, y) = 2 Incorrect: Your answer is incorrect. (larger y-value)
Mathematics
1 answer:
Neporo4naja [7]3 years ago
6 0

Answer:

Tangent at (1,2) has value \frac{15}{4}. And (-2, 4) is the point of intersection of horizontal tangent.

Step-by-step explanation:

Given curve equation,

y^2=x^3+3x^2\hfill (1)

To find tangent at (x,y)=(1,2) and point of intersection of horizontal tangent, differentiate (1) withrespect to x we get,

2y\frac{dy}{dx}=x^3+3x^2

\implies \frac{dy}{dx}=\frac{3(x^2+2x)}{2y}

At (1, 2), tangent line is,

\frac{dy}{dx}|_{(1,2)}}=\frac{15}{4}

To find point of intersection of horizontal tangent we have to do,

\frac{dy}{dx}=0

\implies x(x+2)=0\implies x=0 or -2

Thus,

At x=0, y=0

but snce,

\lim_{y\to 0}\frac{dy}{dx}\to \infty

at (0,0) there exist a vertical tangent. And,

At x=-2, y=4.

Thus (-2, 4) is the point of intersection of horizontal tangent.

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Step-by-step explanation:

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x=6-2y/5

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Each letter of the word ASSESSMENT is written on a card and placed in a bag. You are equally likely to get any card.
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Answer:

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Solve 2/3x> 8 or 2/3x< 4
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Decrease 248 by 30%
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Okay this equation really says is what is 30% of 248.

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8 0
3 years ago
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

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3 years ago
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