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2 years ago

Find the parametric equations for the line that is tangent to the given curve at the given parameter (3cost)i (t^4 -3sint)j

Mathematics

1 answer:

ale4655 [162]2 years ago

5 0

Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

In the given equation, when the point t=0

So,

$\to r(0) = (3 \cos 0)i + (0^4 - 6 \sin 0)j + (2e^{3\times 0})k)$

$= (3 \times 1)i + (0 - 0)j + (2e^{0})k)\\\\ = 3i + 0j + (2 \times 1)k)\\\\ = 3i + 0j + 2k \\$

The value of the coordinates are $3, 0, 2$ . so, the equation of the line is:

$\to \frac{(x-3)}{3 \cos \ t} = \frac{(y-0)}{(t^{4}-6 \sin \ t)} = \frac{(z-2)}{2e^{3t}}=k$

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Please help! A-level maths.

astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.

So $7\le k$ ; in set notation,

$\{k\mid 7\le k$

4 0

3 years ago

I need help finding the slope

Scilla [17]

Answer:

The answer is ✰ 4 ✰

Step-by-step explanation:

:))

8 0

2 years ago

Need help please help me

Ostrovityanka [42]

The solution is the point of intersection between the two equations.

Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.

In the graph the first equation is in blue and the second in red.

The point of intersection = the solution = (-6 , -1)

If you dont have access to a graphing calculator you could draw the graphs by hand;

1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)

y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)

x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3

y = x + 5

x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7

2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph

3) Using a ruler draw a line connection all the dots of one graph and do the same for the other

4) The point of intersection is the solution