Show that the curve x = 5 cos t, y = 4 sin t cos t has two tangents at (0, 0) and find their equations.
$begingroup$
1) As x = 5cos(t); differentiating, dx/dt = - 5sin(t)
2) As y = 4sin(t)cos(t), y = 2sin(2t) [Since 2sin(t)cos(t) = sin(2t)]
==> dy/dt = 4cos(2t)
3) So, dy/dx = (dy/dt)/(dx/dt) = 4cos(2t)/-5sin(t) = -(4/5)[cos(2t)/sin(t)]
4), At x = 0, cos(t) = 0; ==> t = either pi/2 or 3pi/2
As well y = 0, we get t = 0, pi/2, pi, 3pi/2, 2pi
Intersection of both = {pi/2, 3pi/2}
5) By geometrical definition of differentiation, dy/dx is the slope of the tangent at a given point on the curve. Since here for x = 0, there are two values of t,corresponding to which, we get two slopes. Hence there are two tangents at the given point (0,0).
The two slopes are:
At (t = pi/2): dy/dx = (-4/5)[cos(pi)/sin(pi/2)] = (-4/5)[-1/1] = 4/5
At (t = 3pi/2): dy/dx = (-4/5)[cos(3pi)/sin(3pi/2)] = (-4/5)[-1/-1] = -4/5
6) Equations of tangents: At (0,0)
i) Slope = -4/5 (Smaller slope):
==> y - 0 = (-4/5)(x - 0)
==> y = -4x/5
==> 4x + 5y = 0.
ii) Slope = 4/5 {Larger slope):
Proceeding as above, 4x - 5y = 0
The two tangents are: "4x + 5y = 0; and 4x - 5y = 0"
Yet, My answer is wrong?
calculus
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
$endgroup$
add a comment |
$begingroup$
1) As x = 5cos(t); differentiating, dx/dt = - 5sin(t)
2) As y = 4sin(t)cos(t), y = 2sin(2t) [Since 2sin(t)cos(t) = sin(2t)]
==> dy/dt = 4cos(2t)
3) So, dy/dx = (dy/dt)/(dx/dt) = 4cos(2t)/-5sin(t) = -(4/5)[cos(2t)/sin(t)]
4), At x = 0, cos(t) = 0; ==> t = either pi/2 or 3pi/2
As well y = 0, we get t = 0, pi/2, pi, 3pi/2, 2pi
Intersection of both = {pi/2, 3pi/2}
5) By geometrical definition of differentiation, dy/dx is the slope of the tangent at a given point on the curve. Since here for x = 0, there are two values of t,corresponding to which, we get two slopes. Hence there are two tangents at the given point (0,0).
The two slopes are:
At (t = pi/2): dy/dx = (-4/5)[cos(pi)/sin(pi/2)] = (-4/5)[-1/1] = 4/5
At (t = 3pi/2): dy/dx = (-4/5)[cos(3pi)/sin(3pi/2)] = (-4/5)[-1/-1] = -4/5
6) Equations of tangents: At (0,0)
i) Slope = -4/5 (Smaller slope):
==> y - 0 = (-4/5)(x - 0)
==> y = -4x/5
==> 4x + 5y = 0.
ii) Slope = 4/5 {Larger slope):
Proceeding as above, 4x - 5y = 0
The two tangents are: "4x + 5y = 0; and 4x - 5y = 0"
Yet, My answer is wrong?
calculus
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
$endgroup$
add a comment |
$begingroup$
1) As x = 5cos(t); differentiating, dx/dt = - 5sin(t)
2) As y = 4sin(t)cos(t), y = 2sin(2t) [Since 2sin(t)cos(t) = sin(2t)]
==> dy/dt = 4cos(2t)
3) So, dy/dx = (dy/dt)/(dx/dt) = 4cos(2t)/-5sin(t) = -(4/5)[cos(2t)/sin(t)]
4), At x = 0, cos(t) = 0; ==> t = either pi/2 or 3pi/2
As well y = 0, we get t = 0, pi/2, pi, 3pi/2, 2pi
Intersection of both = {pi/2, 3pi/2}
5) By geometrical definition of differentiation, dy/dx is the slope of the tangent at a given point on the curve. Since here for x = 0, there are two values of t,corresponding to which, we get two slopes. Hence there are two tangents at the given point (0,0).
The two slopes are:
At (t = pi/2): dy/dx = (-4/5)[cos(pi)/sin(pi/2)] = (-4/5)[-1/1] = 4/5
At (t = 3pi/2): dy/dx = (-4/5)[cos(3pi)/sin(3pi/2)] = (-4/5)[-1/-1] = -4/5
6) Equations of tangents: At (0,0)
i) Slope = -4/5 (Smaller slope):
==> y - 0 = (-4/5)(x - 0)
==> y = -4x/5
==> 4x + 5y = 0.
ii) Slope = 4/5 {Larger slope):
Proceeding as above, 4x - 5y = 0
The two tangents are: "4x + 5y = 0; and 4x - 5y = 0"
Yet, My answer is wrong?
calculus
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
$endgroup$
1) As x = 5cos(t); differentiating, dx/dt = - 5sin(t)
2) As y = 4sin(t)cos(t), y = 2sin(2t) [Since 2sin(t)cos(t) = sin(2t)]
==> dy/dt = 4cos(2t)
3) So, dy/dx = (dy/dt)/(dx/dt) = 4cos(2t)/-5sin(t) = -(4/5)[cos(2t)/sin(t)]
4), At x = 0, cos(t) = 0; ==> t = either pi/2 or 3pi/2
As well y = 0, we get t = 0, pi/2, pi, 3pi/2, 2pi
Intersection of both = {pi/2, 3pi/2}
5) By geometrical definition of differentiation, dy/dx is the slope of the tangent at a given point on the curve. Since here for x = 0, there are two values of t,corresponding to which, we get two slopes. Hence there are two tangents at the given point (0,0).
The two slopes are:
At (t = pi/2): dy/dx = (-4/5)[cos(pi)/sin(pi/2)] = (-4/5)[-1/1] = 4/5
At (t = 3pi/2): dy/dx = (-4/5)[cos(3pi)/sin(3pi/2)] = (-4/5)[-1/-1] = -4/5
6) Equations of tangents: At (0,0)
i) Slope = -4/5 (Smaller slope):
==> y - 0 = (-4/5)(x - 0)
==> y = -4x/5
==> 4x + 5y = 0.
ii) Slope = 4/5 {Larger slope):
Proceeding as above, 4x - 5y = 0
The two tangents are: "4x + 5y = 0; and 4x - 5y = 0"
Yet, My answer is wrong?
calculus
calculus
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
edited Nov 23 '15 at 3:02
Kay K.
6,9401337
6,9401337
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
asked Nov 23 '15 at 1:55
BenCarson2016BenCarson2016
1732616
1732616
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your answer is correct.
The graph looks like this.
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
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oldest
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$begingroup$
Your answer is correct.
The graph looks like this.
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
add a comment |
$begingroup$
Your answer is correct.
The graph looks like this.
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
add a comment |
$begingroup$
Your answer is correct.
The graph looks like this.
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
$endgroup$
Your answer is correct.
The graph looks like this.
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
answered Nov 23 '15 at 2:54
Kay K.Kay K.
6,9401337
6,9401337
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
add a comment |
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
How come the program I am using is stating it is wrong?
$endgroup$
– BenCarson2016
Nov 23 '15 at 3:01
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
In the title, you had written y = 6 sin t cos t, but in the body you used y = 4 sin t cos t. So I suggested to edit the one in the title. Maybe you mixed those two also in your program..?
$endgroup$
– Kay K.
Nov 23 '15 at 3:15
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
$begingroup$
No the title is correct. I made a mistake with the body. I will try to crunch the numbers again. Sorry.
$endgroup$
– BenCarson2016
Nov 23 '15 at 4:16
add a comment |
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