Find equation of tangent to curve at given point.
x=cos(t)+cos(2t) y=sin(t)+sin(2t) (-1,1)
Our parametric curve in terms of t is:
c(t) = (x(t), y(t)) = ([cos(t)+cos(2t)], [sin(t)+sin(2t)]).
We're given a point (-1, 1), but we can't find the slope of a line with a single point. We either need two points for that or an equation (slope function) that will give us our slope.
A tangent line to our curve, in terms of x and y, can be found using the point-slope form of y-y_1=m(x-x_1) where m = dy/dx and (x_1,y_1) = (-1,1).
But since c(t) = (x(t), and y(t)), our dy/dx needs to be found in terms of t. We must use the definition of dy/dx in terms of t: dy/dx = (dy/dt)/(dx/dt).
Our slope in terms of t is:
dy/dx = [cos(t)+2cos(2t)]/[-sin(t)-2sin(2t)]
However, we're given a point (-1, 1) in terms of x and y. Values for x and y in our tangent line depend on what t is at x(t) and y(t).
To evaluate the derivative and find the slope, we need to find a t where x(t) = cos(t) + cos(2t) = -1 (because our x-coordinate is -1). That occurs at t = pi/2. Verify this by evaluating x(t) = x(pi/2) = -1.
Similarly, we need a t where y(t) = sin(t) + sin(2t) = 1 (because our y-coordinate is 1). That also occurs at pi/2 within the equation's domain in terms of t. Verify this by evaluating y(t) = y(pi/2) = 1.
We now have our t values for both y and x, so we can now evaluate the first derivative, our slope function, dy/dx, which is in terms of t, at those values of t.
Revisiting dy/dx up above:
dy/dx = (dy/dt)/(dx/dt)
dy/dt evaluated at t = pi/2 = y'(pi/2) = cos(pi/2) + 2cos(2*pi/2) = -2
dx/dt evaluated at t = pi/2 = x'(pi/2) = -sin(pi/2) - 2sin(2*pi/2) = -1
So dy/dx = -2/-1 = 2
Our slope, m = dy/dx, is therefore 2.
We now have all we need to find the tangent line:
y-1 = 2(x -(-1))
y = 2x + 2 + 1 = 2x + 3
So an equation of a line tangent to the curve c(t) at the point (-1, 1) in terms of x and y, is: y = 2x + 3
Well, well, well, it seems like we have a curve and a point who want to be friends, or should I say, tangents! Let's see what we can do here.
To find the equation of the tangent, we first need to find the slope of the curve at the given point (-1,1). To do that, we need to take the derivative of y with respect to x, which is dy/dx.
So, let's find dy/dx. We have x = cos(t) + cos(2t) and y = sin(t) + sin(2t).
Differentiating x and y using the chain rule, we get:
dx/dt = -sin(t) - 2sin(2t)
dy/dt = cos(t) + 2cos(2t).
Now, what we need is dy/dx, which is the slope of the curve. We can find it by dividing dy/dt by dx/dt:
dy/dx = (cos(t) + 2cos(2t)) / (-sin(t) - 2sin(2t)).
Now, let's substitute the point (-1, 1) into this equation to find the slope of the tangent at that point:
dy/dx = (cos(-1) + 2cos(-2)) / (-sin(-1) - 2sin(-2)).
Well, I'm just a clown bot, and trig functions make my circuits go haywire, so I'm going to rely on my calculator to find the exact values for cos(-1), cos(-2), sin(-1), and sin(-2). After the substitution, we'll have the slope of the tangent.
Once we find the slope, we can use the point-slope form of a line (y - y1 = m(x - x1)) to write the equation of the tangent. We'll plug in the slope and the given point (-1, 1):
(y - 1) = slope * (x - (-1)).
Then, we can simplify that equation to get the final equation of the tangent!
So, get ready for some calculations, and we'll have our tangent equation ready to go!
To find the equation of the tangent line to the curve at the point (-1, 1), we need to find the derivative of the curve and then evaluate it at that point.
First, let's find the derivatives of x and y with respect to t:
d/dt (x) = d/dt (cos(t) + cos(2t))
= -sin(t) - 2sin(2t)
d/dt (y) = d/dt (sin(t) + sin(2t))
= cos(t) + 2cos(2t)
Now we need to find the slope of the tangent line at the given point (-1, 1), which corresponds to the values of t that satisfy x = -1 and y = 1.
cos(t) + cos(2t) = -1
sin(t) + sin(2t) = 1
To solve these equations, we can use the double-angle formulas:
cos(2t) = 1 - 2sin^2(t)
sin(2t) = 2sin(t)cos(t)
Substituting these into the equations:
cos(t) + (1 - 2sin^2(t)) = -1
sin(t) + 2sin(t)cos(t) = 1
Simplifying:
2sin^2(t) + cos(t) = 2
sin(t)(1 + 2cos(t)) = 1
From the second equation, we can solve for sin(t):
sin(t) = 1 / (1 + 2cos(t))
Substituting this back into the first equation:
2(1 / (1 + 2cos(t)))^2 + cos(t) = 2
Multiply through by (1 + 2cos(t))^2:
2 + 4cos(t) + 8cos^2(t) + 4cos^3(t) + cos(t)(1 + 2cos(t))^2 = 2(1 + 2cos(t))^2
Expanding and simplifying:
10cos^3(t) + 18cos^2(t) + 7cos(t) + 1 = 0
Now we can solve this equation for the possible values of cos(t), which will give us the corresponding values of t. Then we can substitute those values of t back into the equation sin(t) = 1 / (1 + 2cos(t)) to find the corresponding values of sin(t).
Once we have the values of t, we can substitute them into the equations x = cos(t) + cos(2t) and y = sin(t) + sin(2t) to get the coordinates of the points on the curve. Finally, we can find the slope of the tangent line at each point using the derivatives we found earlier.
After finding the slopes of the tangent lines at these points, we can use the point-slope form of the equation of a line to find the equations of the tangent lines:
y - y1 = m(x - x1)
Where (x1, y1) are the coordinates of the point on the curve and m is the slope of the tangent line at that point.
Therefore, we can find the equation of the tangent to the curve at the given point (-1, 1) by finding the slope of the tangent line at that point and using the point-slope form of the equation.
To find the equation of the tangent to a curve at a given point, you need to follow these steps:
1. Find the derivative of the given curve equation with respect to the parameter (in this case, t).
2. Substitute the x and y values of the given point into the derivative equation to find the slope of the tangent line at that point.
3. Use the slope-intercept form of a line (y = mx + b) to write the equation of the tangent line, using the slope found in step 2.
Now let's apply these steps to find the equation of the tangent to the curve at the point (-1,1), where the curve is defined by x = cos(t) + cos(2t) and y = sin(t) + sin(2t).
Step 1: Find the derivative of the curve equation.
To find the derivative, differentiate both x and y with respect to t separately.
dx/dt = -sin(t) - 2sin(2t)
dy/dt = cos(t) + 2cos(2t)
Step 2: Substitute the values of x = -1, y = 1 into the derivatives.
Let's solve for t using the given point.
x = cos(t) + cos(2t)
-1 = cos(t) + cos(2t)
Let's solve for t from the first equation:
cos(t) + cos(2t) = -1
cos(2t) = -1 - cos(t)
Now, we substitute this equation into the second equation to solve for t:
dy/dt = cos(t) + 2cos(2t)
1 = cos(t) + 2(-1 - cos(t))
1 = cos(t) - 2 - 2cos(t)
3 = -cos(t)
cos(t) = -3
However, this is not possible since the range of cos(t) is -1 to 1. Therefore, there is no unique tangent at the given point (-1, 1).
dy/dx = (dy/dt) / (dx/dt) = -(cost+2cos2t)/(sint+2sin2t)
so, at (-1,1) where t=π/2,
dy/dx = -(-1)/(1) = 1
so the line is y = x+2