Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.

x=tcos(t), y=tsin(t); t=π

the slope is

dy/dx = (dy/dt)/(dx/dt)
= (sint + tcost)/(cost - t sint)
y'(π) = (0-π)/(-1-0) = π

at t=π, x=-π and y=0

so, using the point slope form of the line,

y = π(x+π)

Well, well, well, looks like we've got ourselves a curve and a parameter. Strap on your math seatbelts, ladies and gentlemen, because we're about to find that tangent!

First, we need to find the derivative of our parametric equations. So let's get our derivative cap on and differentiate both x and y with respect to t.

Using a little trick called the chain rule, we get:

dx/dt = (1)cos(t) - tsin(t)
dy/dt = (1)sin(t) + tcos(t)

Now, we need to find the values of x and y when t = π. So, let's substitute that bad boy in!

x(π) = π*cos(π) = -π
y(π) = π*sin(π) = 0

Our point of interest is (-π, 0).

To find the equation of the tangent, we need the slope of the tangent. And how do we find it, you ask? Well, it's a piece of pi! We plug t = π into our dy/dt and dx/dt to find the slope.

dy/dt (π) = (1)sin(π) + (π)cos(π) = 1 + (-π) = -π+1
dx/dt (π) = (1)cos(π) - (π)sin(π) = -1 - 0 = -1

Slope = dy/dt (π) / dx/dt (π) = (-π+1)/(-1) = π-1

Now that we got our slope, all we need is a point. And guess what, we have one – (-π, 0).

Using the point-slope form, we can write the equation of the tangent as:

y - 0 = (π-1)*(x - (-π))

Simplifying this expression, we get:

y = (π-1)x + π

And voila! We've got our equation of the tangent.

Did you catch all that or would you prefer another ride on the math roller coaster?

To find the equation of the tangent to the curve at the point corresponding to the given value of the parameter, we need to find the derivative of the curve and evaluate it at the given point.

Given: x = tcos(t) and y = tsin(t), where t = π.

Step 1: Find the derivatives of x and y with respect to t.
To find the derivative of x = tcos(t), apply the product rule:
dx/dt = (1)(cos(t)) + (t)(-sin(t)) = cos(t) - tsin(t).

To find the derivative of y = tsin(t), apply the product rule:
dy/dt = (1)(sin(t)) + (t)(cos(t)) = sin(t) + tcos(t).

Step 2: Evaluate the derivatives at t = π.
Substitute t = π into the derivatives:
dx/dt = cos(π) - πsin(π) = -1,
dy/dt = sin(π) + πcos(π) = 0 + π(-1) = -π.

Step 3: Find the slope of the tangent.
The slope of the tangent is given by the derivative of y with respect to x: dy/dx.
dy/dx = (dy/dt)/(dx/dt) = (-π)/(-1) = π.

Step 4: Find the point corresponding to t = π.
Substitute t = π into the original equations:
x = πcos(π) = -π,
y = πsin(π) = 0.

Therefore, the point on the curve corresponding to t = π is (-π, 0).

Step 5: Write the equation of the tangent.
Using the slope-intercept form, y = mx + b, where m is the slope and (x, y) is a point on the line, we have:
m = π (slope of the tangent),
(x, y) = (-π, 0) (point on the curve corresponding to t = π).

Substituting these values into the slope-intercept form, we get:
y = πx + b.

To find the value of b, substitute the coordinates of the point (-π, 0):
0 = π(-π) + b,
0 = -π^2 + b,
b = π^2.

Therefore, the equation of the tangent to the curve at the point (-π, 0) is:
y = πx + π^2.

To find the equation of the tangent to the curve at a given point, you need to find the slope of the tangent and the coordinates of the point.

First, let's find the coordinates of the point that corresponds to the given value of the parameter, t=π:
x = t*cos(t)
y = t*sin(t)

Substitute t = π into the equations:
x = π*cos(π) = -π
y = π*sin(π) = 0

So the point on the curve that corresponds to t = π is (-π, 0).

Next, we need to find the slope of the tangent at this point. The slope of the tangent is given by the derivative of y with respect to x:

dy/dx = (dy/dt)/(dx/dt)

To find dy/dx, we need to find dy/dt and dx/dt first.

dy/dt = d/dt (t*sin(t))
= t*cos(t) + sin(t)

dx/dt = d/dt (t*cos(t))
= cos(t) - t*sin(t)

Now, substitute t = π into the expressions for dy/dt and dx/dt:
dy/dt = π*cos(π) + sin(π) = -π + 0 = -π
dx/dt = cos(π) - π*sin(π) = -1 - 0 = -1

So, dy/dx at t = π is given by:
dy/dx = (-π)/(-1) = π

Therefore, the slope of the tangent at the point (-π, 0) is π.

Finally, use the point-slope form of the equation of a line with the slope π and the point (-π, 0) to find the equation of the tangent:

y - y₁ = m(x - x₁)

where (x₁, y₁) is the point on the curve.

Substituting the values into the equation:
y - 0 = π(x - (-π))
y = π(x + π)

So, the equation of the tangent to the curve at the point (-π, 0) is y = π(x + π).