## find the slope of the tangent line at the point p0

We need to find the slope of the tangent line at the point p0. We can do this by finding the derivative of the function at the point p0. The derivative of the function at the point p0 will give us the slope of the tangent line at that point.

### take the derivative of the equation of the curve

The equation of the curve is y=rt. To find the slope of the tangent line at the point p0, we need to take the derivative of the equation of the curve. The derivative of y=rt is dy/dx=r. Therefore, the slope of the tangent line at p0 is r.

### plug in the coordinates of the point p0 into the derivative equation

To find the slope of the tangent line at the point p0, we need to take the derivative of the equation for rt. This will give us the instantaneous rate of change of rt at the point p0. To find the equation of the tangent line, we will then plug in the coordinates of p0 into this derivative equation.

## find the equation of the line tangent to the graph of rt at the point p0

### plug in the coordinates of the point p0 and the slope into the equation of a line

To find a vector equation of the line tangent to the graph of rt at the point p0 on the curve, we need to plug in the coordinates of the point p0 and the slope into the equation of a line. The slope of the line tangent to the graph of rt at the point p0 is given by:

rt'(p0) = r(t0) * t'(t0)

where t0 is the coordinate of p0 on the t-axis.

plugging in the values, we get:

rt'(p0) = r(t0) * t'(t0)

= (3,4) * 1

= 3

## find a vector equation of the line tangent to the graph of rt at the point p0 on the curve

To find a vector equation of the line tangent to the graph of rt at the point p0 on the curve, we need to find the derivative of the function at the point p0. The derivative of the function at the point p0 will give us the slope of the tangent line. Once we have the slope, we can use the point-slope form of a line to find the equation of the tangent line.

## take the gradient of the equation of the curve

To find a vector equation of the line tangent to the graph of r(t) at the point P0 on the curve, we take the gradient of the equation of the curve.

r'(t) = (x'(t), y'(t)) = (cos t, sin t)

At P0, we have:

t = 0 ⇒ r'(0) = (cos 0, sin 0) = (1, 0)

Thus, the equation of the line tangent to the graph of r(t) at P0 is:

x = x(0) + t·cos 0 ⇒ x – x(0) = t·1 ⇒ x – 1 = t

y = y(0) + t·sin 0 ⇒ y – y(0) = t·0 ⇒ y – 2 = 0

## plug in the coordinates of the point p0 into the gradient equation

The gradient of the line tangent to the graph of r(t) at the point p0 on the curve is given by:

gradient = (dy/dx)|_(t=t0)

= (dr/dt)|*(t=t0) * (1/((dx/dt)|*(t=t0)))

plug in the coordinates of the point p0 into the gradient equation to get:

gradient = (dr/dt)|*(t=2, r=3) * (1/((dx/dt)|*(t=2, r=3))) = 36 * (1/-6) = -6