if x=e^t and y=(t-3)^2 , find an equation y=mx+b of the tangent to the curve at (1,9).

dy/dx=2(t-3)dt/dx

but
dx/dt=e^t, so

dy/dx=2(t-3)e^-t
so m= that. Now when x=1, (x=e^t), t must be zero, so

m=2(-3)=-6

y=-6x+b

now, at point 1,9
9=-6+b,or b=15, so equation for line must be

y=-6x+15

check that.

I just figured it out! thank you!

To find the equation of the tangent to the curve at (1,9), we need to find the slope of the tangent line (m) and the y-intercept (b).

1. Let's start by finding the derivative of y with respect to t. We'll need it to find the slope (m) of the tangent line.

Given: y = (t-3)^2

To differentiate y with respect to t, you can use the chain rule:

dy/dt = 2(t-3) * d(t-3)/dt

Simplifying, we have:

dy/dt = 2(t-3) * 1

dy/dt = 2(t-3)

2. Next, we need to find the value of t when x = e^t is equal to 1. We'll use this value of t to find the value of y.

Given: x = e^t

Setting x = 1, we have:

1 = e^t

To solve for t, take the natural logarithm (ln) of both sides:

ln(1) = ln(e^t)

0 = t

So, t = 0 when x = 1.

3. Now we have the value of t, let's determine the value of y at t = 0.

Given: y = (t-3)^2

Plugging in t = 0, we get:

y = (0 - 3)^2

y = 9

Therefore, when t = 0, the corresponding coordinates on the curve are (0,9).

4. Now we have two points on the tangent line: (1,9) and (0,9). We can use these points to find the slope (m) of the tangent line.

The slope (m) can be calculated using the formula:

m = (y2 - y1)/(x2 - x1)

Plugging in the coordinates (0,9) and (1,9), we have:

m = (9 - 9)/(1 - 0)

m = 0/1

m = 0

Therefore, the slope of the tangent line is 0.

5. Finally, we can write the equation of the tangent line using the slope (m = 0) and the point (1,9).

We have y = mx + b, where m is the slope and b is the y-intercept.

Plugging in m = 0 and (x,y) = (1,9), we get:

9 = 0 * 1 + b

9 = b

Therefore, the equation of the tangent line is y = 9.

So, the equation of the tangent to the curve at (1,9) is y = 9.