# Thread: coordinates of P - parametric equations

1. ## coordinates of P - parametric equations

Hi

The parametric equations are x= e^t -1 and y=2e^t +3t

So dx/dt = e^t and dy/dt = 2e^t + 3

Let P be a point of curve C. Suppose the tangent line to C at the point P has slope 7. Find the coordinates of P

dy/dx = (2e^t + 3)/e^t = 7 so 7= 2+ 3/e^t and e^t = 3/5

Now I can then figure out that x = -2/5 but how do I figure out what y is?

y = 2e^t + 3t => y = 2(3/5) + 3t

Thanks

Calculus beginner

2. Originally Posted by calcbeg
Hi

The parametric equations are x= e^t -1 and y=2e^t +3t

So dx/dt = e^t and dy/dt = 2e^t + 3

Let P be a point of curve C. Suppose the tangent line to C at the point P has slope 7. Find the coordinates of P

dy/dx = (2e^t + 3)/e^t = 7 so 7= 2+ 3/e^t and e^t = 3/5

Now I can then figure out that x = -2/5 but how do I figure out what y is?

y = 2e^t + 3t => y = 2(3/5) + 3t

Thanks

Calculus beginner
$\displaystyle t = \ln\left(\frac{3}{5}\right)$

3. So y = 6/5 + 3 ln (3/5) but what does that equal to make it a coordinate with x= -2/5?

How do you add 6/5 + 3ln(3/5)?

Thanks

4. Originally Posted by calcbeg
So y = 6/5 + 3 ln (3/5) but what does that equal to make it a coordinate with x= -2/5?

How do you add 6/5 + 3ln(3/5)?

Thanks
you don't ... leave it as is