Question

For `f(t)= (1/(2t-3), t/sqrt(t^2+1) )` what is the distance between `f(0)` and `f(1)`?

Answer 1

`sqrt17/(3sqrt2)`

Explanation:

Find the points, then use the distance formula to calculate the distance between them.

`f(t)=(x(t),y(t))`

The first part the parametric equation gives the `x` coordinate and the second part gives the `y` coordinate.

`f(0)=(1/(2(0)-3),0/sqrt(0^2+1))=(-1/3,0)`

`f(1)=(1/(2(1)-3),1/sqrt(1^2+1))=(-1,1/sqrt2)`

The distance between `(x_1,y_1)` and `(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` so the distance between `f(0)` and `f(1)` is given by:

`d=sqrt((-1-(-1/3))^2+(1/sqrt2-0)^2)`

`d=sqrt((-2/3)^2+(1/sqrt2)^2)`

`d=sqrt(4/9+1/2)=sqrt(17/18)=sqrt17/(3sqrt2)`