Question

Find an equation for the line tangent to the curve at the point defined by the given value of t.

x=t, y=√(2t), t=8

Answer 1

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

Answer 2

what? @sourwing

Answer 3

I just gave you the formula

Answer 4

for what? implicit differentiation. i already knew what that was. tried it, didnt work.

Answer 5

it is true formula but write it as dy/dx=dy/dt*dt/x so you can easily see that dt can be cancelled so it is equal to dy/dx

Answer 6

1/4 x + 2
1/4 x
-1/4 x
-1/4 x - 2

^^ those are the answer chooices.

i didnt get any of them but i got close to the first one

Answer 7

what is your dy/dx equal to?

Answer 8

.23555555 i think

Answer 9

wrong, but how did you get that number?

Answer 10

x=t,\[\frac{ dx }{dt }=1\]
\[y=\sqrt{2t}=\sqrt{2}t ^{\frac{ 1 }{ 2 }},\]
\[\frac{ dy }{dt }=\sqrt{2}\frac{ 1 }{2 }t ^{\frac{- 1 }{ 2 }}=\frac{ 1 }{\sqrt{2t} }\]
\[\frac{ dy }{dx }=\frac{ \frac{ dy }{dt } }{\frac{ dx }{ dt } }=1*\sqrt{2t}=\sqrt{2t}\]
\[at t=8,\frac{ dy }{dx }=\sqrt{2*8}=4,x=8,y=\sqrt{2t}=\sqrt{2*8}=4\]
eq. of tangent at (8,4) is
\[y-4=4(x-8)=4x-32,y=4x-32+4=4x-28\]

Answer 11

x = t, y = sqrt(2t) = (2t)^(1/2), at t=8
dx/dt = 1
dy/dt = 1/2 * (2t)^(-1/2) * 2 = (2t)^(-1/2) = 1/(sqrt(2t))
At t = 8, dx/dt = 1 and dy/dt = 1/sqrt(2*8) = 1/sqrt(16) = 1/4
dy/dx = dy/dt / dx/dt = 1/4

slope m = 1/4

x = t and y = sqrt(2t)
At t = 8, x = 8 and y = sqrt(16) = 4
So the point is (8,4)

The tangent line is: y = mx + b
y = 1/4x + b and it passes through (8,4)
4 = 1/4*8 + b
4 = 2 + b
b = 2

So the equation of the tangent line is: y = 1/4x + 2

Answer 12

sorry i calculated dy/dx as wrong , it is 1/4
thanks for correcting.