# Thread: Equation for line tangent to curve

1. ## Equation for line tangent to curve

Find an equation for the line that is tangent to the curve y= 10t + 15t^2 at the point (2, 80).

I thought I coul use the formula y=mx+b but I would need another coordinate. Is there is apecific to solve a an equation for tangent line. Please help I have to many theories. Maybe im just over thinking it.

2. Originally Posted by asweet1
Find an equation for the line that is tangent to the curve y= 10t + 15t^2 at the point (2, 80).

I thought I coul use the formula y=mx+b but I would need another coordinate. Is there is apecific to solve a an equation for tangent line. Please help I have to many theories. Maybe im just over thinking it.
Find $\frac{\,dy}{\,dt}$ at $t=2$. This will be your $m$.

With this, use the point slope form of the line:

$y-y_0=m(x-x_0)$

Can you take it from here?

3. would the answer be (x-2)/(y-2)

4. Originally Posted by asweet1
Find an equation for the line that is tangent to the curve y= 10t + 15t^2 at the point (2, 80).

I thought I coul use the formula y=mx+b but I would need another coordinate. Is there is apecific to solve a an equation for tangent line. Please help I have to many theories. Maybe im just over thinking it.
Originally Posted by asweet1
That is not correct.

You need to find the derivative of the function at $t=2$.

$y=10t+15t^2\implies\frac{\,dy}{\,dt}=10+30t$. At $t=2$, $m=\left.\frac{\,dy}{\,dt}\right|_{t=2}=10+30(2)=70$.

Thus, the equation of the tangent line at $(2,80)$ is

$y-80=70(x-2)$

I leave it for you to simplify this and rewrite it in $y=mx+b$ form.

Does this make sense?

5. Originally Posted by asweet1