Equation for line tangent to curve

• September 22nd 2009, 09:02 PM
asweet1
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.
• September 22nd 2009, 09:13 PM
Chris L T521
Quote:

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?
• September 22nd 2009, 09:18 PM
asweet1
• September 22nd 2009, 09:28 PM
Chris L T521
Quote:

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.

Quote:

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?
• September 22nd 2009, 09:29 PM
mr fantastic
Quote:

Originally Posted by asweet1