# tangent equations parrallel or perpendicular to ...

• May 17th 2009, 06:07 AM
SirNostalgic
tangent equations parrallel or perpendicular to ...
ok i have two question's i can't answer;

#1) find the equations of the tangents to the curve $y = x^3 -3x^2 -7x +5$ that are parrallel to the line with equation $y = 2x - 3$

i realise that it being parrallel the gradient would be the same. i also know you can solve these algabraicly just lacking the know how :)

#2) find the equation of the tangent line to the curve with equation $y = e^2x-6$ that is perpendicular to the line with equation $y = 5 - \frac {x}{2}$

same sort of deal, i realise it being perpendicular it has the gradient of a normal. don't know how to figure these out algabraicly.

thankin you kind people of math help forum.
• May 17th 2009, 06:31 AM
VonNemo19
We know that the slope of the line $y=2x+3$is 2. So therefore, all we have to do is find the point at which a line tangent to the curve $y=x^3-3x^2-7x+5$, and then determine the equation of this line, knowing that it's slope is 2.

So, how do we do this? we know that the slope of a line that goes throgh any 2 points on the curve is given by $\frac{y_2-y_1}{x_2-x_1}$, or in functional notation $\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}$. But this is only 2 points that goes through the curve, it is not the tangent line. To find the tangent line to a curve at any point on the curve, we'd have to take the limit as $\Delta{x}$ tends to 0. Or

$\lim_{\Delta{x}\to0}\frac{x^3-3x^2-7x+5-f(x)}{\Delta{x}}$.

So, Let's do it.

$\lim_{\Delta{x}\to0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}=\lim_{\Delta{x}\to0}\frac{(x+\Del ta{x})^3-3(x+\Delta{x})^2-7(x+\Delta{x})+5-f(x)}{\Delta{x}}$= $\lim_{\Delta{x}\to0}\frac{(x^3+3x^2\Delta{x}+3x\De lta{x^2}+\Delta{x^3})-3(x^2+2x\Delta{x}+x\Delta{x})-7(x+\Delta{x})+5-f(x)}{\Delta{x}}$

after multiplying and adding we have

$\lim_{\Delta{x}\to0}\frac{3x^2\Delta{x}+3x\Delta{x ^2}+\Delta{x^3}-6x\Delta{x}-3\Delta{x^2}-7\Delta{x}}{\Delta{x}}$

Factoring out $\Delta{x}$

$\lim_{\Delta{x}\to0}3x^2+3x\Delta{x}+\Delta{x^2}-6x-3\Delta{x}-7$

Taking the limit by substituting $0$ for $\Delta{x}$ we get

$\lim_{\Delta{x}\to0}3x^2+3x\Delta{x}+\Delta{x^2}-6x-3\Delta{x}-7=3x^2-6x-7$

and we have a formula for finding the slope at any point on the curve of $f(x)$.

setting our formula equal to 2 allows us to solve for x

$3x^2-6x-7=2$

$x^2-2x=2+\frac{7}{3}$

$x^2-2x+1=3+\frac{7}{3}$

$(x-1)^2=3+\frac{7}{3}$

$x=\pm{\sqrt{3+\frac{7}{3}}}+1$

Now all you have to do is sub $\pm{\sqrt{3+\frac{7}{3}}}+1$
into $f(x)$ to obtain the corresponding values of $y$, and knowing that the slope of the line tangent to the curve at that point is 2, you can write an equation to that line.
• May 17th 2009, 07:34 AM
SirNostalgic
alotta latex there... lol

i understand everything so far, so what next??
• May 17th 2009, 07:57 AM
VonNemo19
You got it?