# Thread: Help on Parametric Curves please

1. ## Help on Parametric Curves please

I have no idea how to do this math problem.

x = (4(t^3) + 1)/4
y = 3(t^5) +1
Find all points (a,b) on this curve such that the tangent line to the curve at (a, b) passes through the point (1/5, 1).

Help?!

2. Originally Posted by raven17
I have no idea how to do this math problem.

x = (4(t^3) + 1)/4
y = 3(t^5) +1
Find all points (a,b) on this curve such that the tangent line to the curve at (a, b) passes through the point (1/5, 1).

Help?!
First we need some info on the point $(a,b)$ putting this into the equation for x gives

$\displaystyle a=t^3+\frac{1}{4} \iff t=\left(a-\frac{1}{4}\right)^{\frac{1}{3}}$

Now putting this into the equation for y gives

$\displaystyle y=3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}+1$

So the ordered pair $\displaystyle \left(a,3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}+1\right)$ must lie on the graph.

Now we can use this generic point to find the slope passing through the point $\displaystyle \left( \frac{1}{5},1\right)$ this gives

$\displaystyle m=\frac{3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}+1-1}{a-\frac{1}{5}}= \frac{3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}}{a-\frac{1}{5}}$

Now we also know the slope at this point must be

$\displaystyle \frac{dy}{dx}\bigg|_{x=a}$ so...

Can you finish from here?

3. Originally Posted by raven17
I have no idea how to do this math problem.

x = (4(t^3) + 1)/4
y = 3(t^5) +1
Find all points (a,b) on this curve such that the tangent line to the curve at (a, b) passes through the point (1/5, 1).

Help?!

$\displaystyle x = {{4\,t^3 + 1}\over4}\ \ \to\ \ x = t^3 + {1\over4}$

$y = 3\,t^5 +1$

$\displaystyle {{dy}\over{dx}}={{dy/dt}\over{dx/dt}}={{15t^4}\over{3t^2}} =5t^2$

Slope from $\displaystyle ({1\over5},\ 1)$ to $(x,\ y)$ must equal $5t^2$.

$\displaystyle {{3\,t^5 +1-1}\over{t^3 + {1\over4}-{1\over5}}} =5t^2$.

Solve for $\displaystyle t$ (There are 2 solutions.) and use $\displaystyle t$ to find $(x,\ y)$.

4. Thanks soo much! That really helped.

5. Thank you too! I understood how to finish up the problem thanks to you.