# Implicit differentiation.

• Nov 10th 2009, 04:02 PM
Rumor
Implicit differentiation.
Here's the problem:

"Consider the curve defined by the implicit equation $\displaystyle y+5cosy=x^2+2.$

a) Find $\displaystyle dy/dx$ in terms of x and y.

b) Find the equation of the tangent line at the point shown where x=2.

c) Find two positive values of x in the portion of the curve shown at which there is a vertical tangent line." (There's a picture here of the graph, given so that one could visually look to see where they think a vertical tangent line might be.) "Use calculus to find the values to 3 decimal places. You can use your calculator to solve the equation that must be true for a vertical tangent line."

Any help on how to solve these, please?
• Nov 10th 2009, 04:19 PM
Quote:

Originally Posted by Rumor
Here's the problem:

"Consider the curve defined by the implicit equation $\displaystyle y+5cosy=x^2+2.$

a) Find $\displaystyle dy/dx$ in terms of x and y.

b) Find the equation of the tangent line at the point shown where x=2.

c) Find two positive values of x in the portion of the curve shown at which there is a vertical tangent line." (There's a picture here of the graph, given so that one could visually look to see where they think a vertical tangent line might be.) "Use calculus to find the values to 3 decimal places. You can use your calculator to solve the equation that must be true for a vertical tangent line."

Any help on how to solve these, please?

For part 'a'

$\displaystyle y+5cosy=x^2+2$

Differentiating:

$\displaystyle y'+5\frac{d}{dx}cos(y)=2x$

The derivative $\displaystyle \frac{d}{dx}cos(y)$ must be evaluated using the chain rule.

$\displaystyle \frac{d}{dx}cos(y)=-y'sin(y)$

So the equation becomes:

$\displaystyle y'-5y'sin(y)=2x$

$\displaystyle y'=\frac{2x}{1-5sin(y)}$
• Nov 10th 2009, 04:23 PM
b) Find the equation of the tangent line at the point shown where x=2.

It seems like there is a problem here. There is no way to isolate $\displaystyle y$ in any of these equations. So it's impossible to find the tangent line. When $\displaystyle x=2$, we have $\displaystyle y+5cos(y)=6$. Do you know what the $\displaystyle y$ coordinate of the point is?
• Nov 10th 2009, 04:42 PM
Rumor
Quote:

It seems like there is a problem here. There is no way to isolate $\displaystyle y$ in any of these equations. So it's impossible to find the tangent line. When $\displaystyle x=2$, we have $\displaystyle y+5cos(y)=6$. Do you know what the $\displaystyle y$ coordinate of the point is?