Parametric Eqns: tangent to the curve given by x=e^(sqrt(t)) y= t - ln(t^2)

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Hippocrates

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So here is the problem I am stuck on.

Find an equation of the tangent to the curve given by x=e^(sqrt(t)) y= t - ln(t^2) at the point corresponding to t=1. Your answer should be in the form of y f(x) without t's.

I keep coming up with an answer of -2/e and I really just feel lost.
 
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Hippocrates said:
So here is the problem I am stuck on.

Find and equation of the tangent to the curve given by x=e^(sqrt(t)) y= t - ln(t^2) at the point corresponding to t=1. Your answer should be in the form of y f(x) without t's.

I keep coming up with an answer of -2/e and I really just feel lost.
Click to expand...

How - please show work.
 
Hippocrates said:
dy/dt = 1 - 2/t
dx/dt = e^sqrt(t) / 2*sqrt(t)

dy/dx = dy/dt / dx/dt = (1 - 2/t) * (2 * sqrt(t) / e^sqrt(t)) = 2sqrt(t)/e^sqrt(t) - 4sqrt(t)/t * e^sqrt(t)
substitute 1 in for t
dy/dx = 2/e - 4/e = -2/e
Click to expand...

Okay so you found the derivative of the function at the given point.

What is the relationship between the derivative and the slope of the tangent line?

What is the slope of the tangent line?

The tangent line would go through which point? What is the co-ordinate of that point?

Now apply all this to the equation of straight line (y = mx + b) and get the equation of the tangent line.
 
Subhotosh Khan said:
Okay so you found the derivative of the function at the given point.

What is the relationship between the derivative and the slope of the tangent line?

What is the slope of the tangent line?

The tangent line would go through which point? What is the co-ordinate of that point?

Now apply all this to the equation of straight line (y = mx + b) and get the equation of the tangent line.
Click to expand...

Thanks so much, I can't believe I was missing the real point of the question. To answer your questions for those who might see this in the future. The derivative is the slope at that point. To find the point use the original equations of x=e^t^(1/2) and y=t - ln(t^2) and plug in 1 for t. This gives you x=e and y=1. Using the line equation 1=(-2/e)e + b. Solve for b=3 and get the answer of y(x) = -2x/e + 3.
 
Hippocrates said:
Thanks so much, I can't believe I was missing the real point of the question. To answer your questions for those who might see this in the future. The derivative is the slope at that point. To find the point use the original equations of x=e^t^(1/2) and y=t - ln(t^2) and plug in 1 for t. This gives you x=e and y=1. Using the line equation 1=(-2/e)e + b. Solve for b=3 and get the answer of y(x) = -2x/e + 3.
Click to expand...

Thanks for showing the complete work for this problem.
 
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