Finding slope at a point in a direction on a 3d surface
Your idea is totally correct. You need to normalize the direction along the plane is travelling. Since the length of $e_1+e_2$ is $\sqrt{2}$ you have $x(t)=1+\sqrt{2}t$ and $y(t)=\frac{1}{2}+\sqrt{2}t$. If you plug that into $z$ and differentiate with respect to $t$ at $t=0$, then you get the desired result.
Gregory Peck
Updated on August 01, 2022Comments
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Gregory Peck over 1 year
This is not a duplicate, I have attempted the question and am not sure why my answer is incorrect.
QUESTION: The surface with equation $z = x^3 +xy^2 $ intersects the plane with equation $ 2x−2y = 1$ in a curve. What is the slope of that curve when $ x=1 $ and $ y= 1/2 $ ?
Since the line $ 2x - 2y = 1 \space $ travels in the direction $ e_{1} + e_{2} $
we can say $ x = 1 + t $ and $ y = \frac{1}{2} + t $
We then substitute this into the equation and differentiate with respect to $t$. Finally we find the limit of this function as $ t \rightarrow 0 $.
So writing the function in terms of $t$ we get:
$ (t+1)^{3} + (t+1)(t+ \frac{1}{2})^{2} $
$ = 2t^{3} + 5t^{2} + \frac{17t}{4} + 5/4 $
Differentiating this wrt $t$ we get:
$ 6t^{2} + 10t + \frac{17}{4} $
Finding the limit as $ t \rightarrow 0 $ we get $ \frac{17}{4} $
However the answer is supposed to be $ \frac{17}{ \frac{4}{\sqrt{2} } } $
I realise to get the directional derivative we must divide the vector by it's length to find the equivalent unit vector. In this case $ e_{1} + e_{2} $ is $ \sqrt{1+1} = \sqrt{2} $
However when I enter any constant times $t$ (i.e- $ x = 1+2t $) I still get $ \frac{17}{4} $ as my final answer.
Can someone tell me where I am going wrong and explain to me why? Thanks.
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Gregory Peck over 8 yearsThanks for the answer! That makes complete sense what you are saying. However the answer i get is $ \frac{17 \sqrt{2}}{4} $ which is not supposed to be right, can you confirm that this is definitely the right answer?
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sranthrop over 8 years$\frac{17\sqrt{2}}{4}=\frac{17}{\frac{4}{\sqrt{2}}}$ :)
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Gregory Peck over 8 yearsWhoops... that was embarrassing
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Frieder over 8 years@John Doe: Would you please be so kind, to write down your complete tangent-vector? Thank you.