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Point Corresponding To θ=π4\theta = \frac { \pi } { 4 }

Question 47

Multiple Choice

 Find dydx and d2ydx2 if possible, and find the slope and concavity (if possible)  at the \text { Find } \frac { d y } { d x } \text { and } \frac { d ^ { 2 } y } { d x ^ { 2 } } \text { if possible, and find the slope and concavity (if possible) at the } point corresponding to θ=π4\theta = \frac { \pi } { 4 } .
x=4cosθy=4sinθ\begin{array} { l } x = 4 \cos \theta \\y = 4 \sin \theta\end{array}


A) dydx=cotθ,d2ydx2=14sin3θ;\frac { d y } { d x } = \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 1 } { 4 \sin ^ { 3 } \theta } ; at θ=π4;\theta = \frac { \pi } { 4 } ; slope 1 and concave down
B) dydx=cotθ,d2ydx2=14sin3θ\frac { d y } { d x } = - \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 1 } { 4 \sin ^ { 3 } \theta } ; at θ=π4:\theta = \frac { \pi } { 4 } : slope 1- 1 and concave down
C) dydx=cotθ,d2ydx2=14sin3θ;\frac { d y } { d x } = - \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 1 } { 4 \sin ^ { 3 } \theta } ; at θ=π4\theta = \frac { \pi } { 4 } : slope 1- 1 and concave up
D) dydx=cotθ,d2ydx2=14sin3θ\frac { d y } { d x } = \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 1 } { 4 \sin ^ { 3 } \theta } ; at θ=π4:\theta = \frac { \pi } { 4 } : slope 1 and concave up
E) dydx=cotθ,d2ydx2=14sin3θ\frac { d y } { d x } = - \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 1 } { 4 \sin ^ { 3 } \theta } ; at θ=π4\theta = \frac { \pi } { 4 } : slope of 1- 1 and concave down

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