y=sin t x=-cos t find dy/dx in terms of X
M mr.burger New member Joined Jun 12, 2005 Messages 21 Jun 12, 2005 #1 y=sin t x=-cos t find dy/dx in terms of X
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jun 12, 2005 #2 Hello, mr.burger! y=sin t, x=-cos t. Find dy/dx in terms of x Click to expand... There are a number of ways to do this . . . [1] Eliminate the parameter. . . . Square: . y<sup>2</sup> .= .sin<sup>2</sup>t . . . . . . . . . . .x<sup>2</sup> .= .cos<sup>2</sup>t . . . . . Add: . x<sup>2</sup> + y<sup>2</sup> .= .sin<sup>2</sup>t + cos<sup>2</sup>t .= .1 . . . Solve for y: . y. = . (1 - x<sup>2</sup>)<sup>1/2</sup> . . . Then differentiate. [2] We have: . y = sin t . . ---> . . dy/dt = cos t . . . . . . . . . . . .x = cos t . . ---> . . dx/dt = -sin t . . . dy . . . . dy/dt . . . . . cos t . . . --- . = . -------- . = . -------- . = . - cot t . . . dx . . . . dx/dt . . . . .-sin t Now we must get this in terms of x (only). . . . We have: . y = sin t . . . . . . . . . . . . x = cos t . . . . . . . . . . . . . . . . . . . . . .y . . . . .sin t . . . Divide the equations: . -- . = . ------- . = . tan t . . . . . . . . . . . . . . . . . . . . . .x . . . . cos t . . . . . . . . . . . . . . . . y . . . . . . . . . . . . . . . . x . . . Since: .tan t .= .--, .then: . - cot t .= .- -- . . . . . . . . . . . . . . . . x . . . . . . . . . . . . . . . . y . . . . . . . . . . . . . . .dy . . . . . x Hence, we have: . --- . = .- -- . . . . . . . . . . . . . . .dx . . . . . y Since: . y .= .(1 - x<sup>2</sup>)<sup>1/2</sup> . . . see part [1] . . . . . . . . . . . . dy . . . . . - x . . . we have: . --- . = . --<u>-------</u>- . . . . . . . . . . . . dx . . . . √1 - x<sup>2</sup>
Hello, mr.burger! y=sin t, x=-cos t. Find dy/dx in terms of x Click to expand... There are a number of ways to do this . . . [1] Eliminate the parameter. . . . Square: . y<sup>2</sup> .= .sin<sup>2</sup>t . . . . . . . . . . .x<sup>2</sup> .= .cos<sup>2</sup>t . . . . . Add: . x<sup>2</sup> + y<sup>2</sup> .= .sin<sup>2</sup>t + cos<sup>2</sup>t .= .1 . . . Solve for y: . y. = . (1 - x<sup>2</sup>)<sup>1/2</sup> . . . Then differentiate. [2] We have: . y = sin t . . ---> . . dy/dt = cos t . . . . . . . . . . . .x = cos t . . ---> . . dx/dt = -sin t . . . dy . . . . dy/dt . . . . . cos t . . . --- . = . -------- . = . -------- . = . - cot t . . . dx . . . . dx/dt . . . . .-sin t Now we must get this in terms of x (only). . . . We have: . y = sin t . . . . . . . . . . . . x = cos t . . . . . . . . . . . . . . . . . . . . . .y . . . . .sin t . . . Divide the equations: . -- . = . ------- . = . tan t . . . . . . . . . . . . . . . . . . . . . .x . . . . cos t . . . . . . . . . . . . . . . . y . . . . . . . . . . . . . . . . x . . . Since: .tan t .= .--, .then: . - cot t .= .- -- . . . . . . . . . . . . . . . . x . . . . . . . . . . . . . . . . y . . . . . . . . . . . . . . .dy . . . . . x Hence, we have: . --- . = .- -- . . . . . . . . . . . . . . .dx . . . . . y Since: . y .= .(1 - x<sup>2</sup>)<sup>1/2</sup> . . . see part [1] . . . . . . . . . . . . dy . . . . . - x . . . we have: . --- . = . --<u>-------</u>- . . . . . . . . . . . . dx . . . . √1 - x<sup>2</sup>
